newb Posted June 12, 2006 Share Posted June 12, 2006 [code]<?php$query = "SELECT * FROM `thumbnails` ORDER BY title ASC";$result = mysql_query( $query ); if ( !$result ) {die($query.'<br />'.mysql_error());while( $row = mysql_fetch_assoc( $result ) ) {echo '<a href="' . $row['link'] . '">' .$row['thumb'].'<br /></a>;$id = $row['id'];?>[/code]i want it to generate [a href=\"http://mysite.com/view.php?id=1\" target=\"_blank\"]http://mysite.com/view.php?id=1[/a] for example but instead it generates [a href=\"http://mysite.com/view.php?id=$id\" target=\"_blank\"]http://mysite.com/view.php?id=$id[/a]btw, data in the row link is "view.php?id=$id" Link to comment https://forums.phpfreaks.com/topic/11766-id-wont-generate-in-the-url/ Share on other sites More sharing options...
.josh Posted June 12, 2006 Share Posted June 12, 2006 uh.. well first off, you don't have $row['id'] anywhere in your link. 2nd, are you saying that if you do this:echo $row['link'];it spits out this?"view.php?id=$id"??i'm gonna go on a hunch and tell you to put the $id = $row['id'] before the echo instead of after it. but that's just a hunch, cuz i can't see why on earth you would setup your db the way i kinda think you set it up... Link to comment https://forums.phpfreaks.com/topic/11766-id-wont-generate-in-the-url/#findComment-44538 Share on other sites More sharing options...
newb Posted June 12, 2006 Author Share Posted June 12, 2006 [!--quoteo--][div class=\'quotetop\']QUOTE[/div][div class=\'quotemain\'][!--quotec--]are you saying that if you do this:echo $row['link'];it spits out this?"view.php?id=$id"??[/quote]yes.[!--quoteo--][div class=\'quotetop\']QUOTE[/div][div class=\'quotemain\'][!--quotec--]i'm gonna go on a hunch and tell you to put the $id = $row['id'] before the echo instead of after it. but that's just a hunch, cuz i can't see why on earth you would setup your db the way i kinda think you set it up...[/quote]ill try that thankxz. Link to comment https://forums.phpfreaks.com/topic/11766-id-wont-generate-in-the-url/#findComment-44701 Share on other sites More sharing options...
poirot Posted June 12, 2006 Share Posted June 12, 2006 First, your code is missing some closing "}" and a single quote, this would give you some errors.Let me see if I got it, you are storing PHP code in the database? If you are, you should use:[code]<?php$query = "SELECT * FROM `thumbnails` ORDER BY title ASC";$result = mysql_query($query);if (!$result) { die($query . '<br />' . mysql_error());}while ($row = mysql_fetch_assoc($result)) { $id = $row['id']; eval("\$link = \"{$row['link']}\";"); echo '<a href="' . $link . '">' . $row['thumb'] . '<br /></a>';}?>[/code]Although I would not store the links, supposing they'll always the same, but only the id will change:[code]<?php$query = "SELECT * FROM `thumbnails` ORDER BY title ASC";$result = mysql_query($query);if (!$result) { die($query . '<br />' . mysql_error());}while ($row = mysql_fetch_assoc($result)) { echo '<a href="view.php?id=' . $row['id'] . '">' . $row['thumb'] . '<br /></a>';}?>[/code] Link to comment https://forums.phpfreaks.com/topic/11766-id-wont-generate-in-the-url/#findComment-44710 Share on other sites More sharing options...
newb Posted June 12, 2006 Author Share Posted June 12, 2006 ah ok.[code] echo '<a href="view.php?id=' . $row['id'] . '">' . $row['thumb'] . '<br /></a>';[/code]that got me what i wanted, thanks alot. Link to comment https://forums.phpfreaks.com/topic/11766-id-wont-generate-in-the-url/#findComment-44740 Share on other sites More sharing options...
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