CODEWITHSUNDEEP
php
user3769723
user3769723

Reputation: 39

How to check insert success or not When i using this code?

How to check insert success or not When i using this code ?

i want to check insert success or not

if success echo "success"; else echo "error";

$link = mysql_connect($host, $user, $pass);
$sql = mysql_query("INSERT INTO `transaction` (txnid, pro_name, price, status) 
           VALUES (
                    '".$data['txn_id']."' ,
                    '".$data['pro_name']."' ,
                    '".$data['price']."' ,
                    '$status'
                  )", $link);
return mysql_insert_id($link);

Upvotes: 0

Views: 84

Answers (3)

Mark Miller
Mark Miller

Reputation: 7447

Simply:

if (!$sql) echo 'error';
else echo 'success';

From the docs on mysql_query:

For ... INSERT ... mysql_query() returns TRUE on success or FALSE on error.

More importantly though, read this: Why shouldn't I use mysql_* functions in PHP?

Upvotes: 1

Shail Paras
Shail Paras

Reputation: 1163

According to mysql_insert_id. it returns new generated id on successful insert.

Check here :

mysql_insert_id

 $link = mysql_connect($host, $user, $pass);
 $sql = mysql_query("INSERT INTO `transaction` (txnid, pro_name, price, status) 
           VALUES (
                    '".$data['txn_id']."' ,
                    '".$data['pro_name']."' ,
                    '".$data['price']."' ,
                    '$status'
                  )", $link);

$lastInsertedId = mysql_insert_id($link);

if (empty($lastInsertedId)) 
   $return = 'error';
else 
   $return = $lastInsertedId;

return $return;

Upvotes: 0

Enkhbayar
Enkhbayar

Reputation: 1

$link = mysql_connect($host, $user, $pass);
$sql = mysql_query("INSERT INTO `transaction` (txnid, pro_name, price, status) 
           VALUES (
                    '".$data['txn_id']."' ,
                    '".$data['pro_name']."' ,
                    '".$data['price']."' ,
                    '$status'
                  )", $link);
//------------Changes----------------
$inserted_id = mysql_insert_id($link);
if(mysql_errno() != 0){ 
   echo "error";
else
   echo "success";
return $inserted_id;

Upvotes: 0

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