Undefined variable in mysql_insert_id()

Question

I'm stuck here for almost 4 hrs can't find what's the error in variable here although I declare correctly the variable.

The error prompt like this :

Notice: Undefined variable: last_id in C:\xampp\htdocs\CIS\addSubscriber.php on line 87

Here is the code:

 if(mysql_query("INSERT INTO subscribers(fName, mName, lName, suffix, globeNumber, emailAddress, contactNumber, 
    address, state, country, virtualNumber, agentCode, ipAddress, dateRegistration, dateStarted, dateExpired, time, 
    transaction_id, accountStatus, subscriptionStatus, status, remarks)
  VALUES ('$fName' , '$mName', '$lName', '$sName', '$globe', '$emailAdd', '$contactNo', '$address', '$state', '$country', 
  '$virtualNum', '$cAgentCode', '$ipAddress', '$dateRegistration', '$dateStarted', '$dateExpired', '$time', 
  '$transaction', '$accStatus', '$subsStatus', '$status', '$remarks' )"))
{ 

    $last_id = mysql_insert_id(); 


}

**//this is line 87** 

$query =  "INSERT INTO `transactions` (`transactionNumber`, `subscriptionStart`, `subscriptionEnd`, `subsStatus`, `cus_id`)
VALUES ('".$transaction."', , '".$dateStarted."', '".$dateExpired."', '".$subsStatus."', '".$last_id."');"; 

  if($query_run = mysql_query($query)){


 ?>
  <script> alert('<?php echo $last_id; ?>'); 
 </script>

    <?php
} else {
    ?>
    <script>alert('Agent Code is not available!');</script>
    <?php
}

}

 echo mysql_error();

I tried this using in other file but it was working I can't find the trigger of error.

Different from basic Undefined Variable error:

Either your query gives bad results or you are using $last_id before the query executed — Maha Dev, Oct 26 '15 at 05:55
@Rahautos: you can check the codes I set it `$last_id = mysql_insert_id();` — Edmhar, Oct 26 '15 at 05:56
@Rahautos actually I tried again use this codes in other file, it's working but not in this file. — Edmhar, Oct 26 '15 at 05:58
@MuthaFury: Got it, i found the error thanks. Kindly post it in answer so I can accept it: — Edmhar, Oct 26 '15 at 06:08

Abdulla Nilam · Answer 1 · 2015-10-26T06:03:56.113

What about this

$query = mysql_query("INSERT INTO subscribers(fName, mName, lName, suffix, globeNumber, emailAddress, contactNumber, 
    address, state, country, virtualNumber, agentCode, ipAddress, dateRegistration, dateStarted, dateExpired, time, 
    transaction_id, accountStatus, subscriptionStatus, status, remarks)
  VALUES ('$fName' , '$mName', '$lName', '$sName', '$globe', '$emailAdd', '$contactNo', '$address', '$state', '$country', 
  '$virtualNum', '$cAgentCode', '$ipAddress', '$dateRegistration', '$dateStarted', '$dateExpired', '$time', 
  '$transaction', '$accStatus', '$subsStatus', '$status', '$remarks' )");

if(!$query)
{ 
    echo mysql_error();
}
else
{
    $last_id = mysql_insert_id(); 
    $query2 =  mysql_query("INSERT INTO `transactions` (`transactionNumber`, `subscriptionStart`, `subscriptionEnd`, `subsStatus`, `cus_id`)
        VALUES ('$transaction', '$dateStarted', '$dateExpired', '$subsStatus', '$last_id')"); 
    if(!$query2)
    { ?>
        <script> 
            alert('<?php echo $last_id; ?>'); 
        </script>
    <?php
    } 
    else {
    ?>
        <script>
            alert('Agent Code is not available!');
        </script>
    <?php
    }
}

MySQL extension was deprecated in PHP 5.5.0, and it was removed in PHP 7.0.0. Instead, the MySQLi or PDO_MySQL extension should be used. S

thanks for the suggestion sir, but I found better answer. I can't vote up your answer due to I'm in maximum maybe later. — Edmhar, Oct 26 '15 at 06:13

score 1 · Accepted Answer · answered Oct 26 '15 at 06:10

1

Try putting

else { die(mysql_error()); }

to find the error.

answered Oct 26 '15 at 06:10

MuthaFury

805
1
7
22

@Edmhar Glad I can help :) – MuthaFury Oct 26 '15 at 06:13

Undefined variable in mysql_insert_id()

2 Answers2