tobimichigan Posted September 29, 2010 Share Posted September 29, 2010 These r my files: Upload.php <?php $db = mysql_connect("localhost","root","") or die ("Unable to connect. Check your connection parameters."); mysql_select_db("moviesite", $db) or die(mysql_error($db)); ?> <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> <title>upload_image</title> </head> <body> <form action="check_image.php" method="post" enctype="multipart/form-data"> <table> <tr> <td> Your Username </td> <td> <input type="text" name="username"/> </td> </tr> <td> Upload Image* </td> <td> <input type="file" name="image_filename"/> </td> </tr> <tr> <td colspan="2"> <small> <em> * Acceptable image formats include: GIF, JPG/JPEG and PNG. </em> </small> </td> </tr> <tr> <td> Image Caption <br/> </td> <td> <input type="text" name="caption" /> </td> </tr > <tr> <td colspan="2" style="text-align: center"> <input type="submit" name="submit" value="Upload"/> </td> </tr> </table> </form> </body> </html> check_image.php <?php $db = mysql_connect("localhost","root","") or die ("Unable to connect. Check your connection parameters."); mysql_select_db("moviesite", $db) or die(mysql_error($db)); //change this path to match your images directory $dir ="C:/xampp/htdocs/WroxPhp6.0/img"; //make sure the uploaded file transfer was successful if ($_FILES["uploadfile"]["error"] != UPLOAD_ERR_OK) { switch ($_FILES["uploadfile"]["error"]) { case UPLOAD_ERR_INI_SIZE: die("The uploaded file exceeds the upload_max_filesize directive " . "in php.ini."); break; case UPLOAD_ERR_FORM_SIZE: die("The uploaded file exceeds the MAX_FILE_SIZE directive that " . "was specified in the HTML form."); break; case UPLOAD_ERR_PARTIAL: die("The uploaded file was only partially uploaded."); break; case UPLOAD_ERR_NO_FILE: die("No file was uploaded."); break; case UPLOAD_ERR_NO_TMP_DIR: die("The server is missing a temporary folder."); break; case UPLOAD_ERR_CANT_WRITE: die("The server failed to write the uploaded file to disk."); break; case UPLOAD_ERR_EXTENSION: die("File upload stopped by extension."); break; } } //get info about the image being uploaded $uploadfile = $_POST["uploadfile"]; $image_caption = $_POST["caption"]; $image_username = $_POST["username"]; $image_date = date("Y-m-d"); list($width, $height, $type, $attr) = getimagesize($_FILES["uploadfile"]["tmp_name"]); // make sure the uploaded file is really a supported image switch ($type) { case IMAGETYPE_GIF: $image = imagecreatefromgif($_FILES["uploadfile"]["tmp_name"]) or die("The file you uploaded was not a supported filetype."); $ext = ".gif"; break; case IMAGETYPE_JPEG: $image = imagecreatefromjpeg($_FILES["uploadfile"]["tmp_name"]) or die("The file you uploaded was not a supported filetype."); $ext = ".jpg"; break; case IMAGETYPE_PNG: $image = imagecreatefrompng($_FILES["uploadfile"]["tmp_name"]) or die("The file you uploaded was not a supported filetype."); $ext = ".png"; break; default: die("The file you uploaded was not a supported filetype."); } //insert information into image table $query = "INSERT INTO images (image_caption, image_username, image_date) VALUES ('$image_caption.', '$image_username', '$image_date')"; $result = mysql_query($query, $db) or die (mysql_error($db)); //retrieve the image_id that MySQL generated automatically when we inserted //the new record $last_id = mysql_insert_id(); //because the id is unique, we can use it as the image name as well to make //sure we don’t overwrite another image that already exists $imagename = $last_id.$ext; // update the image table now that the final filename is known. $query = "UPDATE images SET image_filename = '.