Jump to content

Cazrin

Members
  • Posts

    10
  • Joined

  • Last visited

    Never

Everything posted by Cazrin

  1. Just an idea and not sure if it would solve anything, have you tried it with " " around your _blank? So it would be target="_blank"> That's the only thing which looks odd to me, everything else seems fine.
  2. Unless you used PHP to load every e-mail address in your database into a JavaScript array, you'll have to use AJAX. Having never used AJAX I can't tell you how to do it (sorry ) but I'm sure it'll be pretty straight forward.
  3. I'd remove that attachment asap. You've got the ftp_info.txt in there.
  4. It's a problem with your if() statement I think <?php if (($_FILES['ColourImage']['type'] == "image/gif") && filesize($ColourImage['tmp_name']) > $Max_Size && ($Img_Dimensions[0] > 50) || ($Img_Dimensions[1] > 50)) ?> Try adding some extra brackets in to make sure: <?php if (($_FILES['ColourImage']['type'] == "image/gif") && (filesize($ColourImage['tmp_name']) > $Max_Size) && ($Img_Dimensions[0] > 50) || ($Img_Dimensions[1] > 50)) ?>
  5. Do what adam said to create the shortened version of the news post. As for the link to the rest of the news, create another file which will take the id of the news posts from the URL ($_GET['id'] for example) and then display the news post based on that id. There's another shove in the right direction
  6. No need to have 2 tables to do it, just 1 table and 2 columns. Column 1 = smiley Column 2 = replace Build your 2 arrays based on each column, and then use the same code you've got already.
  7. I guess you could put an if() statement before that, to check which is greater. <?php $a = 400; $b = 300 if($a>$b) { // Swap the values of $a and $b $temp = $b; $b=$a; // Is now 400 $a=$temp; // Is now 300 } $query = "SELECT * FROM `table1` WHERE `col1` BETWEEN ".$a." and ".$b; ?> Haven't used BETWEEN before so not sure if that would work But I'd say give it a go, makes sense to me!
  8. It depends what you're wanting to select from the table. If you're wanting to query a specific row, it's usually best to go by the row ID, so you'd want to keep that in your query. Some more info on what you're wanting to actually query would be great
  9. Hi there, I recently want this on my website and managed to get it to work. I was using DIVs instead of tables but give this a try: $counter = 0; while($output=mysql_fetch_object($result)){ echo "<td><img src=\"$medal[image]\" width=\"99\" height=\"26\" alt=\"$row[reason]\" /></td>n"; $counter++; if($counter == $num_cols) { echo "</tr> <tr>"; } That's the way round I did it, rather than before the line of code. Let me know if that works
  10. I see you're sending this from your own server on xamp. I'm pretty sure that xamp doesn't include an SMTP server as standard, so make sure you get one installed otherwise it isn't going to work at all. I get this same issue when testing scripts that try send e-mails, but they work fine once they're uploaded to my webspace. Maybe someone can recommend a good SMTP server to try, but I'm sure google will give you some free programs out there.
×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.