emvy03

Okay, I think I understand. I have a similar page for a news page so I think I could do that. Would I be able to post any issues I may have with that?

Alternatively, Is there a better way to list images in a table view?

Hi guys, I posted a syntax error I was having a problem with a few hours ago with my fopen script, which was solved, however it turns out the whole script has an overall issue with it. <?php $gallery_name = $_POST['galleryname']; $path="../gallery/$gallery_name"; mkdir("$path", 0777); //Create the gallery php file $ourFileName = "$path/$gallery_name.php"; $ourFileHandle = fopen($ourFileName, 'w') or die("can't open file"); fclose($ourFileHandle); include "loginscript.php"; $sql = mysql_query("INSERT into images (gallery_name) VALUES ('$gallery_name')"); //Write the upload form script to the gallery php file $myFile = "$path/$gallery_name.php"; $fh = fopen($myFile, 'w') or die("can't open file"); $stringData = "Welcome to '$gallery_name'\n <br />"; $stringData.="<?php\n"; $stringData.="include('../../storescripts/connect_to_mysql.php');\n"; $stringData.="\$gallery=\"$gallery_name\";\n"; $stringData.="\$query=mysql_query(\"SELECT image_path FROM images WHERE gallery_name='\$gallery'\");\n"; $stringData.=" if(@mysql_num_rows($query) > 0){\n"; $stringData.="\$image_path=\$row['image_path'];"; $stringData.="echo \"<table>\";\n"; $stringData.="\$count = 0;\n"; $stringData.="while(\$row = mysql_fetch_array(\$query)){\n"; $stringData.="if(\$count == 0){\n"; $stringData.="echo \"<tr><td>\";\n"; $stringData.="}else {\n"; $stringData.="echo \"<td>\";\n"; $stringData.="}\n"; $stringData.='echo "<img src=\'$image_path\' />";'; $stringData.="if (\$count == 3){\n"; $stringData.="echo \"</td></tr>\";\n"; $stringData.="\$count = 0;\n"; $stringData.="}else {\n"; $stringData.="echo \"<td>\";\n"; $stringData.="\$count++;\n"; $stringData.="}\n"; $stringData.="}\n"; $stringData.="\$cells_left = 4 - \$count;\n"; $stringData.="if( \$cells_left > 0){\n"; $stringData.="\$i = 0;\n"; $stringData.="while (\$i <= \$cells_left){\n"; $stringData.="echo \"<td></td>\";\n"; $stringData.="\$i++;\n"; $stringData.="}\n"; $stringData.="echo \"</tr>\";\n"; $stringData.="\n}"; $stringData.="\n"; $stringData.="echo \"</table>\";\n"; $stringData.="} else {\n"; $stringData.="echo \"No Images in the Gallery\";\n"; $stringData.="}\n"; $stringData.="?>\n"; fwrite($fh, $stringData); fclose($fh); ?> <?php echo '<a href="managegallery.php">Go Back</a>'; ?> That is my code which writes a script to a new file to take all the images from a directory and display them in a dynamic table. The table has 4 fixed columns and then a variable number of rows based on the number of images. The issue I am having is once this code writes the gallery page, this page then doesn't display images. I can edit the gallery page so that it displays images but then if I make these changes to the fwrite script I get syntax errors that I am unable to solve. I'm stuck between having a valid fwrite script and display errors or display images with fwrite errors. Sorry, if I've made it sound confusing. Any help would be gratefully appreciated.

Hi, I have some php script which writes code to a given file. The problem I have is I keep getting an error with this one particular line: $stringData.=" '<img src="' . \$row[\'image_path\'] . '"/>'; "; It keeps giving the error: Parse error: syntax error, unexpected T_CONSTANT_ENCAPSED_STRING in ... Does anyone know what the error could be please? Cheers.

Thank you. Everything is now solved!! Muddy_Funster, you sir, are a complete legend!!

Yes!! Thank you so much for all your help. I guess now instead of echo-ing values, I put a simple mysql query there to insert the values?

Hi, Sorry, I'm on holiday at the moment so my ability to access the internet is somewhat temperamental. I applied your suggestion and the following is displayed: Array ( [0] => Small [1] => Medium [2] => Large ) Thanks mate!!

Hi mate. I tried that suggestion but it doesn't seemed to have solved it.

Hi, Thanks so much for the help. I've done that, and added an if(isset()) function in there to solve a foreach error. But, all that happens when I tick a box is a 1 is generated for the small size, regardless of which box I tick.

Hello. Ta for the reply, the only thing is I have the idea of what needs doing but not the know how of how to execute it. So I need a little help.

