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Bug #11427 Duplicate record insertion using SQL
Submitted: 2001-06-12 06:27 UTC Modified: 2001-06-12 12:27 UTC
From: john at neutralize dot com Assigned:
Status: Not a bug Package: MySQL related
PHP Version: 4.0.5 OS: Win NT4 PWS
Private report: No CVE-ID: None
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From: john at neutralize dot com
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 [2001-06-12 06:27 UTC] john at neutralize dot com
The script was fine for an Access database but when I changed it over for a MySQL database the insertion queries sometimes (most of the time but not all the time) adds two records to the database.

Heres my connection:

$db_hostname   = "jd";
$db_username   = "john";
$db_password   = "nirvana";
$db_database   = "neuquote";

$conn = mysql_connect($db_hostname, $db_username, $db_password)
  or die ('Unable to connect to database');
		
$db = mysql_select_db($db_database,$conn)
  or die ('Unable to connect to database');

Heres my queries:

$today = getdate();
$today = $today['mday']."/".$today['mon']."/".$today['year']." ".$today['hours'].":".$today['minutes'].":".$today['seconds'];
						
// Insert new company
$sql="INSERT INTO Company (CompanyName,CompanyAddress1,CompanyAddress2,CompanyAddress3,CompanyAddress4,CompanyArea,CompanyCountryId,CompanyPostalCode,CompanyTel,CompanyFax,CompanyEmail,CompanyURL,CompanyDesc,CompanyStatus,TimeCompanyCreated) ". 
				 "VALUES ('$company_name','$company_address1','$company_address2','$company_address3','$company_address4','$company_area',$company_country,'$company_postal_code','$company_tel','$company_fax','$company_email','$company_url','$company_desc',0,'$today')";

$res_insert_company = mysql_query($sql,$conn);

//create a new hash to insert into the db and the confirmation email
$hash=md5($email.$hidden_hash_var);
$sql="INSERT INTO CompanyContact (ContactTitle,ContactFirstName,ContactSurname,ContactMiddleInitial,ContactAddress1,ContactAddress2,ContactAddress3,ContactAddress4,ContactArea,ContactCountryId,ContactPostalCode,ContactTel,ContactExt,ContactFax,ContactEmail,ContactPositionId,ContactUsername,ContactPassword,ContactRemoteAddress,ContactConfirmHash,ContactStatus,TimeContactCreated,CompanyId) ". 
					"VALUES ($title,'$first_name','$surname','$middle_initial','$address1','$address2','$address3','$address4','$area',$country_code,'$postal_code','$tel','$ext','$fax','$email',$pos,'$user_name','". md5($password1) ."','$GLOBALS[REMOTE_ADDR]','$hash',0,'$today',$company_id)";

$res_insert_contact=mysql_query($sql,$conn);

Most of the time two companies are added to the db and sometimes two companycontacts, this is not consistent though as sometimes it works fine.

I have checked the submit button is not being clicked twice and the code is running one insert for company and one for companycontact.

Is this a common problem with MySQL and PHP?

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 [2001-06-12 12:27 UTC] sniper@php.net
submitted twice

 [2003-07-20 14:29 UTC] helmutkemper at yahoo dot com
I have a same problem using windows me, php 4.0.1 and mysql 
4.23.23

Kemper
 [2004-05-13 22:21 UTC] certaplus at hotmail dot com
I have the same problem. On two different servers, I run Apache on one and IIS on the other. Both are ran on Windows XP with MySQL 4.0.1 and PHP 4.3.4. I also use IE 6.0 as well as Netscape 7.1 to display the pages.

form code :

...</tr>
          </table>
        <form name="form1" method="POST" action="process.php">
         
            <p>Brand 
              <input type="text" name="brand">
            </p>
            <p> Model 
              <input type="text" name="model">
            </p>
            <p> 
              <input type="submit" name="Submit" value="submit">
            </p>
          </form>
        </div>...


PHP code :
<?php


define ('db_user', 'hendrix');
define ('db_password', 'rememberme');
define ('db_host', 'localhost');
define ('db_name', 'products');

$link = mysql_connect (db_host, db_user, db_password)
	or die ("Couldn't connect");


mysql_select_db (db_name)
	or die ("Database couldn't be accessed");


$brand= $_POST['brand'];
$model= $_POST['model'];

$query="INSERT INTO product (id,brand,model) VALUES ('','$brand','$model')";
$result= mysql_query($query,$link);


if(!mysql_query($query,$link)) die(mysql_error());
echo "Your information as been entered";


mysql_close();


?>

Jack
 
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