Comment Box Script

[dbqclassof2012]

Hello I am currently working on making a comment box for my website and have had a lot of progress but this end part i have not been able to get. I have the form and everything done it submits it to the database fine but then bringing the comments back to the page it always comes up with this error

Warning:  mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/dbqclass/public_html/getcomments.php on line 22

I have not been able to fix this and if anyone could help me it would be greatly appreciated!!!!!Heres the code i am working with

<?php
// Open Connection To mySQL.
// Replace 'username' and 'password' with appropiate login values for your server.

$mysqlLink = mysql_connect( 'localhost' , 'username' , 'pass' );

// Select mySQL Database.

mysql_select_db( 'comments' , $mysqlLink );

// Create Query To Create Table.

$sql = 'SELECT `name`, `comment_date`, `comment` FROM `comments`';

// Issue Query.

$result = mysql_query( $sql );

// Loop through results and print them.

while( $row = mysql_fetch_array( $result ) ) {
    echo $row[ 'name' ] . $row[ 'comment_date' ] . $row[ 'comment' ];
}

// Close mySQL Connection.

mysql_close( $mysqlLink );

?>

IF YOU PROVIED ME WITH MY ANSWER I WILL REWARD YOU WITH OVER 200 CREDITS!

[freecrm]

This might be a quick answer.

I'm not sure you have to put ' in the $sql

i.e.

$sql = 'SELECT name, comment_date, comment FROM comments';

[dbqclassof2012]

Ok thanks for the answer but it didnt work so if you have another answer it would be greatly appreciated

[freecrm]

Ok thanks now trying it out

Let me know how you get on

[scopey]

Edit to the following and tell us what you get:

<?php
// Open Connection To mySQL.
// Replace 'username' and 'password' with appropiate login values for your server.

$mysqlLink = mysql_connect( 'localhost' , 'username' , 'pass' );

// Select mySQL Database.

mysql_select_db( 'comments' , $mysqlLink );

// Create Query To Create Table.

$sql = 'SELECT `name`, `comment_date`, `comment` FROM `comments`';

// Issue Query.

$result = mysql_query( $sql ) or die(mysql_error());

// Loop through results and print them.

while( $row = mysql_fetch_array( $result ) ) {
    echo $row[ 'name' ] . $row[ 'comment_date' ] . $row[ 'comment' ];
}

// Close mySQL Connection.

mysql_close( $mysqlLink );

?>

Also, are you sure that both the table and the database have the name of 'comments'.

Usually the database name has a name like 'dbqclassof2012_comments' or something. Check phpMyAdmin for the proper database name. It will show the prefix (ie the 'dbqclassof2012_' bit) above all the database names.

[dbqclassof2012]

I have just tried that but t does not seem to work the error it keeps saying :
Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/dbqclass/public_html/getcomments.php on line 21

so i think it is with the coding but i dont know how to fix it because im not the best coder

[scopey]

So you tried adding in the 'or die(mysql_error());' bit? The error you are getting is usually returned when you are selecting an invalid database or have an error in your syntax. Try adding the 'or die(mysql_error());' code to the end of all your mysql functions.

[freecrm]

I agree with scopey here - all my database names are preceeded by my username.

i.e. me_databasename.

The name "comments" sounds too much like a table.

The line we are refering to is mysql_select_db( 'comments' , $mysqlLink );

If you are doing multiple pages, it may be worth setting up a connection file rather than putting the connection script into each page. This keeps secure info seperate and easier to manage.

This is done by setting all the variables in one connection.php file, usually put in a seperate folder in the root and would look something like this

<?php
# FileName="Connection_php_mysql.htm"
# Type="MYSQL"
# HTTP="true"
$hostname_connect= "localhost";
$database_connect= "databasename";
$username_connect= "username";
$password_connect= "password";
$connect = mysql_pconnect($hostname_connect, $username_connect, $password_connect) or die(mysql_error());
?>

then each file would just have one line at the top to reference this file

<?php require_once('Connections/exampleconnection.php'); ?>

And the select db would be

mysql_select_db($databasename, $connect);

[dbqclassof2012]

I'm sorry but i have not been able to get this to work i will give you the whole code everything but the password and could you please change it to what it needs to be because fro some reason i cannot get.

<?php
// Open Connection To mySQL.
// Replace 'username' and 'password' with appropiate login values for your server.

$mysqlLink = mysql_connect( 'localhost' , 'dbqclass' , 'Password' );

// Select mySQL Database.

mysql_select_db( 'dbqclass_comments' , $mysqlLink );

// Create Query To Create Table.

$sql = 'SELECT `name`, `comment_date`, `comment` FROM `dbqclass_comments`';

// Issue Query.

$result = mysql_query( $sql );

// Loop through results and print them.

while( $row = mysql_fetch_array( $result ) ) {
    echo $row[ 'name' ] . $row[ 'comment_date' ] . $row[ 'comment' ];
}

// Close mySQL Connection.

mysql_close( $mysqlLink );

?>

So the username is dbqclass if you didn't notice above

Thanks

[leafypiggy]

for your username, and database name, the prefix, like stated before is:

yourcpanelname_name

so, your cPanel name corresponds to the location of your site, like, my website's root would be,

/home/leafy/public_html/


since you are on a shared server, it must be preceeded with the name, so that there isn't two people with the database "forums" and writing information over each other.

have you tried that?



EDIT: also, try this script in any file, or by itself, like, connectiontest.php

<?php
# Purpose: Establish a connection to the database
#
$dbname = 'cpanelname_dbname';
$dbhost = 'localhost';
$dbusername = 'cpanelname_dbuname';
$dbpassword = 'dbpass';
#
$dbconnection = @mysql_connect($dbhost, $dbusername, $dbpassword);
 
if ($dbname != '' && !@mysql_select_db($dbname)) 
 
{
 
echo 'I could NOT connect to the database! Check the username, password, etc.';
}
 
 else 
{
echo 'I connected. The connection is: ' . $dbconnection;
}
?>