$imagename.' WHERE image_id = '' . $last_id; $result = mysql_query($query, $db) or die (mysql_error($db)); //save the image to its final destination switch ($type) { case IMAGETYPE_GIF: imagegif($image, $dir . '/' . $imagename); break; case IMAGETYPE_JPEG: imagejpeg($image, $dir . '/' . $imagename, 100); break; case IMAGETYPE_PNG: imagepng($image, $dir . '/' . $imagename); break; } imagedestroy($image);" ?> <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> <title>Here's your pic</title> </head> <body> <h1> So how does it feel to be famous? </h1 > <p> Here is the picture you just uploaded to our servers: </p > <img src="images/ <?php echo $imagename; ?>" style="float:left;"> <table> <tr> <td> Image Saved as: </td> <td> <?php echo $imagename; ?> </td> </tr> <tr> <td> Image Type: </td> <td> <?php echo $ext; ?> </td> </tr> <tr> <td> Height: </td> <td> <?php echo $height; ?> </td > </tr> <tr> <td> Width: </td> <td> <?php echo $width; ?> </td > </tr> <tr> <td> Upload Date: </td > <td > <?php echo $image_date; ?> </td > </tr> </table> </body> </html> This is my table: <?php $db = mysql_connect("localhost", "root", "") or die ("Unable to connect. Check your connection parameters."); mysql_select_db("moviesite", $db) or die(mysql_error($db)); //create the images table $query = 'CREATE TABLE images ( image_id INTEGER NOT NULL AUTO_INCREMENT, image_caption VARCHAR(255) NOT NULL, image_username VARCHAR(255) NOT NULL, image_filename VARCHAR(255) NOT NULL DEFAULT "", image_date DATE NOT NULL, PRIMARY KEY (image_id) ) ENGINE=MyISAM'; mysql_query($query, $db) or die (mysql_error($db)); echo "Images table successfully created."; ?> Now, the problem I'm encountering is this: 1. When I execute this script, all the table fields are filled but the image_filename is empty. 2. On the html output, an "x" displays rather than the actual image uploaded. What am I doing wrong now? Please can some1 tell me? Be so grateful. Quote Link to comment https://forums.phpfreaks.com/topic/214722-image-upload-in-php-60/ Share on other sites More sharing options...
tobimichigan Posted September 29, 2010 Author Share Posted September 29, 2010 Oh by the way, I reading one of wrox's books..I wonder why the codes in some of these texts don't run at all.. Quote Link to comment https://forums.phpfreaks.com/topic/214722-image-upload-in-php-60/#findComment-1117166 Share on other sites More sharing options...
litebearer Posted September 29, 2010 Share Posted September 29, 2010 1. try echoing out the filename JUST BEFORE you attempt to insert into the table - this will tell you if the variable has what you are expecting it to contain. 2. the x is the result of (A) no filename being in the table, OR (B) no image found by that name Quote Link to comment https://forums.phpfreaks.com/topic/214722-image-upload-in-php-60/#findComment-1117170 Share on other sites More sharing options...
PFMaBiSmAd Posted September 29, 2010 Share Posted September 29, 2010 The UPDATE query in check_image.php is incorrectly written. The closing double-quote is at the end of some php code (making the php code part of the query string.) The closing double-quote on that query should be at the end of the query. And I see how this might have occurred. There should be a double-quote in the query, right before the dot in: . $last_id; There are two single-quotes instead. Change the two single-quotes to a double-quote and remove the double-quote that you added after the imagedestroy($image);" Quote Link to comment https://forums.phpfreaks.com/topic/214722-image-upload-in-php-60/#findComment-1117175 Share on other sites More sharing options...
