problems

Hi list
Hope somebody can help me.

when i run this code
<?php

require ("header.php");
echo "<h2>Add contact</h2>";

echo "
<table border=1>

<tr>


<form name=custom action=".$_SERVER['PHP_SELF']." method=POST>
<input type=hidden name=dirty value=0>
<td>First name</td><td><input type=text
name=txtFirstName></td><tr>
<td>Last name </td><td><input type=text
name=txtLastName></td><tr>
<td>Phone </td><td><input type=text name=txtPhone></td><tr>
<td>Mobil </td><td><input type=text name=txtMobilPhone></td><tr>
<td>Email </td><td><input type=text
name=txtEmailAddress></td><tr>
<td>Contact type </td><td><input type=text
name=txtContactType></td></tr>

</form>
</table>
<input type=submit onclick=\"custom.dirty.value=1\" value=\"Create new
contact\">";

if ( $_POST['dirty'] == 0 )
{
echo "her skal der kaldes kode til indsætte u mysql";
}
?>


I get the following message

Notice: Undefined index: dirty in d:\program files\apache
group\apache\htdocs\shj\add_contact.php on line 25

I have tried many dirrefent thingsnow but i can't get through....
please !!

regards
Simon

simse wrote on Thursday 11 September 2003 15:15:

<snip>
> if ( $_POST['dirty'] == 0 )
> {
> echo "her skal der kaldes kode til indsætte u mysql";
> }
> ?>

<snip>

> Notice: Undefined index: dirty in d:\program files\apache
> group\apache\htdocs\shj\add_contact.php on line 25

Change the if condition to

isset($_POST['dirty'])

What is happening is that first time when you access the page, the "dirty"
form element has not been submitted and, therefore, does not exist in the
$_POST array. Hence, the message.

Having said that, the message is just a notice, it's not an error. You can
set what types of messages should display in your php.ini file.

Using isset() function should get rid of the notice.

--
Business Web Solutions
ActiveLink, LLC
www.active-link.com/intranet/

False as PHP create the variables with default value on first use if not
defined before.

"Zurab Davitiani" <agt@mindless.com> wrote in message
news:0_68b.3424$kT.1675@newssvr25.news.prodigy.com ...
> simse wrote on Thursday 11 September 2003 15:15:
>
> <snip>
> > if ( $_POST['dirty'] == 0 )
> > {
> > echo "her skal der kaldes kode til indsætte u mysql";
> > }
> > ?>
>
> <snip>
>
> > Notice: Undefined index: dirty in d:\program files\apache
> > group\apache\htdocs\shj\add_contact.php on line 25
>
> Change the if condition to
>
> isset($_POST['dirty'])
>
> What is happening is that first time when you access the page, the "dirty"
> form element has not been submitted and, therefore, does not exist in the
> $_POST array. Hence, the message.
>
> Having said that, the message is just a notice, it's not an error. You can
> set what types of messages should display in your php.ini file.
>
> Using isset() function should get rid of the notice.
>
> --
> Business Web Solutions
> ActiveLink, LLC
> www.active-link.com/intranet/

On Sun, 21 Sep 2003 09:25:03 -0400, "Savut" <webki@hotmail.com> wrote:

>"Zurab Davitiani" <agt@mindless.com> wrote in message
>news:0_68b.3424$kT.1675@newssvr25.news.prodigy.co m...
>> simse wrote on Thursday 11 September 2003 15:15:
>>
>> <snip>
>> > if ( $_POST['dirty'] == 0 )
>> > {
>> > echo "her skal der kaldes kode til indsætte u mysql";
>> > }
>> > ?>
>>
>> <snip>
>>
>> > Notice: Undefined index: dirty in d:\program files\apache
>> > group\apache\htdocs\shj\add_contact.php on line 25
>>
>> Change the if condition to
>>
>> isset($_POST['dirty'])
>>
>> What is happening is that first time when you access the page, the "dirty"
>> form element has not been submitted and, therefore, does not exist in the
>> $_POST array. Hence, the message.
>>
>> Having said that, the message is just a notice, it's not an error. You can
>> set what types of messages should display in your php.ini file.
>>
>> Using isset() function should get rid of the notice.

Yes - use isset(); don't just hide the warning by reducing the error_reporting
level in php.ini.

>False as PHP create the variables with default value on first use if not
>defined before.

No, it does not. Zurab is correct.

Please demonstrate the default values used for variables of each type if you
want to back up this statement.

<pre>
<?php
var_dump($_GET['notset']);
var_dump($_GET['notset']);
?>
</pre>

Outputs:

Notice: Undefined index: notset in /home/andyh/public_html/test.php on line 3
NULL

Notice: Undefined index: notset in /home/andyh/public_html/test.php on line 4
NULL

If your statement were correct, there would only be one warning. Use of a
variable (i.e. read) in PHP does not define it. Only writing to a variable
does.

--
Andy Hassall (andy@andyh.co.uk) icq(5747695) (http://www.andyh.co.uk)
Space: disk usage analysis tool (http://www.andyhsoftware.co.uk/space)