Bug #5648 Updated: INSERT statements no longer work using mysql_query()

From: Date: Sun, 30 Jun 2002 08:37:25 +0000
Subject: Bug #5648 Updated: INSERT statements no longer work using mysql_query()
References: 1  Groups: php.bugs 
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ID: 5648 Updated by: akujin11@yahoo.com Reported By: alanflum@aksi.net Status: Closed Bug Type: MySQL related Operating System: Debian GNU Linux "Woody" Kernel PHP Version: 4.0.1pl2 New Comment: i have been experianceing the exact same problem. i require the use of the INSERT SQL command and i cannot seem to get it to work. the command SEEMS to process correctly, with one error. the data is never added to the database. the manual does NOT give the required information, and i need a solution. if you can not give me what i want, can you direct me to someone that can? the Akujin Previous Comments: ------------------------------------------------------------------------ [2000-08-20 06:18:52] sniper@php.net Read the manual: http://www.php.net/manual/function.mysql-query.php --Jani ------------------------------------------------------------------------ [2000-07-17 22:20:10] alanflum@aksi.net We recently upgraded to MySQL 3.23.21-beta and php 4.01pl2 (Apache 1.3.12, Debian v.10.8 Debian GNU Linux "Woody"). INSERT statements no longer work using the mysql_query() . The return value is always 0. UPDATE, and SELECT seem to work fine. In addition, INSERT seems to work OK using mysql_db_query() .Others have reported this same exact problem on phpbuilder.com mailing list archive. In addition, all php scripts on the upgraded machine using INSERT and mysql_query experience the same problem (for example phpMyAdmin). Here is some sample code: $user= "demo"; $host= "localhost"; $password= "mypassword"; $database= "mydatabase"; $connection = mysql_connect($host,$user,$password)or die("No Connection"); mysql_select_db($database, $connection)or die("Can't Select Database"); $addrow = "INSERT INTO test VALUES ('$name1', '$name2')"; $result= mysql_query($addrow, $connection) or die("Can't Add Selections to Test Table"); If I remove the die() in the last statement and check the value of $result, it always returns a 0. ------------------------------------------------------------------------ -- Edit this bug report at http://bugs.php.net/?id=5648&edit=1

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