I'm trying to grab a value from a mySQL database and put that value on the end of a url path

e.g. http://url.com/$valuefromdb

Is there any tutorials to how to do this or can anybody recommend how to tackle this?

Kind Regards


JayJ

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I've tried various things now to attempt to get this to work - all to no avail.

My current is

// Define variables
$filename = mysql_query("SELECT log_filename FROM uploads_log");
$issueno = mysql_query("SELECT log_issue FROM uploads_log");
$issuemonth = mysql_query("SELECT log_month FROM uploads_log");

print "<a href=www.url/uploads/$filename><?php echo Issue $issueno.", ".$issuemonth; ?></a>";
<?php

$xconn =mysql_pconnect("localhost","username","userpassword") or die("unable to connect to dbengine".mysql_error());// assuming a connection to mysql db
$qry = "select field from table" ; //your sql statement


$exec_qry = mysql_query($qry) or die ("query failed".mysql_query());

$pulling =  mysql_fetch_array( $exec_qry,MYSQL_ASSOC);

foreach($pulling as $valuefromdb);
printf('%s','<a href="targetlink.ext". "?parameter= ". $valuefromdb . " '  ) ;

mysql_close($xconn);

?>

I hope this helps, Goodluck.

Thanks.

I've tried that and got the following errors:

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in table2.php on line 7

Warning: Invalid argument supplied for foreach() in table2.php on line 9

Hi...
Try this....

<?php

// Define variables
$result1 = mysql_query("SELECT log_filename FROM uploads_log");
$filename = mysql_fetch_array($result1);
$result2 = mysql_query("SELECT log_issue FROM uploads_log");
$issueno = mysql_fetch_array($result2);
$result3 = mysql_query("SELECT log_month FROM uploads_log");
$issuemonth = mysql_fetch_array($result3);

print "<a href='www.url/uploads/".$filename."'>Issue ".$issueno.", ".$issuemonth."</a>"; 

?>
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