tobimichigan Posted September 29, 2010 Author Share Posted September 29, 2010 The UPDATE query in check_image.php is incorrectly written. The closing double-quote is at the end of some php code (making the php code part of the query string.) The closing double-quote on that query should be at the end of the query. And I see how this might have occurred. There should be a double-quote in the query, right before the dot in: . $last_id; There are two single-quotes instead. Change the two single-quotes to a double-quote and remove the double-quote that you added after the imagedestroy($image);" You seem to be close to the answer pf, I adjusted the code thus and there was a slight improvement: check_image.php <?php $db = mysql_connect("localhost","root","") or die ("Unable to connect. Check your connection parameters."); mysql_select_db("moviesite", $db) or die(mysql_error($db)); //change this path to match your images directory $dir ="C:/xampp/htdocs/WroxPhp6.0/img"; //make sure the uploaded file transfer was successful if ($_FILES["uploadfile"]["error"] != UPLOAD_ERR_OK) { switch ($_FILES["uploadfile"]["error"]) { case UPLOAD_ERR_INI_SIZE: die("The uploaded file exceeds the upload_max_filesize directive " . "in php.ini."); break; case UPLOAD_ERR_FORM_SIZE: die("The uploaded file exceeds the MAX_FILE_SIZE directive that " . "was specified in the HTML form."); break; case UPLOAD_ERR_PARTIAL: die("The uploaded file was only partially uploaded."); break; case UPLOAD_ERR_NO_FILE: die("No file was uploaded."); break; case UPLOAD_ERR_NO_TMP_DIR: die("The server is missing a temporary folder."); break; case UPLOAD_ERR_CANT_WRITE: die("The server failed to write the uploaded file to disk."); break; case UPLOAD_ERR_EXTENSION: die("File upload stopped by extension."); break; } } //get info about the image being uploaded $uploadfile = $_POST["uploadfile"]; $image_caption = $_POST["caption"]; $image_username = $_POST["username"]; $image_date = date("Y-m-d"); list($width, $height, $type, $attr) = getimagesize($_FILES["uploadfile"]["tmp_name"]); // make sure the uploaded file is really a supported image switch ($type) { case IMAGETYPE_GIF: $image = imagecreatefromgif($_FILES["uploadfile"]["tmp_name"]) or die("The file you uploaded was not a supported filetype."); $ext = ".gif"; break; case IMAGETYPE_JPEG: $image = imagecreatefromjpeg($_FILES["uploadfile"]["tmp_name"]) or die("The file you uploaded was not a supported filetype."); $ext = ".jpg"; break; case IMAGETYPE_PNG: $image = imagecreatefrompng($_FILES["uploadfile"]["tmp_name"]) or die("The file you uploaded was not a supported filetype."); $ext = ".png"; break; default: die("The file you uploaded was not a supported filetype."); } //insert information into image table $query = "INSERT INTO images (image_caption, image_username, image_date) VALUES ('$image_caption.', '$image_username', '$image_date')"; $result = mysql_query($query, $db) or die (mysql_error($db)); //retrieve the image_id that MySQL generated automatically when we inserted //the new record $last_id = mysql_insert_id(); //because the id is unique, we can use it as the image name as well to make //sure we don’t overwrite another image that already exists $imagename = $last_id.$ext; // update the image table now that the final filename is known. $query = "UPDATE images SET image_filename = '.$imagename.' WHERE image_id = '$last_id'"; $result = mysql_query($query, $db) or die (mysql_error($db)); //save the image to its final destination switch ($type) { case IMAGETYPE_GIF: imagegif($image, $dir . '/' . $imagename); break; case IMAGETYPE_JPEG: imagejpeg($image, $dir . '/' . $imagename, 100); break; case IMAGETYPE_PNG: imagepng($image, $dir . '/' . $imagename); break; } imagedestroy($image); ?> <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> <title>Here's your pic</title> </head> <body> <h1> So how does it feel to be famous? </h1 > <p> Here is the picture you just uploaded to our servers: </p > <img src="images/ <?php echo $imagename; ?>" style="float:left;"> <table> <tr> <td> Image Saved as: </td> <td> <?php echo $imagename; ?> </td> </tr> <tr> <td> Image Type: </td> <td> <?php echo $ext; ?> </td> </tr> <tr> <td> Height: </td> <td> <?php echo $height; ?> </td > </tr> <tr> <td> Width: </td> <td> <?php echo $width; ?> </td > </tr> <tr> <td> Upload Date: </td > <td > <?php echo $image_date; ?> </td > </tr> </table> </body> </html> Now, the image_filename is no longer empty on the table..its now actually showing the image_filename, but the only problem here now is displaying on the html page..its still showing the "X" thing rather than the real image uploaded to the folder. What am I getting wrong this time? Quote Link to comment https://forums.phpfreaks.com/topic/214722-image-upload-in-php-60/#findComment-1117198 Share on other sites More sharing options...
