PHP Forum

**Tomos** · 08-14-2005, 04:10 PM

I've just started to run PHP and MySQl with an apache server on my home PC. I've created a PHP program to communicate with the MySQL server to access some information on the test table. However, I constantly receive this line when accessing the page:

Fatal error: Call to undefined function mysql_connect() in C:\Program Files\Apache Group\Apache2\htdocs\query_test.php on line 7

This is the code I have used (Note, the XXXXXXXXX is the password and this line:

$connection=mysql_connect("localhost","root","XXXX XXX")

is the 7th line on my editor, and so therefore is the problematic line)

PHP Code:


			
<html>

<head>

<title>Query Test</title>

</head>

<body>

<?php

$connection=mysql_connect("localhost","root","XXXXXXX")

or die ("Server is down right now, due to a problem with the PHP documents communicating with the server. We should be back up soon!");

$database = "Members";

$db = mysql_select_db($database,$connection)

or die ("Server is down right now, due to a problem with the PHP documents communicating with the server. We should be back up soon!");

$query = "SELECT * FROM test";

$result = mysql_query($query)

or die ("Couldn't execute query");

mysql_close($connection);

?>

</body> 

</html>

Can anyone explain to me what I'm doing wrong? I really want to be able to understand but I can't see what I'm doing wrong? Am I missing out a part of code that I need to put in?

**pushedtomb** · 08-14-2005, 04:39 PM

try

PHP Code:


			
 $connection = @mysql_connect("localhost", "root", "XXX") or die ("Could not connect due to".mysql_error());

$database= "Members";

$db = @mysql_select_db($database, $connection) or die ("Could not open database due to".mysql_error());

//rest of code....

**Tomos** · 08-14-2005, 04:42 PM

I no longer get the error message. Now I need to find out how to display the tables.

Can you explain why that worked compared to mine for future knowledge?

Thanks!

**pushedtomb** · 05:06 PM

i think that the @ suppresses the errors, im not sure though :S... to pull things out of the database use the SELECT statement.

PHP Code:


			
 //connection info here
 $query = "SELECT * FROM table";
 $result =  mysql_query("query") or die ("Could not select info from database due to".mysql_error());
 while ($row = mysql_fetch_array($result) ) {
 
 $id = $row['id'];
 $title = $row['title'];
 //other fields you want must be set out the same
 print ('<a href=\'?id='.$id.'>'.$title.'</a><br />');
 //rest of coding and close database connection

**Tomos** · 08-14-2005, 05:05 PM

Thanx

**asm** · 08-14-2005, 05:35 PM

i think you should take the @ out of the script as you will never find the problem with it left in because this will suppress any error and therefore you will not no what is wrong due to the lack of output from the php parser.

**Skorch1** · 08-14-2005, 06:53 PM

Good point ASM! Imagine looking through all your lines of looking for a simple syntax error, not even knowing where to look for who knows what error.

**0beron** · 08-15-2005, 10:13 AM

If it is php5 you are using, this may be of interest:
http://www.php.net/manual/en/faq.dat...ses.mysql.php5

**Tomos** · 08-15-2005, 12:16 PM

Thanx guys! I'll try that now!