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grozanc

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Everything posted by grozanc

  1. Wow, talk about not thinking things through. I originally tried $string_extracredit = 0; as I used that in other spots and stopped the error but didn't think to use $string_extracredit = ''; Anyway, thanks for the help. You solved the problem!
  2. Hello Everyone, I've got the following snippet of code that is use to retrieve student's grades when they log into the class website. I've been using the bit of code for years, but I'm switching to a new CMS and for the first time I get the "Notice: Undefined variable error". After a lot of searching I found a solution that I can't figure out how to implement because my variables are coming from an array and not a single instance. Can anyone show me how I should use isset() or empty() to fix the undeclared variable? The undeclared variable error in the following code is $string_extracredit and the error is appearing on line 5. Thanks, Gary $query_extracredit = "SELECT title, points FROM $section WHERE oasis_id='$oasis_id' && extracredit='1' ORDER BY date ASC"; $result_extracredit = mysql_query($query_extracredit) or die('Error, query failed'); if (mysql_num_rows($result_extracredit) > 0) { while(list($extracredit_title, $extracredit_points) = mysql_fetch_array($result_extracredit)) $string_extracredit .= "".$extracredit_title.": ".$extracredit_points." Points<br/>"; } echo "$string_extracredit";
  3. OK, I've tried the following and nothing seems to work. Can anyone spot what I'm doing wrong. I'm a newb when it comes to PHP so I don't expect that I'm doing this right. Please let me know if I'm not giving you enough information to be able to help. $prod_str = $order->products[$i]['qty'] . " x " . $order->html_entity_decode(products[$i]['name']); Page Load Error $prod_str = $order->products[$i]['qty'] . " x " . html_entity_decode($order->products[$i]['name']); Page renders but ascii character aren't converted. html_entity_decode($prod_str = $order->products[$i]['qty'] . " x " . $order->products[$i]['name']); Page renders but ascii character aren't converted. $prod_str = $order->products[$i]['qty'] . " x " . $order->products[$i]['name']; $prod_str = html_entity_decode($prod_str); Page renders but ascii character aren't converted.
  4. Anyone have any ideas or do I need to supply more info
  5. Hello Everyone, I have a pdf that is automatically generated, unfortunately it's pulling ascii codes from the database and displaying them as ascii. I want to use the html_entity_decode() command to make the ascii display as the actual symbol (in my case a trade mark symbol instead of ™). Here is the line of code the outputs the information from the database $prod_str = $order->products[$i]['qty'] . " x " . $order->products[$i]['name']; My question is where would I wrap the html_entity_decode() around to change products[$i]['name'] from ascii to text? Thanks, Gary
  6. I'm trying to dump three pieces of information into a database from a form, two of the pieces are coming from an array. I can get one piece of information from an array in, but I can't get the second. My latest attempt is to create the entry with the first array and then update that entry with the second array but that isn't working either. I'm a noob when it comes to this stuff so if this is explained somewhere please point me in the right direction. I've listed what I have below. The first foreach array works and the values are dumped into the database. I can't get the second foreach array to update the previous filed. I'm sure there is a better way to do this, but I don't have a clue. Any help would be appreciated, and yes, I did search google, but I guess I'm not using the right key words! Thanks, Gary <SELECT NAME="oasis_id[]"> <OPTION VALUE=""> </OPTION> <OPTION VALUE=\"".$oasis_id.".Present\"> Present </OPTION> <OPTION VALUE=\"".$oasis_id.".Absent\"> Absent </OPTION> </SELECT> <INPUT TYPE="TEXT" NAME="particpation_points[]" > <INPUT TYPE="SUBMIT" VALUE="Submit" > $class = $_POST['class']; $attendance_date = $_POST['attendance_date']; foreach($_POST['oasis_id'] as $oasis_id) { $query = "INSERT INTO $class (attendance_date, attendance) values ('$attendance_date','$oasis_id')"; mysql_query($query) or die('Error, query failed : ' . mysql_error()); } foreach ($_POST['particpation_points'] as $particpation_points) { $query2 = "UPDATE $class SET particpation_points='$particpation_points' WHERE oasis_id='$oasis_id' && attendance_date='$attendance_date'"; mysql_query($query2) or die('Error, query failed : ' . mysql_error()); }