Hi. For instance, I want a column in my database table for small, medium and large. Then, based on whether a user ticks the check box for the respective size a 1 or 0 would be inserted into the table. I was thinking I would need an if statement that says; if the box is ticked then insert a 1 into the table, else put a 0?

Hi, A few of you may remember but I recently solved a problem I was having in uploading check box data to MySQL database, which I managed to solve. At the time all I wanted to do was take check box data and put it into a table, and done so using an array which took the selected check boxes and put it onto separate rows. I.e. If i selected 4 check boxes, then 4 rows would be created. What I now want to do is create a separate column in my table for each check box and have a 0 or 1 inserted into there based on whether the box was ticked. My code at the moment is: <?php include "storescripts/connect_to_mysql.php"; if(isset($_POST['size'])) { $size = $_POST['size']; $n = count($size); $i = 0; while ($i < $n) { mysql_query("INSERT INTO events (value) VALUES ('$language[$i]')") or die(mysql_error()); $i++; } echo "</ol>"; } ?> <form method="post" action="<?php echo $_SERVER['PHP_SELF'];?>"> Select the size you would like<br> <input name="size[]" type="checkbox" value="Small"> Small<br> <input name="size[]" type="checkbox" value="Medium"> Medium<br> <input name="size[]" type="checkbox" value="Large"> Large<br> <input name="send" type="submit" id="send" value="Upload"> </form>

Hi all, I've made a basic cms using php sessions which worked perfectly ok when testing offline, but since uploading live I'm getting this error message when ever I go to a page once logged in. Notice: A session had already been started - ignoring session_start() in /homepages/18/d374931144/htdocs/storeadmin/loginscript.php on line 2. I will the whole php script if necessary but the only code on the problematic line is: session_start(); I was thinking maybe writing an 'if' condition but I am unsure and struggling Cheers.

Solved, just moved the position of the variables Just need to hope the upload function works when the site goes live.

Hi, Sorry guys. The code is (I changed it a little bit from the previous post): <?php function getExtension($str) { $i = strrpos($str,"."); if (!$i) { return ""; } $l = strlen($str) - $i; $ext = substr($str,$i+1,$l); return $ext; } $errors=0; if(isset($_POST['Submit'])) { $image=$_FILES['image']['name']; if ($image) { $filename = stripslashes($_FILES['image']['name']); $extension = getExtension($filename); $extension = strtolower($extension); if (($extension != "jpg") && ($extension != "jpeg") && ($extension != "png") && ($extension != "gif")) { echo '<h1>Unknown extension!</h1>'; $errors=1; } else { $size=filesize($_FILES['image']['tmp_name']); if ($size > MAX_SIZE*1024) { echo '<h1>You have exceeded the size limit!</h1>'; $errors=1; } $image_name=time().'.'.$extension; $gallery_name = ($_POST['galleryname']); $path="gallery/$gallery_name/"; $newname="$path".$image_name; $copied = copy($_FILES['image']['tmp_name'], $newname); if (!$copied) { echo '<h1>Copy unsuccessfull!</h1>'; $errors=1; }}}} if(isset($_POST['Submit']) && !$errors) { echo "<h1>File Uploaded Successfully! Try again!</h1>"; } ?> <?php include "storescripts/connect_to_mysql.php"; $sql = mysql_query("INSERT into images (gallery_name,image_path) VALUES ('$gallery_name','$image_name')"); ?> Giving error message: Notice: Undefined variable: gallery_name in C:\Program Files\xampp\htdocs\ppyfc\addImageCode.php on line 69 Notice: Undefined variable: image_name in C:\Program Files\xampp\htdocs\ppyfc\addImageCode.php on line 69

Hi, thanks for your reply. I did both of your suggestions, but I'm still getting the same error.

Cheers guys, that worked a treat. Just having a little trouble with the image upload. <?php function getExtension($str) { $i = strrpos($str,"."); if (!$i) { return ""; } $l = strlen($str) - $i; $ext = substr($str,$i+1,$l); return $ext; } $errors=0; if(isset($_POST['Submit'])) { $image=$_FILES['image']['name']; if ($image) $filename = stripslashes($_FILES['image']['name']); $extension = getExtension($filename); $extension = strtolower($extension); if (($extension != "jpg") && ($extension != "jpeg") && ($extension != "png") && ($extension != "gif")) { echo '<h1>Unknown extension!</h1>'; $errors=1; } else { $size=filesize($_FILES['image']['tmp_name']); $image_name=time().'.'.$extension; $gallery_name = $_POST['galleryselect']; $newname=$gallery_name.'/'.$image_name; $copied = copy($_FILES['image']['tmp_name'], $newname); if (!$copied) { echo '<h1>Copy unsuccessfull!</h1>'; $errors=1; }}}} if(isset($_POST['Submit']) && !$errors) { echo "Upload Successful."; } include "storescripts/connect_to_mysql.php"; $sql = mysql_query("INSERT into images (image_path) VALUES ($newname)"); ?> Keeps giving the error 'Unknown variable newname'. I've tested this code in a simpler system where the user just uploads a picture so I don't think there is a fundamental issue, just little nagging errors I think.

