How can I get a result from mysql_fetch_row (or mysql_fetch_array)

Question

$sqlQuery = "SELECT * FROM allowedUsers WHERE UserID = '" . $kUserID . "'";
$result=mysql_query($sqlQuery, $db);
if(!result)
{
    echo "Error running query <br>" . mysql_error();
    exit;
}
while($row = mysql_fetch_array($result))
{
    echo $row[2];
}

I run the SQLQuery in phpMyAdmin and I am getting a valid result (1 row) the table (allowedUsers) has 6 fields I can't get anything out of the DB.

Any help is appreciated.

The code looks fine. Is there really no output in $row? Are you sure the user ID exists? — Pekka
– Pekka, Commented Feb 24, 2010 at 14:55

thetaiko · Accepted Answer · 2010-02-24 15:08:48Z

1

if(!result) should be if(!$result)

answered Feb 24, 2010 at 15:08

thetaiko

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Comments

Jack M. · Accepted Answer · 2010-02-24 14:58:14Z

0

According to PHP.net's documentation, you don't need to pass $db to mysql_query(). Take a look at the example code:

<?php
    mysql_connect("localhost", "mysql_user", "mysql_password") or
        die("Could not connect: " . mysql_error());
    mysql_select_db("mydb");

    $result = mysql_query("SELECT id, name FROM mytable");

    while ($row = mysql_fetch_array($result, MYSQL_NUM)) {
        printf("ID: %s  Name: %s", $row[0], $row[1]);  
    }

    mysql_free_result($result);
?>

It may be helpful to see your connection code, ensure you've selected a database, etc.

answered Feb 24, 2010 at 14:58

Jack M.

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1 Comment

Jack M. Over a year ago

Take a look at @thetaiko's answer, as well. He has a good point which may show you the actual error you're encountering.

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