PFMaBiSmAd Posted September 29, 2010 Share Posted September 29, 2010 Is the image getting saved to the - $dir ="C:/xampp/htdocs/WroxPhp6.0/img"; location? Quote Link to comment https://forums.phpfreaks.com/topic/214722-image-upload-in-php-60/#findComment-1117201 Share on other sites More sharing options...
tobimichigan Posted September 29, 2010 Author Share Posted September 29, 2010 I just went to that directory pf, yes it is appearing on the images folder. What do think might be wrong? Quote Link to comment https://forums.phpfreaks.com/topic/214722-image-upload-in-php-60/#findComment-1117214 Share on other sites More sharing options...
PFMaBiSmAd Posted September 29, 2010 Share Posted September 29, 2010 <img src="images/ <?php echo $imagename; ?>" style="float:left;"> ^^^ The URL in the src="..." attribute must be to where the image files are located. If that script is in your WroxPhp6.0 folder, then the above code should be - <img src="img/<?php echo $imagename; ?>" style="float:left;"> Quote Link to comment https://forums.phpfreaks.com/topic/214722-image-upload-in-php-60/#findComment-1117227 Share on other sites More sharing options...
tobimichigan Posted September 29, 2010 Author Share Posted September 29, 2010 Here's what I tried Pf: <img src="img/<?php echo $imagename; ?>" style="float:left;"> and <img src="img/<?php echo $imagename; ?>" style="float:left;"> and finally, <img src="WroxPhp6.0/img/<?php echo $imagename; ?> Wow, men, if these 3 codes rn't wking then I wonder which would get the image displayed. All of these seem to display 2 "X" while the rest of the information of the image displays intact..do u have any better ideas Pf? Quote Link to comment https://forums.phpfreaks.com/topic/214722-image-upload-in-php-60/#findComment-1117245 Share on other sites More sharing options...
PFMaBiSmAd Posted September 29, 2010 Share Posted September 29, 2010 I tried the check_image.php code with the suggested change and it worked as expected. There is however another error in the UPDATE query. There are two extra dots, one before and one after the $imagename that is being stored in the table. Quote Link to comment https://forums.phpfreaks.com/topic/214722-image-upload-in-php-60/#findComment-1117329 Share on other sites More sharing options...
ignace Posted September 29, 2010 Share Posted September 29, 2010 PHP 6.0 Quote Link to comment https://forums.phpfreaks.com/topic/214722-image-upload-in-php-60/#findComment-1117343 Share on other sites More sharing options...
tobimichigan Posted October 3, 2010 Author Share Posted October 3, 2010 ok pf here's my final code: upload_image.php <?php $db = mysql_connect("localhost","root","") or die ("Unable to connect. Check your connection parameters."); mysql_select_db("moviesite", $db) or die(mysql_error($db)); ?> <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> <title>upload_image</title> </head> <body> <form action="check_image.php" method="post" enctype="multipart/form-data"> <table> <tr> <td> Your Username </td> <td> <input type="text" name="username"/> </td> </tr> <td> Upload Image* </td> <td> <input type="file" name="uploadfile"/> </td> </tr> <tr> <td colspan="2"> <small> <em> * Acceptable image formats include: GIF, JPG/JPEG and PNG. </em> </small> </td> </tr> <tr> <td> Image Caption <br/> </td> <td> <input type="text" name="caption" /> </td> </tr > <tr> <td colspan="2" style="text-align: center"> <input type="submit" name="submit" value="Upload"/> </td> </tr> </table> </form> </body> </html> check_image.php <?php $db = mysql_connect("localhost","root","") or die ("Unable to connect. Check your connection parameters."); mysql_select_db("moviesite", $db) or die(mysql_error($db)); //change this path to match your images directory $dir ="C:/xampp/htdocs/WroxPhp6.0/img"; //make sure the uploaded file transfer was successful if ($_FILES["uploadfile"]["error"] != UPLOAD_ERR_OK) { switch ($_FILES["uploadfile"]["error"]) { case UPLOAD_ERR_INI_SIZE: die("The uploaded file exceeds the upload_max_filesize directive " . "in php.ini."); break; case UPLOAD_ERR_FORM_SIZE: die("The uploaded file exceeds the MAX_FILE_SIZE directive that " . "was specified in the HTML