Question

Here is the code………

<?php
include('../inc/php/inc/dbc.php');
$query = "SELECT * FROM available_fsv WHERE a_status = '1'";
$result_query = mysql_query($query);

while($row = mysql_fetch_assoc($result_query)){ 

        $billingid = $row['billingid'];

$query = "UPDATE available_fsv SET b_status = '1' WHERE billingid = '$billingid'";
$result_query = mysql_query($query);
echo $result_query; 
}
?>

The error i am getting………..

Warning: mysql_fetch_assoc() expects parameter 1 to be resource, boolean given in C:\wamp\www\php\fsv_shutdown_cron.php on line 6

Database structure is like–

____________________________________________________
|   id   |   a_status | b_status    | billingid    |
|--------|------------|-------------|--------------|
|    1   |    1       |     0       |      1       |
|--------|------------|-------------|--------------|
|    2   |    0       |     0       |      12      |
|--------|------------|-------------|--------------|
|    3   |    0       |     0       |      9       |
|--------|------------|-------------|--------------|
|    4   |    1       |     0       |      3       |
|________|____________|_____________|______________|

What i wanna do is if a_status is 1 then update b_status to 1.

I am learning php and I know this is a stupid question but please help me. Thanks in advance.. 🙂

Answer 1

This call mysql_query($query) returned FALSE, instead of result set, which means that your query had errors in it. Use the following code to see the error:

$result_query = mysql_query ($query);
if ($result_query === FALSE)
{
    echo (mysql_error ());
    die (1);
}

Looks like I was wrong. You really overwrite original value of $result_query when you execute your update query.