Member Avatar for Mr.Kazaam

I was following this tut. on how to install Apache, PHP, Mysql, and PHPmyadmin. I need a bit of help! I get a "Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in C:\Program Files\Apache Software Foundation\Apache2.2\htdocs\mysql_test.php on line 15" everytime I test the mysql thing.

What ive done so far is :

[U]Configuring PHP to work with MySQL:[/U] Now that both PHP and MySQL are installed, we have to configure them to work together. 1. Open up your php.ini file (C:/WINDOWS/php.ini) and find the line: ;extension=php_mysql.dll To enable the MySQL extension, delete the semi-colon at the beginning of that line. 2. Next we must add the PHP directory to the Windows PATH. To do this, click: Start > My Computer > Properties > Advanced > Environment Variables. Under the second list (System Variables), there will be a variable called "Path". Select it and click "Edit". Add ";C:\php" to the very end of the string and click "OK". 3. Restart your computer for the changes to take effect. 4. Create a new file in your "htdocs" directory called "mysql_test.php". 5. Copy the following code into "mysql_test.php" and click save. (Make sure to replace the MYSQL_PASS constant with the MySQL Password you specified during the MySQL installation). <?php # Define MySQL Settings define("MYSQL_HOST", "localhost"); define("MYSQL_USER", "root"); define("MYSQL_PASS", "password"); define("MYSQL_DB", "test"); $conn = mysql_connect("".MYSQL_HOST."", "".MYSQL_USER."", "".MYSQL_PASS."") or die(mysql_error()); mysql_select_db("".MYSQL_DB."",$conn) or die(mysql_error()); $sql = "SELECT * FROM test"; $res = mysql_query($sql); while ($field = mysql_fetch_array($res)) { $id = $field['id']; $name = $field['name']; echo 'ID: ' . $field['id'] . '<br />'; echo 'Name: ' . $field['name'] . '<br /><br />'; } ?> 6. Open up Internet Explorer and type in "http://localhost/mysql_test.php". If the "mysql_test.php" page returns something similiar to: ID: 1 Name: John

The author of the tutorial told me to write and save this in the htdocs.

<?php # Define MySQL Settings define("MYSQL_HOST", "localhost"); define("MYSQL_USER", "root"); define("MYSQL_PASS", "rashad"); define("MYSQL_DB", "test"); $conn = mysql_connect("".MYSQL_HOST."", "".MYSQL_USER."", "".MYSQL_PASS."") or die(mysql_error()); mysql_select_db("".MYSQL_DB."",$conn) or die(mysql_error()); $sql = "SELECT * FROM test"; $res = mysql_query($sql); while ($field = mysql_fetch_array($res)) { $id = $field['id']; $name = $field['name']; echo 'ID: ' . $field['id'] . '<br />'; echo 'Name: ' . $field['name'] . '<br /><br />'; } ?>

replacing the " password " with my password?


The tutorial link is here : http://www.bicubica.com/apache-php-mysql/index.php

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  • 2 Years Discussion Span
  • Latest Post Latest Post by R1Programmer
Member Avatar for Will Gresham

Do you have a table called 'test' within the database 'test'?

change $res = mysql_query($sql); to $res = mysql_query($sql) or die(mysql_error()); Does this return an error?

Member Avatar for Mr.Kazaam

Yes. I have the test table. I did what the instructions said and did the use test; thing. Whenever I try the code you gave me, it says "Table 'test.test' doesn't exist"...

Member Avatar for Will Gresham

can you use this code and let us know what it returns:

<?php mysql_connect("localhost","root","password"); $tables = mysql_list_tables("test"); while (list($table) = mysql_fetch_row($tables)) { echo "$table <br />"; } ?>

obviously change the password to your password.

Member Avatar for Mr.Kazaam

Okay I did as you said. In return I got " name ".

Member Avatar for Riukeen

Hi Guys...


This is problem..

$sql = "SELECT * FROM test";


change table name .. "name" (=

Member Avatar for Andrieux

The problem is that your database is titled test, but the table inside that databse is named "name".

So when you use $sql = "SELECT * FROM test"; , it's trying to find the table "test" within the database that you connected to at the top of your file. (By the way, that's not a great way to connect to a database via PHP.

//define variables $dbhost = "localhost"; $dbuser = "andrieux"; $dbpass = "pass"; $dbdatabase = "andrieux"; //connect $db = mysql_connect($dbhost, $dbuser, $dbpass); mysql_select_db($db, $dbdatabase);

You don't need error fallbacks for something as simple as connecting to localhost, so you have alot of unneeded code.

Member Avatar for Riukeen

Hi Guys...


This is problem..

$sql = "SELECT * FROM test";


change table name .. "name" (=

$sql = "SELECT * FROM name";

Member Avatar for prabakar..

<?php

# Define MySQL Settings
define("MYSQL_HOST", "localhost");
define("MYSQL_USER", "root");
define("MYSQL_PASS", "password");
define("MYSQL_DB", "test");

$conn = mysql_connect("".MYSQL_HOST."", "".MYSQL_USER."", "".MYSQL_PASS."") or die(mysql_error());
mysql_select_db($db,"".MYSQL_DB."") or die(mysql_error());

$sql = "SELECT * FROM test";
$res = mysql_query($sql);

while ($field = mysql_fetch_array($res))
{
$id = $field;
$name = $field;

echo 'ID: ' . $field . '<br />';
echo 'Name: ' . $field . '<br /><br />';
}

?>


In the 9th line fatal error begins,
help me.. wat i want to do..

Member Avatar for Hypalink

Replace your code in the beginning to this; in otherwords don't define - just declare them as variables. It's just simpler i think.

$dbhost = "your host"; $dbuser = "your username"; $dbpassword = "your password"; $dbdatabase = "your database"; $con = mysql_connect("$dbhost","$dbuser","$dbpassword"); if (!$con) { die('Could not connect: ' . mysql_error()); }
Member Avatar for R1Programmer

thanks guys. I have same problem. It ´s help me. :)

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