yddib Posted September 10, 2008 Share Posted September 10, 2008 <?php mysql_connect("", "", "") or die(mysql_error()); //connects to the server mysql_select_db(""); //selects the database $result = @mysql_query("SELECT picgal FROM gradinfo where u_id ='$id'"); if (!$result) { exit('<p>Error performing query: ' . mysql_error() . '</p>'); } while ($row = mysql_fetch_array($result)) echo $row['picgal']; ?> Where picgal is =' http://www.example.net/things/photos/xphotos/index.html' echo $row['picgal'] brings out the link but i want the page. Can anyone help?? Thanks Link to comment https://forums.phpfreaks.com/topic/123588-get-html-link-out-of-database/ Share on other sites More sharing options...
JonnoTheDev Posted September 10, 2008 Share Posted September 10, 2008 You mean grab the contents of the page from the external link? Link to comment https://forums.phpfreaks.com/topic/123588-get-html-link-out-of-database/#findComment-638218 Share on other sites More sharing options...
yddib Posted September 10, 2008 Author Share Posted September 10, 2008 Well its a link to a file on the host!! Does that make sense!? Link to comment https://forums.phpfreaks.com/topic/123588-get-html-link-out-of-database/#findComment-638219 Share on other sites More sharing options...
JonnoTheDev Posted September 10, 2008 Share Posted September 10, 2008 Doesnt really matter if on same host because you are using a URL rather than a filepath. If URL file opening is allowed on your server you can use: $handle = fopen($row['picgal'], "r"); while (!feof($handle)) { $contents .= fread($handle, 8192); } fclose($handle); echo $contents; Other than that you can use a standard include(); so: include($row['picgal']); If fopen on URLS is not allowed then you would need to use CURL http://uk2.php.net/manual/en/book.curl.php Link to comment https://forums.phpfreaks.com/topic/123588-get-html-link-out-of-database/#findComment-638222 Share on other sites More sharing options...
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