Simple If statement wont work (MS Excel)

[Help!php]

=IF(F29>=400,"A",IF(F29>=200,"B",IF(F29>=100,"C","Fail")))

 

Why wont this work. Any number I enter it displays A.

[trq]

it wont work because it not valid php.

[KevinM1]

It won't work because that's not valid PHP.  You need to do:

 

if ($F29 >= 400) {
   echo "A";
} elseif ($F29 >= 200) {
   echo "B";
} elseif ($F29 >= 100) {
   echo "C";
} else {
   echo "Fail";
}

 

For more info on conditionals in PHP, look at: http://php.net/manual/en/control-structures.if.php

[Help!php]

I am trying this on Excel before I move on to PHP and it still wont work. I just want to get this running on excel. and  It wont display the result

[KevinM1]

Why are you asking for help in Excel in the PHP Coding Help section of the forum?  Moved.

[Psycho]

That works fine for me in Excel. Are you sure the value you are referencing is really in cell F29 and that is a number type?

[Help!php]

Yeh it is. I managed to get this done .. Thanks though