form."); break; case UPLOAD_ERR_PARTIAL: die("The uploaded file was only partially uploaded."); break; case UPLOAD_ERR_NO_FILE: die("No file was uploaded."); break; case UPLOAD_ERR_NO_TMP_DIR: die("The server is missing a temporary folder."); break; case UPLOAD_ERR_CANT_WRITE: die("The server failed to write the uploaded file to disk."); break; case UPLOAD_ERR_EXTENSION: die("File upload stopped by extension."); break; } } //get info about the image being uploaded $uploadfile = $_POST["uploadfile"]; $image_caption = $_POST["caption"]; $image_username = $_POST["username"]; $image_date = date("Y-m-d"); list($width, $height, $type, $attr) = getimagesize($_FILES["uploadfile"]["tmp_name"]); // make sure the uploaded file is really a supported image switch ($type) { case IMAGETYPE_GIF: $image = imagecreatefromgif($_FILES["uploadfile"]["tmp_name"]) or die("The file you uploaded was not a supported filetype."); $ext = ".gif"; break; case IMAGETYPE_JPEG: $image = imagecreatefromjpeg($_FILES["uploadfile"]["tmp_name"]) or die("The file you uploaded was not a supported filetype."); $ext = ".jpg"; break; case IMAGETYPE_PNG: $image = imagecreatefrompng($_FILES["uploadfile"]["tmp_name"]) or die("The file you uploaded was not a supported filetype."); $ext = ".png"; break; default: die("The file you uploaded was not a supported filetype."); } //insert information into image table $query = "INSERT INTO images (image_caption, image_username, image_date) VALUES ('$image_caption.', '$image_username', '$image_date')"; $result = mysql_query($query, $db) or die (mysql_error($db)); //retrieve the image_id that MySQL generated automatically when we inserted //the new record $last_id = mysql_insert_id(); //because the id is unique, we can use it as the image name as well to make //sure we don’t overwrite another image that already exists $imagename = $last_id.$ext; // update the image table now that the final filename is known. $query = "UPDATE images SET image_filename = '$imagename' WHERE image_id = '$last_id'"; $result = mysql_query($query, $db) or die (mysql_error($db)); //save the image to its final destination switch ($type) { case IMAGETYPE_GIF: imagegif($image, $dir . '/' . $imagename); break; case IMAGETYPE_JPEG: imagejpeg($image, $dir . '/' . $imagename, 100); break; case IMAGETYPE_PNG: imagepng($image, $dir . '/' . $imagename); break; } imagedestroy($image); ?> <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> <title>Here's your pic</title> </head> <body> <h1> So how does it feel to be famous? </h1 > <p> Here is the picture you just uploaded to our servers: </p > <table> <tr> <td> Image Saved as: </td> <td> <img src="img/<?php $imagename; ?>" style="float:left;"> </td> </tr> <tr> <td> Image Type: </td> <td> <?php echo $ext; ?> </td> </tr> <tr> <td> Height: </td> <td> <?php echo $height; ?> </td > </tr> <tr> <td> Width: </td> <td> <?php echo $width; ?> </td > </tr> <tr> <td> Upload Date: </td > <td > <?php echo $image_date; ?> </td > </tr> </table> </body> </html> I really dont get why this image is showing x. I have removed the dots at the update image (like u indicated) but the image is still not displaying. What is wrong with this modified code? Quote Link to comment https://forums.phpfreaks.com/topic/214722-image-upload-in-php-60/#findComment-1118516 Share on other sites More sharing options...
litebearer Posted October 3, 2010 Share Posted October 3, 2010 perhaps change this <tr> <td> Image Saved as: </td> <td> <img src="img/<?php $imagename; ?>" style="float:left;"> </td> </tr> to this <tr> <td> Image Saved as: </td> <td> <img src="img/<?php echo $imagename; ?>" style="float:left;"> </td> </tr> Quote Link to comment https://forums.phpfreaks.com/topic/214722-image-upload-in-php-60/#findComment-1118537 Share on other sites More sharing options...
tobimichigan Posted October 3, 2010 Author Share Posted October 3, 2010 Lite, I hope you re-read this post after posting cos u absolutely posted nothing on the screen..Maybe you could re-post what u actually intended. Quote Link to comment https://forums.phpfreaks.com/topic/214722-image-upload-in-php-60/#findComment-1118546 Share on other sites More sharing options...