  7. Hello Again, I don't want to mark this as "SOLVED" because the change I made doesn't seem like it should have worked. But, changing $i = 0; to $i = 1; allowed the last image name in the array to be inserted into the database. If anyone can say for certain that this code is correct, please let me know! Thanks, Gary
  8. Thanks for the help, but unfortunately it didn't work completely. Now, if I upload one image, the image name doesn't get stored in the database, if I upload two images only the first image name goes into the database. If I upload three images, the first two names get stored, and the last images name doesn't. So basically the name for the last file uploaded is getting lost. Any suggestions? // Post Variables $challenge = $_POST['challenge']; $insight = $_POST['insight']; $oasis_id = $_POST['oasis_id']; $filename=serialize($_POST['filename']); // Upload Images $i = 0; while(list($key,$value) = each($_FILES['image']['name'])) { $_name = 'filename' . $i++; $$_name = ''; if(!empty($value)) { $$_name = $value; $add = 'upimage/'.$$_name; copy($_FILES['image']['tmp_name'][$key], $add); chmod($add, 0777); } } // Insert into database $query = "INSERT INTO table ( oasis_id, challenge, insight, uploadedfile1, uploadedfile2, uploadedfile3 )". "values ( '$oasis_id', '$challenge', '$insight', '$filename1', '$filename2', '$filename3')"; mysql_query($query) or die('Error, query failed : ' . mysql_error());
  9. Hello All, First, I'm new to php and I have spent a lot of time searching forums and google for the answer to my question with no luck, so if it's a topic already covered, PLEASE reply with a link or point me in the right direction. OK, I have a form that let's users upload multiple images. The upload portion works fine. What I can't figure out how to do is put the image name of each file that gets uploaded all into the same table record. I also need each file name to be put into it's own column in the table. For example, file1, file2 and file3 will get dumped into the same record and into it's respective column (uploaded_image1, uploaded_image2, uploaded_image3). I've included the working upload code and the code that inserts some of the information into the database (just not the image names). echo "<textarea name=challenge rows=15 cols=60> </textarea><br/><br/>" echo "<textarea name=insight rows=15 cols=60> </textarea><br/><br/>" for($i=1; $i<=$max_no_img; $i++){ echo "Image $i <input type=file name='image[]' ><br/>"; } echo "<input type=submit value=Submit>"; // Connect to DB $host = 'xxx'; $user = 'xxx'; $pass = 'xxx'; $db = 'xxx'; mysql_connect($host,$user,$pass) or die(mysql_error()); mysql_select_db($db) or die(mysql_error()); // Post Variables $challenge = $_POST['challenge']; $insight = $_POST['insight']; $filename=serialize($_POST['filename']); // Upload Images while(list($key,$value) = each($_FILES[image][name])) { if(!empty($value)) { $filename = $value; $add = "upimage/$filename"; copy($_FILES[image][tmp_name][$key], $add); chmod("$add",0777); } } // Insert into database $query = "INSERT INTO table ( challenge, insight, uploaded_image1, uploaded_image2, uploaded_image3,)". "values ( '$challenge', '$insight', '$filename' '$filename2' '$filename3')"; mysql_query($query) or die('Error, query failed : ' . mysql_error()); I know just enough about php that the reason the above wont work, is because $filename2 and $filename3 haven't been given a value. Unfortunately, I have no idea how to separate that information in the array. If anyone can point me in the right direction I'd be forever grateful! Thanks, Gary
  10. Hello All, Thanks for the help. I didn't find a solution, but I did realize that my line of questioning wasn't going to solve this problem. Again, thanks!
  11. "empty" is the actual text information is the table column.
