Notice: Undefined index: sendit in C:\...

[Wickham]

I've got a test form working to show the data on the same page after processing and it works well but I've had to insert

error_reporting (E_ALL ^ E_NOTICE);

to get rid of this notice:-

Notice: Undefined index: sendit in C:\wamp\www\test\test-form4a.php on line 10

which is if($_POST['sendit'])

 

The php on first loading the page ignores the echos to display the data and uses the "else" to display the form. When the submit button is pressed the page loads again and $_POST['sendit'] then has some data to process, but on the first pass the sendit is undefined.

 

I've tried making it a variable like

if($_POST[$sendit])

or

if($_POST['$sendit'])

with

$sendit = $_POST['sendit'];

but the notice still shows and the form doesn't process.

 

I've also tried putting the variable is different places.

 

Is there a way to get it right without the exception code?

 

>




PHP test



This is testing WampServer2.
error_reporting (E_ALL ^ E_NOTICE);
if($_POST['sendit'])
{
$rating = $_POST['rating'];
$item = $_POST['item'];
echo  "You rated {$item} as {$rating} out of 10.";
echo "
Thank You.";
}
else 
{ echo "Please submit form";
?>
</pre>
<form action="test-form4a.php" method="post">

Song 1
Song 2

Rating (out of 10 please):  

</form>
<br><br>}<br>?><br><br><br

[falkencreative]

Instead of

 

if($_POST['sendit'])

 

I usually use

 

if(isset($_POST['sendit']))

[Wickham]

Thanks Ben.. I had tried

if(!isset($_POST['sendit']))

which has ! before isset as shown on this webpage

http://www.dmxzone.com/ShowDetail.asp?NewsId=13811

but it didn't work.

 

Why did that web answer include the !

[falkencreative]

the "!" stands for "not", so that statement would read "if $_POST['sendit'] isn't set, then..."

 

basically, to use the above statement, you would have to switch around your logic. If it isn't set, display the form, otherwise, process the form.

[Wickham]

Uh! I could have guessed that! :/