Newbie - Guess its a permissions problem.

kathekas

I have a box running Vsta Home Premium (IIS 7) with PHP and MySQL installed.

From within my own intranet if run the page with

<?php
phpinfo();
?>

it returns all the info fine.

But if I run a test page with some code, it doesn't work.

So I guess this is a permissions problem, but as this is my first time with PHP I don't know where to start.

CoderDan

when you say 'it doesn't work' - what do you mean exactly? Is there a PHP error, or does it display the php source code?
If it's the latter, you could a) have the wrong fire extension, or b) be using <? instead of <?php and have short_tags disabled.

kathekas

Thanks CodeDan - your response was quite thought provoking.

I list the code below.

If I take all the PHP, leaving just the HTML it work.

Put the PHP back in and it just returns a blank page

<?php
$page_name = "test1";

        $db = @mysql_connect("localhost", "username", "password");
        @mysql_select_db("kathikas",$db);

        $sql = "UPDATE test_track1 SET hits = hits + 1 WHERE page_name ='$page_name'";
        @mysql_query($sql);

<HTML>
<HEAD>
<TITLE>Test Hit Count Page hitcount2.php </TITLE>
</HEAD>
<BODY>
<H1 align=centre>You have been counted! too</H1>
</BODY>
</HTML>

CoderDan

change your @ function calls and change them to or die's, ie. instead of:

@mysql_connect("localhost", "username", "password")

use

mysql_connect("localhost", "username", "password") or die( mysql_error() )

The @ suppresses errors, and the most likely problem is an error, so we want to see them - the or die() allows us to stop teh script with an optional message, and in this case, if a mysql fun fails, mysql_error() returns the reason it failed, so it is ideal.
And it's amazingly simple for even the best php coders to typo a username or password, or forget which database they mean to use, or similar 🙂 (although, in the code above, that shouldn't prevent the html output, but it is a good place to start)
also, adding a line like:

error_reporting(E_ALL);

error_reporting(E_ALL ^ E_NOTICE);

can help also in figuring out errors assuming it hides them by default.

Bjom

or even better, don't use

...or die(mysql_error())

, but use

...or trigger_error(mysql_error(), E_USER_ERROR)

since the latter reacts to your error_reporting levels while the former won't.

kathekas

Thanks CoderDan Bjom.

Basic, stupid mistake, the sql select statement left the fields statement *

After I had got the original code working I altered the number of fields in the table.

So it failed due to the number of values not matching the number of data entries

Not sure why the page did not display.

Thanks very much for your help.

Its amazing how error reporting helps solve problems !!!!!!!

CoderDan

glad you got it sorted out, also, Bjom is correct that for production aimed code, it'd be better to use trigger_error with a level of E_USER_ERROR