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MySQL Query result

Posted on 2012-03-30
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Last Modified: 2012-04-04
Hello Experts;

I have a MySQL query to echo a result from a single cell in a column named 'status' based on an email address in column 'email'. The email variable is passed from a previous page but the query does not return the cell value in the 'status' column defined by the email address; the code snippet is;

<?php
session_start();
include_once("config.php");
$email=$_SESSION["email"];

mysql_select_db("$dbname", $link) || die("Couldn't open db: $dbname. Error if any was: ".mysql_error() );
$result = mysql_query("SELECT status FROM users WHERE email='$email'");
$row = mysql_fetch_array($result);

$status = $row["status"];

echo $dbname; //test database connect
echo $email; //check variable was passed
mysql_close($link);
?>

Can someone identify what I'm doing wrong and help in correcting the code to return the cell value in the 'status' column  based on the email address variable?

Thank you,
visionetv
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Question by:visionetv
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4 Comments
 
LVL 84

Expert Comment

by:Dave Baldwin
ID: 37788550
The code looks ok assuming you haven't misspelled anything.  Are you getting an error message or anything?  I don't see that you are echoing $status so you can check it.
0
 

Accepted Solution

by:
visionetv earned 0 total points
ID: 37788998
Dave;

Thanks for pointing out a silly mistake... tough to know the code is working when I didn't echo the result.

Thank,
visoinetv
0
 

Author Comment

by:visionetv
ID: 37789040
The 100 points should be awared to Dave Baldwin.
 visionetv
0
 

Author Closing Comment

by:visionetv
ID: 37805196
Overlooked echoing result from query to confirm php code functioning properly
0

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