[idhrahil]

I am trying to run a script that produces a name from a query to a MySQL database:

PHP Code:

<?php

$sessionid = 'ne8295jskeuv03582je8ao2lf99sm3ng';

$query =
"SELECT users.First_name
FROM sessions
INNER JOIN users
ON sessions.UserID = users.UserID
WHERE sessions.SessionID = $sessionid ";

$con = mysql_connect ("mysql7.000webhost.com", "DELETED BY MODERATOR", "DELETED BY MODERATOR");
$db = mysql_select_db("DELETED BY MODERATOR",$con);

$result = mysql_query ($query);

while($row = mysql_fetch_array($result))
  {
  echo $row['First_name'];
  }

?>

It involves using a php variable within the SQL query. When I replace this with the string itself it works fine, but when I try to do it with the variable in its place it says:

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/a5938807/public_html/test.php on line 25

What's going on? How do I put a variable into a query so that it is picked up properly as a string?

[dx4]

Try assigning the query variable in a single line.

[idhrahil]

Nope, still doesn't work. And that wouldn't explain why it works when I put in the data myself. It's only when I try to use the variable $sessionid instead of directly inputting a string that it doesn't work.

[d3iti]

If the problem is with the name of the session, you can try to put single quotes in the query to the variable $sessionid:

PHP Code:

$query =
"SELECT users.First_name
FROM sessions
INNER JOIN users
ON sessions.UserID = users.UserID
WHERE sessions.SessionID = '$sessionid' "; 


[idhrahil]

That seems to work - thanks!

[d3iti]

No problem.