[RESOLVED] query with INSERT going wrong

[thegoldenchild]

Hi,

Im trying to input text in to a db using a form. But i cant get it to work. It says the query wont work.

$dblogin = "xxxx";
$dbpassword = "xxxx";
$host = "xxxx";
$dbname = "xxxx";

$connection = "mysql_connect($dblogin, $dbpassword, $host)" or die("connection klopt niet");
$db = "mysql_select_db($dbname, $connection)" or die("db klopt niet");

$query = "INSERT INTO contact2 (naam, email, bericht) VALUES (test,[e-mail],test)" or die("werkt niet");
$result = mysql_query($query) or die("werkt niet 2");

[rincewind456]

Your query need some quotes!

$query = "INSERT INTO contact2 (naam, email, bericht) VALUES ('test', '[e-mail]', 'test')" or die("werkt niet"); 

[thegoldenchild]

It still won't work, i've tried it with and without the quotes.

$result = mysql_query($query) or die("werkt niet 2");

is the sentence that gives the error..

I mean that if i execute it, it get "werkt niet 2", so i dies becos it can't execute. I don't get any other errors.

Im trying to do it with variables but that's not working either.

$query = "INSERT INTO contact2 (naam, email, bericht) VALUES ('$naam','$email','$bericht')" or die("werkt niet");

[rincewind456]

This

or die("werkt niet"); 

on the query is incorrect as the error handling is done on the following line.
The variable $query should be

$query = "INSERT INTO contact2 (naam, email, bericht) VALUES ('test', '[e-mail]', 'test')"; 

or if you want to use variables, then like this

$query ="INSERT INTO contact2 (naam, email, bericht) VALUES ('$naam', '$email', '$bericht')"; 

[thegoldenchild]

Nope it doesn't work..

I still get the same error..

I've copied exactly as u said..

U can see the result by entering this form and clicking on Verzenden..

http://82.171.153.236/psptrace/contactadmin.php

srry that it's not in english.. :o

[rincewind456]

Ok check that the table names in the insert query "naam, email, bericht" are exactly the same as in the databse table, then use this line

$result = mysql_query($query) or die(mysql_error("werkt niet 2"));

and see what error is thrown out.

[thegoldenchild]

I've checked all names with the DB double time and this is the error i get:

Warning: mysql_error(): supplied argument is not a valid MySQL-Link resource in E:\xampp\htdocs\psptrace\contactadminsubmit.php on line 53

[rincewind456]

Have you tried running the query through phpMyadmin or the command line to check it works.

[thegoldenchild]

it does work, but i get the variables as the result, so $naam, $email and $bericht end up in the db

[rincewind456]

does it work through the page form or through phpMyadmin or the command line

[thegoldenchild]

Ye i used the interface of PHPMyAdmin with the SQL tab

[rincewind456]

Then the only thing i can think is that the connection is not happening try this

$connection = "mysql_connect($dblogin, $dbpassword, $host)" or die(mysql_error("connection klopt niet")); 
$db = "mysql_select_db($dbname, $connection)" or die(mysql_error(("db klopt niet")));

$query ="INSERT INTO contact2 (naam, email, bericht) VALUES ('$naam', '$email', '$bericht')"; 
$result = mysql_query($query) or die(mysql_error("werkt niet 2"));

[thegoldenchild]

Tried it, gave the same error as before:

Warning: mysql_error(): supplied argument is not a valid MySQL-Link resource in E:\xampp\htdocs\psptrace\contactadminsubmit.php on line 53

Nothing else, but it still doesn't work nor it does add anything to the db

[rincewind456]

Can you show your db table structure

[thegoldenchild]

Ye sure,

i've made a screenie out of it. Don't know if i missed anything though..

http://www.tgcdesign.nl/psptrace/screen.jpg

[rincewind456]

try this, it may need the id being inserted as well

$query ="INSERT INTO contact2  VALUES ('$naam', '$email', '$bericht','')"; 

[thegoldenchild]

I've added it, and it doesn't make any diffirence..🙁

[rincewind456]

Here this works all tested

<?php
$dblogin = "xxxx";
$dbpassword = "xxxx";
$host = "xxxx";
$dbname = "xxxx";

$connection = mysql_connect($host, $dblogin, $dbpassword) or die(mysql_error("connection klopt niet"));
$db = mysql_select_db($dbname, $connection) or die(mysql_error(("db klopt niet")));

$query ="INSERT INTO contact2 (naam , email , bericht )
VALUES ('$naam', '$email', '$bericht')";
mysql_query($query) or die(mysql_error()."werkt niet 2");
?>

But you will need to sanitize the input so that any quotes in the comments don't stuff up the database

[thegoldenchild]

Yes i know, thnQ..am only doin php for 3 weeks..done lot's of things but this thing get's me really crazy..im gonna try it..

To clean the input im using strip_tags..

[thegoldenchild]

Thanx a lot, it does work.

I only don't understand why it didn't work with the $result variable..

Im used to working with that tag..pff gonna be a hard one to remember..but thanks a lot!

Greetz!