CODEWITHSUNDEEP
phpmysql
user3412075
user3412075

Reputation: 129

PHP while loop issue won't overwrite variable

I have this while loop code below written for php, I got no idea why the loop keep repeating endlessly so I put a counter to break it out.

Upon debug the result that echo was, it seems my variable $loop_userid won't overwrite.

Referral is 1021 Referral is 1021 Referral is 1021 Referral is 1021 Referral is 1021 Referral is 1021

I have a list of sql records. for account 1021, the referral was 1000

The Idea of this is

3 Accounts

1008 > 1021 > 1000

But my loop stuck at 1021

I wonder why my $loop_userid won't overwrite with the new value that is drawn out from the database.

From mysql database select

account_id  referral

1000
1008        1021
1021        1000

If i echo my sql select statement

select * from account where account_id='1008'
select * from account where account_id='1021'
select * from account where account_id='1021'
select * from account where account_id='1021'
select * from account where account_id='1021'
select * from account where account_id='1021'

Below is my code

$payment_owe_loop = "inside";
//initial userid is my own userid
$loop_userid = $UserId;

//UserId is 1008 at this point

while($payment_owe_loop=="inside")
{

$sql_select = "select * from account where account_id='$loop_userid'";
$result = mysql_query($query);

//echo $sql_select;

$loop_userid = "";
$count = mysql_num_rows($result);

if($count>0)
{
while($rowINNER = mysql_fetch_array($result))
{
$loop_userid = $rowINNER['referral'];
$my_rebate = $rowINNER['rebate'];
$my_cost = (100 - $my_rebate) * $total_cost * 0.01;
}//end while fetch inner row
}//if there rows return

echo " Referral is " . $loop_userid;

if($count==0)
{
$payment_owe_loop = "outside";
}

if($counter>4)
{
$payment_owe_loop = "outside";
}

$counter++;

unset($rowINNER);

}//end while payment loop

Issue solve. I found that I query the wrong sql statement

$sql_select = "select * from account where account_id='$loop_userid'";
$result = mysql_query($query);

It suppose be $sql_select instead of $query

Blinded by the sql statement, I miss out checking the variable that I key in wrongly

Upvotes: 1

Views: 507

Answers (1)

Michael Petch
Michael Petch

Reputation: 47573

The only thing I can fathom here is that since your account database uses a string for account_id that some account_id's have trailing spaces. So you get your referral for 1008 which is 1021 but I'm not sure 1021 is in the database maybe it was entered as "1021 " (some trailing spaces)

I'd be curious to know what you get as a result of this query directly on your database (just do it directly via mysql client)

select * from account where account_id='1021';

Does that actually return a row or not? If it doesn't then I suspect trailing spaces that would have to be trimmed. If they need trimming then this line of code might have to e modified from:

$sql_select = "select * from account where account_id='$loop_userid'";

to:

$sql_select = "select * from account where TRIM(account_id)='$loop_userid'";

On a side note: Although I am sure this isn't related to your problem I would consider amending this code a bit:

if($count>0)
{
    while($rowINNER = mysql_fetch_array($result))
    {
        $loop_userid = $rowINNER['referral'];
        $my_rebate = $rowINNER['rebate'];
        $my_cost = (100 - $my_rebate) * $total_cost * 0.01;
    }//end while fetch inner row
}//if there rows return

And at least change to:

if($count>0)
{
    rowINNER = mysql_fetch_array($result)
    $loop_userid = $rowINNER['referral'];
    $my_rebate = $rowINNER['rebate'];
    $my_cost = (100 - $my_rebate) * $total_cost * 0.01;
}//if there rows return

I am assuming that account.account_id is a unique key and that you can't have more than 1 record. Either $count is 0 (there is no such account_id) or it is 1. If you have duplicate account_id's then that would cause catastrophic problems.

Upvotes: 3

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