  12. Hello, I have a field in a MySQL table that defaults to "empty" unless it's written over. I'm using the following code to assign a variable based on what's written in that MySQL field. $query2 = "SELECT * FROM $table"; $row = mysql_fetch_assoc(mysql_query($query2)); if ($row['name'] == 'empty') { $joblink = "no download"; } else { $joblink = "download"; }; However, it doesn't seem to recognize the if statement and defaults to else part of the if statement regardless of what actually in the filed. Any ideas? Thanks, Gary
  13. Hello All, I'm working on a web site that needs to list all the rows in a table. This is how I have been doing it. # $query = "SELECT link, page_title, page_content, path FROM $table WHERE approved ORDER BY page_title ASC"; # $result = mysql_query($query) or die('Error, query failed'); # if (mysql_num_rows($result) > 0 ) { # while(list($link, $page_title, $page_content, $path ) = mysql_fetch_array($result)) # $string .= "<a href=".$link." class=maplinkrev target=_blank>".$page_title."</a><br>".nl2br($row['page_content'])."<br><br>"; # } # echo "$string"; This works fine. However, not every entry will always have a link and I need to figure out how to use an if statement to display the link only when it's in the table row but still display the page title. I have no idea of where to begin to look to find the answer to this! Any help would be appreciated. Regards, Gary
  14. Hello, I'm trying to use an if statement to either display a jpg or flv image that has the link stored in a database. The file names will vary, but the .jpg or .flv will always be the same. How do I tell the if statement to ignore everything before the .jpg or .flv file? Below is an example of what I'm using. if ($row['image1'] == jpg) { echo "<img name=mainimage src='".$row['image1']."'><br>"; } Basically, what should I put in place of "jpg" so it will distinguish between file types? Thanks, Gary
  15. Hello All, I'm new to all this and have run into a problem I can't figure out or find any resources/help when I do searches. I have a page where users can go and retrieve several different entries from the database. I can get the results to display on the screen via echo. However, I want the results emailed and the most I can manage is to get one row of results not all of them. How can I get the results of the query displayed in the body of the email message? Any help or pointing in the right direction would be appreciated. You can see the code below. <form method="POST" action=""> <font class="forms">Enter Email Address:</font> <input type="text" name="email" class="story"><br><br> <input type="submit" value="retrieve" class="btn"><br><br> </form> </font> <? if($_SERVER['REQUEST_METHOD'] == 'POST') { $host = 'localhost'; $user = 'xxxxxx; $pass = 'xxxxxx'; $db = 'xxxxxx'; mysql_connect($host,$user,$pass) or die(mysql_error()); mysql_select_db($db) or die(mysql_error()); $email=$_POST['email']; $query = "SELECT first_name, entry_type, entry_id, password FROM uploads_entryid WHERE email = '$email'"; $result = mysql_query($query) or die('Error, query failed'); while(list($first_name, $entry_type, $entry_id, $password) = mysql_fetch_array($result)) echo "Hello $first_name,<br><li>$entry_type:<br>Entry ID: $entry_id<br>Password: $password</li>"; } ?> <?php // Email Username, Password and Story ID $to = "$email"; $from = "xxxxxx"; $subject = "Your Pavement Memoirs Username & Password"; $message = <<<EOF <html> <body> Hello $first_name,<br> <br> Thank you for your submission! Feel free to use the following entry id’s, and passwords when logging in to Pavement Memoirs to make adjustments to your post. <br><br> <br><br> Keep in mind each time you edit your post, the new submission will not be viewable until it’s reviewed to keep the site PG rated. <br><br> </body> </html> EOF; $headers = "From: $from\r\n"; $headers .= "Content-type: text/html\r\n"; $to = "$to, $from"; mail($to, $subject, $message, $headers); ?>
  16. I'm trying to include the last of the three pieces of code into the second. I also tried posting the id and that didn't work either. It's frustrating, that it works when you simply upload a new file but it doesn't when you edit and old upload unless I use the actual primary id instead of having the variable pass.
  17. I tried removing the $id=$_GET['id']; as well. It didn't work and I have no idea why?!
  18. I tried to use the header instead of include, but since this is several parts of different code and I'm a newbie, I couldn't get this upload code to work if I used a header instead of the include. Anyone else have any ideas!
  19. I tried passing the variable through the link, but since it is an include and not a redirect it wouldn't work. Any other suggestions?