PFMaBiSmAd Posted October 3, 2010 Share Posted October 3, 2010 I can see his post OK (using Firefox - latest). I'm guessing that the hacker who broke into the forum and modified it forgot that this is a programming help forum and being able to actually see what was posted is kind of important to asking for and giving help. His post is pointing out that you removed the echo statement that was before the $imagename; in the src="..." attribute. Why did you do that? Quote Link to comment https://forums.phpfreaks.com/topic/214722-image-upload-in-php-60/#findComment-1118551 Share on other sites More sharing options...
tobimichigan Posted October 3, 2010 Author Share Posted October 3, 2010 Hey, pf glad to have u back men.. here's my check_image.php: <?php $db = mysql_connect("localhost","root","") or die ("Unable to connect. Check your connection parameters."); mysql_select_db("moviesite", $db) or die(mysql_error($db)); //change this path to match your images directory $dir ="C:/xampp/htdocs/WroxPhp6.0/images"; //make sure the uploaded file transfer was successful if ($_FILES["uploadfile"]["error"] != UPLOAD_ERR_OK) { switch ($_FILES["uploadfile"]["error"]) { case UPLOAD_ERR_INI_SIZE: die("The uploaded file exceeds the upload_max_filesize directive " . "in php.ini."); break; case UPLOAD_ERR_FORM_SIZE: die("The uploaded file exceeds the MAX_FILE_SIZE directive that " . "was specified in the HTML form."); break; case UPLOAD_ERR_PARTIAL: die("The uploaded file was only partially uploaded."); break; case UPLOAD_ERR_NO_FILE: die("No file was uploaded."); break; case UPLOAD_ERR_NO_TMP_DIR: die("The server is missing a temporary folder."); break; case UPLOAD_ERR_CANT_WRITE: die("The server failed to write the uploaded file to disk."); break; case UPLOAD_ERR_EXTENSION: die("File upload stopped by extension."); break; } } //get info about the image being uploaded $uploadfile = $_POST["uploadfile"]; $image_caption = $_POST["caption"]; $image_username = $_POST["username"]; $image_date = date("Y-m-d"); list($width, $height, $type, $attr) = getimagesize($_FILES["uploadfile"]["tmp_name"]); // make sure the uploaded file is really a supported image switch ($type) { case IMAGETYPE_GIF: $image = imagecreatefromgif($_FILES["uploadfile"]["tmp_name"]) or die("The file you uploaded was not a supported filetype."); $ext = ".gif"; break; case IMAGETYPE_JPEG: $image = imagecreatefromjpeg($_FILES["uploadfile"]["tmp_name"]) or die("The file you uploaded was not a supported filetype."); $ext = ".jpg"; break; case IMAGETYPE_PNG: $image = imagecreatefrompng($_FILES["uploadfile"]["tmp_name"]) or die("The file you uploaded was not a supported filetype."); $ext = ".png"; break; default: die("The file you uploaded was not a supported filetype."); } //insert information into image table $query = "INSERT INTO images (image_caption, image_username, image_date) VALUES ('$image_caption.', '$image_username', '$image_date')"; $result = mysql_query($query, $db) or die (mysql_error($db)); //retrieve the image_id that MySQL generated automatically when we inserted //the new record $last_id = mysql_insert_id(); //because the id is unique, we can use it as the image name as well to make //sure we don’t overwrite another image that already exists $imagename = $last_id.$ext; // update the image table now that the final filename is known. $query = "UPDATE images SET image_filename = '$imagename' WHERE image_id = '$last_id'"; $result = mysql_query($query, $db) or die (mysql_error($db)); //save the image to its final destination switch ($type) { case IMAGETYPE_GIF: imagegif($image, $dir . '/' . $imagename); break; case IMAGETYPE_JPEG: imagejpeg($image, $dir . '/' . $imagename, 100); break; case IMAGETYPE_PNG: imagepng($image, $dir . '/' . $imagename); break; } imagedestroy($image); ?> <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> <title>Here's your pic</title> </head> <body> <h1> So how does it feel to be famous? </h1 > <p> Here is the picture you just uploaded to our servers: </p > <img src="images/<?php echo $imagename; ?>" style="float:left;"> <table> <tr> <td> Image Saved as: </td> <td><?php echo $imagename; ?></td> </tr> <tr> <td> Image Type: </td> <td> <?php echo $ext; ?> </td> </tr> <tr> <td> Height: </td> <td> <?php echo $height; ?> </td > </tr> <tr> <td> Width: </td> <td> <?php echo $width; ?> </td > </tr> <tr> <td> Upload Date: </td > <td > <?php echo $image_date; ?> </td > </tr> </table> </body> </html> As u can see I'm using </td> <td><?php echo $imagename; ?></td> </tr> and its still not dispaying any other thing than "X"..any other suggestions? Quote Link to comment https://forums.phpfreaks.com/topic/214722-image-upload-in-php-60/#findComment-1118565 Share on other sites More sharing options...
