fetch_assoc Help ....

Andyr6381

I am trying to write my first script and Im wanting to fetch a string 'home' from a field in my database. But i keep getting this error on my fetch record. Here is the script, anyone know where im going wrong???

<?php

FileName="Connection_php_mysql.htm"

Type="MYSQL"

HTTP="true"

$hostname_test = "localhost";
$database_test = "bb4b";
$username_test = "root";
$password_test = "";

$test = mysql_pconnect($hostname_test, $username_test, $password_test) or die(mysql_error());
?>
<?
mysql_select_db($database_test, $test);
$query_Recordset1 = "SELECT COUNT(*) FROM registration WHERE (brenchley = 'home')";

$Recordset1 = mysql_query($query_Recordset1, $test) or die(mysql_error());
$row_Recordset1 = mysql_fetch_assoc($Recordset1);
$totalRows_Recordset1 = mysql_num_rows($Recordset1);
?>

<html>
<head>
<title>Untitled Document</title>
<meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1">
</head>
<?
echo ($Recordset1);
?>

This is the error

Fatal error: Call to undefined function: mysql_fetch_assoc() in c:\program files\apache group\apache\htdocs\test1.php on line 17

Can any1 help. thanks
Andy
BB4B

PHPdev

Andy,

First I must STRONGLY SUDGEST you remove or at least blank out your database information, when posting... Maybe Im paranoid, bu that is just me...

Second, try using mysql_fetch_array() instead of assoc() see if that works.

The only other thing that I can think of is the version of PHP you are running... I could not imagine it is before 4.0.3

www.php.net may be able to give more info on this function...

PHPdev

Andyr6381

Thanks for the tips,

the script is from my test server so that doesnt matter,

are you saying i should directly replace fetch with array?

Regards
Andy

Andyr6381

doh, read the whole Post lol. sos

Andyr6381

I tried replacing assoc with array. and the script seems to run now, but it returns sumthing about resorseid#2 for recordset1, any1 got any ideas on this 1.?

[deleted]

" but it returns sumthing about resorseid#2 for recordset1, any1 got any ideas on this 1.?"

You'll have to post what you've got now, it looks like you're just printing the wrong vars.

Oh, and mysql_fetch_assoc was added in PHP 4.0.3, so your server seems to be running an ancient (read: way out of f*cking date, must upgrade asap) version of PHP.

You can emulate mysql_fetch_assoc by using mysql_fetch_array($resultvar, MYSQL_ASSOC)

And pconnect is not a good choice, it wastes a lot of resources on the server and could/will sooner or later cause the server to run out of available databases connections.

Andyr6381

I would give you what I am outputting but I had a bugger of a problem yeturday with winxp deciding that it would want to die after a power faliure.

Bleedin MS, great until it goes wrong.

I'm just downloading the latest server, I should be up and running over the weekend, am moving house at the same time, absolute nightmare.

cheers for the pointers will be lookin into them as soon as I can.

Andy

Andyr6381

Hey.

I am trying to write a script to output the number of times the string home appears in the field brenchley, this is the scripts I am using

<?php require_once('../Connections/bb4b.php'); ?>
<?php
mysql_select_db($database_bb4b, $bb4b);
$query_home = "SELECT COUNT(*) FROM registration WHERE (brenchley = 'home') ";
$home = mysql_query($query_home, $bb4b) or die(mysql_error());
$row_home = mysql_fetch_assoc($home);
$totalRows_home = mysql_num_rows($home);
?>

<?php
echo ("mysql_free_result = ");
echo mysql_free_result($home);
echo ("<p> </p>");
echo ("home = ");
echo ($home);

The output results i am getting are:

mysql_free_result = 1

home = Resource id #2

I know that there are 2 homes in the field brenchley and that the query works, I have updated my php and this makes no differance.

Cheers Andy