mysql Duplicate error handling - php

Question

I'm trying to use PHP to enter data from a form. When I try to enter duplicate data a bad message pops like
Something went wrong with this:
INSERT INTO customer VALUES('jamie9422','Jamie Lannister','sept of baelor','jamie#cersei.com',9422222222,0) Duplicate entry 'jamie9422' for key 'PRIMARY' "
Instead, I want to display a clean error message. How can I do that. Here's my code I've written so far...
<?php
include_once "dbConnect.php";
$connection=connectDB();
if(!$connection)
{
die("Couldn't connect to the database");
}
$tempEmail = strpos("{$_POST["email"]}","#");
$customer_id=substr("{$_POST["email"]}",0,$tempEmail).substr("{$_POST["phone"]}",0,4);
//$result=mysqli_query($connection,"select customer_id from customer where customer_id='$customer_id' ");
//echo "customer_id is".$result;
$query = "SELECT * FROM CUSTOMER WHERE CUSTOMER_ID='$customer_id'";
$customer_idcip = $customer_id-1;
echo $customer_idcip;
if ( mysql_query($query)) {
echo "It seems that user is already registered";
} else {
$command = "INSERT INTO customer VALUES('{$customer_id}','{$_POST["name"]}','{$_POST["address"]}','{$_POST["email"]}',{$_POST["phone"]},0)";
$res =$connection->query($command);
if(!$res){
die("<br>Something went wrong with this:{$command}\n{$connection->error}");
}
echo "Welcome ".$_POST["name"]." \nCongratulations on successful Registration. Refill your Wallet here";
//$cutomerRetrival = mysql_query("select from customer where customer_id='$customer_id'");
echo "<br>Please note your customer ID :".$customer_id;
}
/*if($result)
{
echo "Query Fired";
$dupentry = mysqli_num_rows($result);
if($dupentry==1)
{
echo "You are already Registered";
exit;
}
}*/
?>

Answer 1

The error code (number) is 1022.
You can e.g. define a constant for that (so that somebody else in x months has a chance to understand the code) like
define('ER_DUP_KEY', 1022);
and then do something like
if(!$res){
if ( <error code>==ER_DUP_KEY ) {
handleDuplicateEntryError();
}
else {
die("<br>Something went wrong with this:{$command}\n{$connection->error}");
}
}
since I don't know how $res =$connection->query($command); works (and what $connection is I can't tell you exactly how to implement <error code>==ER_DUP_KEY, could be by using mysql_errno.
But it seems to be somehow intermingled with mysql_query($query), i.e. the old, deprecated mysql_* extension and some custom class. You might want to fix that first.... ;-)
see http://docs.php.net/manual/en/mysqlinfo.api.choosing.php

Answer 2

Your code doesn't check for existing record properly
Change
if (mysql_query($query)) {
echo "It seems that user is already registered";
}
to
$result = mysql_query($query);
if (mysql_num_rows($result)) {
echo "It seems that user is already registered";
}
Also, PLEASE do not use $_POST variables without escaping them first, use something like mysql_real_escape_string() to escape each variable passed from the user, otherwise your website will be hacked really fast with SQL Injection.

Answer 3

Make some update into your and then try to get error message 'customer already registered.'
$query = "SELECT * FROM CUSTOMER WHERE CUSTOMER_ID='$customer_id'";
$res= mysql_query($query);
$customer_count = mysql_num_rows($res);
$customer_idcip = $customer_id-1;
echo $customer_idcip;
if ( $customer_count > 0 ) {
echo "It seems that user is already registered";
} else {
...................................

Answer 4

Thank you all.
Actually I was using mysqli API in my connectDB.php file..
Hence I needed to call functions on mysqli.
Instead I was calling mysql. i.e I was creating a new connection, thus the query wasn't getting fired at all.
Changed to mysqli->query($result) that is object oriented style
and it worked fine....

Answer 5

Use Try Catch instead.
try{
$res =$connection->query($command);
}catch(Exception $e){
die( "Write your error appropriate message here");
}