Stupid question

[Elhombrebala]

I have a table that stores sessions.
Every session is unique and different so if I query:

$pegar2 = "SELECT Inmo FROM $tabla2 WHERE (Sesion='$session')";

It should give me only 1 record.
How can I echo this?

[portaloo]

$pegar2 = "SELECT Inmo FROM $tabla2 WHERE Sesion='$session'";
$result = mysql_fetch_array($pegar2);
$Inmo = $result["Inmo"];
echo $Inmo;

or if used somewhere in the page:

<? echo $Inmo?>

note you do not need the semicolon if used this way

[Elhombrebala]

I get an error:

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /chs/p1/costa4seasons.com/home/html/uploader6.html on line 43

[portaloo]

ok first off i would not have a table called $tabla2, actually i'm not sure if you are actually allowed $ in front of the table name but hey i could be wrong.

can you post line 43 so i can have a look at it?

lastly ensure that the field names correspond to what the fields are called in the db; ie are Inmo and Sesion the same names in the db and are they capitalised?