Here we are. (galleryname is the name of the textfield that the user writes the name into) <?php $gallery_name = mysql_real_escape_string($_POST['galleryname']); ?> <?php mkdir("gallery/$gallery_name"); //Create the gallery php file $ourFileName = "gallery/$gallery_name.php"; $ourFileHandle = fopen($ourFileName, 'w') or die("can't open file"); fclose($ourFileHandle); //Write the upload form script to the gallery php file $myFile = "gallery/$gallery_name.php"; $fh = fopen($myFile, 'w') or die("can't open file"); $stringData = "Welcome to '$gallery_name'"; fwrite($fh, $stringData); $stringData = ' <?php include "../storescripts/connect_to_mysql.php"; $query = mysql_query("SELECT image_path FROM images WHERE gallery_name = $gallery_name"); while($row = mysql_fetch_array($query,MYSQL_ASSOC)){ $image_path = $row["image_path"]; print "<img src=$image_path /> <br />"; } ?> '; fwrite($fh, $stringData); fclose($fh); ?> I'm thinking it must be something to do with the quote and double quotes in there but I don't know of a work around. :/

Hi, I think I've written the mysql code but I'm having the same issue where the variable '$gallery_name' isn't being transferred (for want of a better word) so when the galley file is being written instead of writing in the gallery name it literally writes '$gallery_name'. The thing is though, when writing a title for the page it manages to grab the variable with no problems at all.

Cheers mate. I'm okish with mysql I think, I've managed to do a simple mysql databse with update, delete and insert functions so far but thank you all the same. I'm going to give the mysql code a go and see where I get with it. Thanks.

Hi, Thanks for your replies guys. With regards, to the mysql solution. I guess then that a new line would be created for a new image source. I'm assuming a new page is created for each gallery that grabs the gallery name? Thanks.

Hi, I'm trying to write some code for a gallery where a user types a name into a box and submits it, which then creates a gallery page to hold images. So far I have written the code that takes the name typed in, creates a directory to hold the images and also creates a php file with the same name that is the actual gallery page. Using the fopen and fwrite functions I have got so far as writing the php gallery code, where a welcome line uses the gallery name. I have also written some code to display all the images in the given directory- this is where I am having issues, as the code doesn't use the name variable I have set and thus doesn't open the directory. Perhaps showing the code may give a better idea. <?php //Define the name variable $gall = mysql_real_escape_string($_POST['galleryname']); //Create a folder with the given name mkdir("gallery/$gall"); echo "Successfully Created. Go <a href='imageupload.php'>Back</a>"; ?> <?php //Create the gallery php file $ourFileName = "gallery/$gall.php"; $ourFileHandle = fopen($ourFileName, 'w') or die("can't open file"); fclose($ourFileHandle); ?> <?php //Write the gallery script to display images in given directory $myFile = "gallery/$gall.php"; $fh = fopen($myFile, 'w') or die("can't open file"); $stringData = "<h1>Welcome to the gallery $gall</h1>"; fwrite($fh, $stringData); $stringData = ' <?php //Take the gall variable and remove the .php suffix to give the directory $gall = $_POST["galleryselect"]; $final = substr($gall", 0, -4); if ($opendir = opendir($final)){ while(($file = readdir($opendir)) !== FALSE) { if($file!="."&&$file!=".."){ echo "<img id='dispimg' src ='$dir/$file'/><br>"; } } } ?> '; fwrite($fh, $stringData); fclose($fh); ?> Thanks for reading!!

Ta very much for all your help. Managed to find a way to do it. Just having issues with another piece of code now to do with 'fwrite' function. May start a new topic. Cheers

Hi, many thanks for the reply. Yea, by executing the query directly into mysql it works perfectly and the Moon T-Shirt goes in fine. Yet, when the php tries to do the query I get the error message. The second double quote is in there to accompany the double quote just before the 'INSERT ...' P.S. My error message is now (You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '' at line 2)

I don't follow I see there are php variables in there, '$product_name' etc, but I get how to publish the 'raw query'.