how to capture undefined error in php when selecting from different tables which do not have all the required data using and operator

Question

Hello Developers.

Am working on an application where i have to select data from different tables and views. However some views may not have the corresponding data i have queried yet the other tables have the required data. This gives me undefined error response for the views where the data for a selected variables is not found in the view. How best can i write my code. see below.

error returned

Notice: Undefined variable: fname in C:\wamp\www\Kesa\web\add_shares.php on line 175 Notice: Undefined variable: mname in C:\wamp\www\Kesa\web\add_shares.php on line 175 Notice: Undefined variable: lname in C:\wamp\www\Kesa\web\add_shares.php on line 175 Notice: Undefined variable: passbook_no in C:\wamp\www\Kesa\web\add_shares.php on line 176 Notice: Undefined variable: share_amount in C:\wamp\www\Kesa\web\add_shares.php on line 177

if(isset($_POST['button2'])){

    $id = mysql_real_escape_string($_POST['slt_member_id']);

    //selecting employee details
    $query = "SELECT member_tbl.fname AS fname,
        member_tbl.mname AS mname,
        member_tbl.lname AS lname, 
        member_tbl.member_id AS member_id, 
        member_tbl.Pasbook_no AS passbook_no, 
        share_view.share_amount AS share_amount 
        FROM member_tbl, share_view 
        WHERE member_tbl.member_id = '$id' 
        AND member_tbl.member_id = share_view.member_id";

    $result = mysql_query($query, $connection);
    while($res=mysql_fetch_array($result)){

        $share_amount = $res['share_amount'];

        if(isset($share_amount) || !empty($share_amount)){
            $member_id = $res['member_id'];
            $fname = $res['fname'];
            $mname = $res['mname'];
            $lname = $res['lname'];
            $passbook_no = $res['passbook_no'];
            $share_amount = $res['share_amount'];

        }elseif(!isset($share_amount) || empty($share_amount)){
            $member_id = $res['member_id'];
            $fname = $res['fname'];
            $mname = $res['mname'];
            $lname = $res['lname'];
            $passbook_no = $res['passbook_no'];
            $account_type = 'User has no shares';
            $share_amount = 'User has no shares';

        }else{
            $member_id = '';
            $fname = 'No user selected';
            $mname = '';
            $lname = '';
            $passbook_no = 'No user selected';
            $account_type = 'No user selected';
            $share_amount = 'No user selected';
            }
        }//closing selection from members tbl

}else {
      $member_id = '';
      $fname = 'No user selected';
      $mname = '';
      $lname = '';
      $passbook_no = 'No user selected';
      $account_type = 'No user selected';
      $share_amount = 'No user selected';
}

I think the code you posted here is not that which is giving error, Post the view code here — Pradeep Singh, Jan 19 '18 at 11:46
This has nothing at all to do with CakePHP and the mysql_ functions are *deprecated* since years now. There is even a huge red box on top of the manual pages... http://php.net/manual/en/function.mysql-real-escape-string.php Also *if* this code is used inside the CakePHP framework, well you actually managed to ignore the framework completely and might want to do some basic php and CakePHP tutorials. — floriank, Jan 19 '18 at 12:09
Possible duplicate of [Notice: Undefined variable](https://stackoverflow.com/questions/12769982/reference-what-does-this-error-mean-in-php/12778634#12778634) — floriank, Jan 19 '18 at 12:11
Hello pradeep, here is the view code. however it's just echoing the variables i shared in the first code. i believe the main issue is on the code i first share that's the one i need assistance with. but please advise. code here. — Brian O, Jan 19 '18 at 18:26

score 0 · Answer 1 · answered Jan 19 '18 at 12:37

0

Fetch your result as an associative array with fetch_assoc(). Then you can access the result via the column name. Currently you are fetching the result as an indexed array by fetch_array(). So you will have to access the values as $result[1].

answered Jan 19 '18 at 12:37

Kishor Malakar

81
1
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how to capture undefined error in php when selecting from different tables which do not have all the required data using and operator

1 Answers1