  20. Hello All, I have a script that lets people go back and edit an upload to my site. The path of the file isn't getting dumped into the database. There are three php files that do the work. It works fine for the original upload, but the variable ($id primary id that points to the record in the database) gets lost in the version where they do the edit. I've manually entered the variable and it puts the path to the file into the database. I'm including the script for all three pages. The variable isn't getting passed properly to the third script. If anyone can point me in the right direction, I'd appreciate it! Code for Form: <form action="edit_videoscript.php " method="post" enctype="multipart/form-data" name="uploadform"> <font class="forms">First Name:<br> <input type="text" value="<?=$row['first_name']?>" name="first_name" id="first_name" class="story"/> <br><br> E-Mail:<br> <input value="<?=$row['email']?>" type="text" name="email" id="email" class="story"/> <br><br> <a href="javascript:openCityStatePhotoWindow();" class="linkpopup">City & State:<br></a> <input value="<?=$row['city']?>" type="text" name="city" id="city" class="storycity"/> <? include 'library/state1edit.php'; ?> <br><br> <a href="javascript:openLinksWindow();" class="linkpopup">Web Link:<br></a> <input value="<?=$row['link']?>" type="text" name="link" id="link" class="story"/> <br><br> <a href="javascript:openVideoWindow();" class="linkpopup">Video:<br></a> <input name="userfile" type="file" id="userfile" class="storyphoto"> <br><br> <a href="javascript:openCaptionVideoWindow();" class="linkpopup">Video Caption:<br></a> </font> <textarea name="caption" id="caption" wrap="hard" class="storycaption"><?=$row['caption']?></textarea><br> <br><br> <input type="hidden" name="thanks" value="7"> <input type="hidden" name="id" value="<?=$row['id']?>"> <input name="upload" type="submit" id="upload" value="Upload" class="storybtn" onClick="YAHOO.example.container.wait.show();"> </form> Code for the script that starts the upload and puts all but the path into the database <?php $host = 'localhost'; $user = 'xxxxxx'; $pass = 'xxxxxx'; $db = 'xxxxxx'; mysql_connect($host,$user,$pass) or die(mysql_error()); mysql_select_db($db) or die(mysql_error()); // Set variables $hasPortrait = false; if ($_FILES['userfile']['name'] != "") { $hasPortrait = true; } if (!$hasPortrait) { // Set variables $id=$_POST['id']; $first_name = $_POST['first_name']; $email = $_POST['email']; $city = $_POST['city']; $state = $_POST['state']; $link = $_POST['link']; $caption = $_POST['caption']; $thanks = $_POST['thanks']; // Insert into database $query = "UPDATE uploads_content SET first_name = '$first_name', email = '$email', city = '$city', state = '$state', link = '$link', caption = '$caption', approved = '0' WHERE id = '$id' "; mysql_query($query) or die('Error, query failed : ' . mysql_error()); if($query) {header( "Location: http://www.pavementmemories.com/thanks.php?first_name=$first_name&thanks=$thanks" );} else {header( "Location: http://www.pavementmemories.com/oops_contact.php" );} } else { $temp_video = '/home/grozanc/public_html/pavementmemories.com/videos/temp/'; $video_processor = '/home/grozanc/public_html/pavementmemories.com/process_video2.php'; if(isset($_POST['upload'])) { $first_name = $_POST['first_name']; $email = $_POST['email']; $city = $_POST['city']; $state = $_POST['state']; $link = $_POST['link']; $caption = $_POST['caption']; $thanks =$_POST['thanks']; $fileName = $_FILES['userfile']['name']; $tmpName = $_FILES['userfile']['tmp_name']; $fileSize = $_FILES['userfile']['size']; $fileType = $_FILES['userfile']['type']; // get the file extension first $ext = substr(strrchr($fileName, "."), 1); // generate the random file name $randName = md5(rand() * time()); if(!get_magic_quotes_gpc()) { $fileName = addslashes($fileName); $filePath = addslashes($filePath); } if( is_uploaded_file($_FILES['userfile']['tmp_name']) ) { $newfile = $randName . '.' . $ext; move_uploaded_file($_FILES['userfile']['tmp_name'], $temp_video . $newfile); } $cmd = '/usr/bin/php ' . $video_processor . ' "' . $temp_video . $newfile .'" ' . $randName . ' ' . $insert_id; proc_close( proc_open( $cmd . " >> /home/grozanc/public_html/logs/videoprocess.log &", array(), $foo ) ); //exec('/usr/bin/php ' . $video_processor . ' "' . $temp_video . $newfile .'" ' . $randName . ' ' . $insert_id .' &', $out,$rval); file_put_contents('/home/grozanc/public_html/logs/videocommands.log',"\n" . $cmd,FILE_APPEND); //header("Location: http://www.pavementmemories.com/thanks.php"); //exit; $first_name = str_replace("'","’",$first_name); $city = str_replace("'","’",$city); $caption = str_replace("'","’",$caption); $query = "UPDATE uploads_content SET first_name = '$first_name', email = '$email', city = '$city', state = '$state', link = '$link', caption = '$caption', name = '$fileName', approved = 0, featured = 0 WHERE `id` = '{$id}' "; mysql_query($query) or die('Error, query failed : ' . mysql_error()); $insert_id = mysql_insert_id(); include ( "thanks.php" ); exit; } } ?> This last script gets called from the previous one to process the upload and isn't getting the $id variable in the query. If I enter an id of an existing record instead of the $id, it runs the update. <?php require_once('xxxxxx'); $id=$_GET['id']; $base_dir = '/home/grozanc/public_html/'; $web_dir = $base_dir . 'pavementmemories.com/'; $log_dir = $base_dir . 'logs/'; $videos_dir = 'videos/'; $sample_rates = array(44100, 22050, 1102); $ffmpeg_binary = "/usr/bin/ffmpeg"; $flvtool2 = "/usr/bin/flvtool2"; $width = '320'; $height = '240'; $srcfile = $argv[1]; $destfile = $videos_dir . $argv[2] . '.flv'; $fulldest = $web_dir . $destfile; $ffmpegObj = new ffmpeg_movie($srcfile); $srcFPS = $ffmpegObj->getFrameRate(); $srcAB = intval($ffmpegObj->getAudioBitRate()/1000); $srcAR = $ffmpegObj->getAudioSampleRate(); if( !in_array($srcAR,$sample_rates) ) $srcAR = 44100; //exec($ffmpeg_binary . ' -i ' . $srcfile . ' ' . $destfile . " | " . $flvtool2 . " -U stdin " . $destfile,$output,$retval); $vid_command = $ffmpeg_binary . " -i " . $srcfile . " -ar " . $srcAR . " -ab " . $srcAB . " -f flv -s " . $width . "x" . $height . " " . $fulldest; file_put_contents($log_dir . 'ffmpegcmd.log', "\n" . $vid_command, FILE_APPEND); exec($vid_command,$output,$retval); /* echo "Ret Val: $retval<br>"; echo "<pre>"; print_r($output); echo "</pre>"; */ $logdata = date("m-d-Y") . "\n" . var_export($output,true) . "\n\n"; file_put_contents($log_dir . 'videooutput.log', $logdata,FILE_APPEND); $destfile = $_POST['path']; mysql_query(" UPDATE uploads_content SET video_processed = 1, path = '$destfile' WHERE `id` = '$id' "); echo mysql_error(); ?>
  21. Thanks Ken, That solved the problem!
  22. Any ideas on how to fix this or can someone see the problem?
  23. Hello All, I'm trying to create a page that lets people who submitted photos to a website go back in and edit the information they submitted. I got the script to go back in and successfully change the submitted information if they also submit a new photo. If they change everything but the photo the script errors out. My thinking was to use the if statement to have one action happen if the file input field is blank, and a second action if the file input field has information. I think the problem is how I'm trying to triggr the if statement. I've also included the form code as well. Any ideas? where I think the problem is $hasPortrait = false; if ($_FILES['userfile']['name'] != "") { $hasPortrait = true; } if ($hasPortrait = false) { action #1} else {action #2} fullcode <?php $host = 'localhost'; $user = 'xxxxxx'; $pass = 'xxxxxx'; $db = 'xxxxxx'; mysql_connect($host,$user,$pass) or die(mysql_error()); mysql_select_db($db) or die(mysql_error()); // Set variables $hasPortrait = false; if ($_FILES['userfile']['name'] != "") { $hasPortrait = true; } // you can change this to any directory you want // as long as php can write to it $uploadDir = 'photos/'; $home_path = '/home/grozanc/public_html/pavementmemories/'; // Set maximum sizes for resized images. $maxWidth = 375; $maxHeight = 250; $maxWidth2 = 75; $maxHeight2 = 75; $fileName = $_FILES['userfile']['name']; if ($hasPortrait = false) { // Set variables $id=$_POST['id']; $first_name = $_POST['first_name']; $email = $_POST['email']; $city = $_POST['city']; $state = $_POST['state']; $link = $_POST['link']; $caption = $_POST['caption']; $thanks = $_POST['thanks']; // Insert into database $query = "UPDATE uploads_content SET first_name = '$first_name', email = '$email', city = '$city', state = '$state', link = '$link', caption = '$caption', approved = '0' WHERE id = '$id' "; mysql_query($query) or die('Error, query failed : ' . mysql_error()); } else { $id=$_REQUEST['id']; if(isset($_POST['upload'])) { $first_name = $_POST['first_name']; $email = $_POST['email']; $city = $_POST['city']; $state = $_POST['state']; $link = $_POST['link']; $caption = $_POST['caption']; $thanks = $_POST['thanks']; $fileName = $_FILES['userfile']['name']; $tmpName = $_FILES['userfile']['tmp_name']; $fileSize = $_FILES['userfile']['size']; $fileType = $_FILES['userfile']['type']; // get the file extension first $ext = substr(strrchr($fileName, "."), 1); // generate the random file name $randName = md5(rand() * time()); // and now we have the unique file name for the upload file $filePath = $uploadDir . $randName . '.' . $ext; $filePathResized = $uploadDir . $randName . '.sized.' . $ext; $filePathThumb = $uploadDir . $randName . '.thumb.' . $ext; // move the files to the specified directory // if the upload directory is not writable or // something else went wrong $result will be false $result = move_uploaded_file($tmpName, $filePath); if (!$result) { echo "Error uploading file"; exit; } else { $size=getimagesize($filePath); if ($size[0]>$maxWidth or $size[1]>$maxHeight) { $result2 = @exec("/usr/bin/convert -geometry '" . $maxWidth . "x" . $maxHeight . ">' $filePath $filePathResized"); # echo $result2; } { $result3 = @exec("/usr/bin/convert -geometry '" . $maxWidth2 . "x" . $maxHeight2 . ">' $filePath $filePathThumb"); # echo $result3; } } if(!get_magic_quotes_gpc()) { $fileName = addslashes($fileName); $filePath = addslashes($filePath); } $query2 = 'UPDATE uploads_content SET first_name = "' . $first_name . '" , email = "' . $email . '" , city = "' . $city . '" , state = "' . $state . '" , caption = "' . $caption . '" , name = "' . $fileName . '" , size = "' . $fileSize . '" , type = "' . $fileType . '" , path = "' . $filePath . '" , sized = "' . $filePathResized . '" , thumb = "' . $filePathThumb . '" , approved = "' . 0 . '" where id=' . $id; mysql_query($query2) or die('Error, query failed : ' . mysql_error()); } include 'thanks.php'; } ?> Form Code <form action="edit_photoscript.php " method="post" enctype="multipart/form-data" name="uploadform"> <font class="forms">First Name:<br> <input value="<?=$row['first_name']?>" type="text" name="first_name" id="first_name" class="story"/> <br><br> E-Mail:<br> <input value="<?=$row['email']?>" type="text" name="email" id="email" class="story"/> <br><br> <a href="javascript:openCityStatePhotoWindow();" class="linkpopup">City & State:<br></a> <input value="<?=$row['city']?>" type="text" name="city" id="city" class="storycity"/> <? include 'library/state1edit.php'; ?> <br><br> <a href="javascript:openLinksWindow();" class="linkpopup">Web Link:<br></a> <input value="<?=$row['link']?>" type="text" name="link" id="link" class="story"/> <br><br> <a href="javascript:openPhotoWindow();" class="linkpopup">Photo:<br></a> <img src='<?=$row['thumb']?>'><br> <input name="userfile" type="file" id="userfile" class="storyphoto"> <br><br> <a href="javascript:openCaptionPhotoWindow();" class="linkpopup">Photo Caption:<br></a> </font> <textarea name="caption" id="caption" wrap="hard" class="storycaption"><?=$row['caption']?></textarea><br> <br><br> <input type="hidden" name="thanks" value="6"> <input type="hidden" name="id" value="<?=$row['id']?>"> <input name="upload" type="submit" id="upload" value="Upload" class="storybtn" onClick="YAHOO.example.container.wait.show();"> </form>
  24. Thanks Everyone! The last comment solved the problem. The $id variable wasn't being passed from the form properly. Now that I fixed that everything works fine!
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