litebearer Posted October 3, 2010 Share Posted October 3, 2010 can read my post using ie 8 Quote Link to comment https://forums.phpfreaks.com/topic/214722-image-upload-in-php-60/#findComment-1118567 Share on other sites More sharing options...
litebearer Posted October 3, 2010 Share Posted October 3, 2010 What does your source say? (use the page/view source in your browser) Quote Link to comment https://forums.phpfreaks.com/topic/214722-image-upload-in-php-60/#findComment-1118568 Share on other sites More sharing options...
tobimichigan Posted October 3, 2010 Author Share Posted October 3, 2010 What does your source say? (use the page/view source in your browser) Here's what it says: <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> <title>Here's your pic</title> </head> <body> <h1> So how does it feel to be famous? </h1 > <p> Here is the picture you just uploaded to our servers: </p > <img src="images/" style="float:left;"> <table> <tr> <td> Image Saved as: </td> <td>4.jpg</td> </tr> <tr> <td> Image Type: </td> <td> .jpg </td> </tr> <tr> <td> Height: </td> <td> 400 </td > </tr> <tr> <td> Width: </td> <td> 620 </td > </tr> <tr> <td> Upload Date: </td > <td > 2010-10-03 </td > </tr> </table> </body> </html> Quote Link to comment https://forums.phpfreaks.com/topic/214722-image-upload-in-php-60/#findComment-1118569 Share on other sites More sharing options...
tobimichigan Posted October 3, 2010 Author Share Posted October 3, 2010 Using <p> Here is the picture you just uploaded to our servers: </p > <img src="images/<?php echo $imagename; ?>" style="float:left;">, in dreamweaver it then shows the code below in the view source. <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> <title>Here's your pic</title> </head> <body> <h1> So how does it feel to be famous? </h1 > <p> Here is the picture you just uploaded to our servers: </p > <img src="images/5.jpg" style="float:left;"> <table> <tr> <td> Image Saved as: </td> <td>5.jpg</td> </tr> <tr> <td> Image Type: </td> <td> .jpg </td> </tr> <tr> <td> Height: </td> <td> 400 </td > </tr> <tr> <td> Width: </td> <td> 620 </td > </tr> <tr> <td> Upload Date: </td > <td > 2010-10-03 </td > </tr> </table> </body> </html> Quote Link to comment https://forums.phpfreaks.com/topic/214722-image-upload-in-php-60/#findComment-1118572 Share on other sites More sharing options...
tobimichigan Posted October 3, 2010 Author Share Posted October 3, 2010 Why is this code still displaying "x"? Quote Link to comment https://forums.phpfreaks.com/topic/214722-image-upload-in-php-60/#findComment-1118574 Share on other sites More sharing options...
litebearer Posted October 3, 2010 Share Posted October 3, 2010 try (as a test) hard coding <IMG SRC="images/5.jpg"> See if it is displayed. If NOT, either the path or name is wrong Quote Link to comment https://forums.phpfreaks.com/topic/214722-image-upload-in-php-60/#findComment-1118576 Share on other sites More sharing options...
tobimichigan Posted November 19, 2010 Author Share Posted November 19, 2010 Can any 1 come up with a real solution to this? Quote Link to comment https://forums.phpfreaks.com/topic/214722-image-upload-in-php-60/#findComment-1136615 Share on other sites More sharing options...
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