Return error in PHP when mysql duplicate entry is found

[benoit1980]

Hello all.

 

I have only started php a week ago and require your help as this problem is driving me crazy......

 

I have a form and the customers just enter their phone and email.

When a duplicate of any of them is found I wish to show the message "Your phone or Mobile is already registered"

 

Now when no duplication is found, I wish to echo something else.

 

I have done this code below but I cannot get it to work....I created the database, everything is going well in without problem when I remove the statement

<?php
if(mysql_affected_rows() == 0){    echo '<div align="center" class="text-not-correct">Your Phone Number or Email or both are already registered!</div>';?>   

 

Here is part of my code, I only need in this tiny area please.

 

<?php$link = mysqli_connect("localhost", $username, $password, $db_name)or die("cannot connect");  $mres_email = mysqli_real_escape_string($link, $_POST['email_form2']);$mres_phone = mysqli_real_escape_string($link, $_POST['phone_form2']);$mres_code = mysqli_real_escape_string($link, $_POST['code_form2']);       $query = "INSERT INTO $tbl_name VALUES (NULL, '$dat_date', '$mres_phone', '$mres_email', '$mres_code')";mysqli_query($link, $query);if(mysql_affected_rows() == 0){    echo '<div align="center" class="text-not-correct">Your Phone Number or Email or both are already registered!</div>';     mysqli_close($link); }else{header("Refresh: 5;url=http://pwebsite.com/vio/page1.php");echo '<div align="center" class="text-correct">Congratulation you are now registered!</div></br>';echo '<div align="center"><img src="loading.gif" width="100" height="100" /></div>';    ?>   

 

Remember, the script is working fine but not when I try to check for duplicate. please help :-))))

 

 

Thank you!

 

 

Benoit

Edited June 2, 2013 by benoit1980

[jcbones]

I would check the mysqli_errno to see if it is returning a duplicate key error. Not sure of number for that specific error, but you can find it in the mysqld_error.txt file in the docs folder of your installation.

[benoit1980]

Thanks jcbones.

 

I actually used this:

 

if(mysqli_affected_rows($link) <= 0){

    

echo 'something here...';

 

It works very well :-))0 Hope this help someone.

 

Bye all and thank you for your help.

 

Ben

[litebearer]

Also look into unique constraints for your table fields - http://dev.mysql.com/doc/refman/5.0/en/create-table.html