PHP Assign a value to a var from IF statement variable

Question

I have something like this:

$msg    = '';
$var1   = 'image';
$var2   = 'class';
$var3   = 'This '.$var1.' has this class assigned:'.$var2;

if($E == 0) {
    $msg   = $var3;
} else {
    $var1  = 'no image'; 
    $var2  = 'no class'; 
    $msg   = $var3;
};

echo $msg;

I want to be able to show $var3 with the modified results but it does not work... Thank you for your answers but i think i must clarify

Sorry about the confusion..

The problem is not with the value to test ($E == 0).. My problem is that in both case the result will be the same and $var1 and $var2 will not change...

result in both cases will be the same: echo $msg; will produce "This image has this class assigned: class "

I think is because of the composing of $var3 outside the if and cannot be changed from inside IF statement..

if you are checking to see if number has a value in the POST then you need to use empty() — KJYe.Name
– KJYe.Name, Commented Mar 7, 2011 at 16:16

smottt · Accepted Answer · 2011-03-07 16:18:00Z

3

Maybe something like this:

if($_POST['number']) {
    $var1 = 'image';
    $var2 = 'class';
} else {
    $var1 = 'no image';
    $var2 = 'no class';
}

echo 'This ' . $var1 . ' has this class assigned: ' . $var2;

answered Mar 7, 2011 at 16:18

smottt

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2 Comments

Bogdan Over a year ago

Thank you this will work but not as i want. I need to use a 3rd variable and to reduce the ammount of code

smottt Over a year ago

Just replace echo with $var3 like this -> $var3 = 'This ' . $var1 . ' has this class assigned: ' . $var2; and it should work like you want.

The Bndr · Accepted Answer · 2011-03-09 12:18:17Z

2

you have to write a function

function var3($v1,$v2)
{
   return 'This '.$v1.' has this class assigned:'.$v2;
}

use this function that way:

$msg='';
$var1='image';
$var2='class';
$var3=var3($var1,$var2);
if($_POST['number']) {$msg=$var3;} 
    else 
        {
         $var1='no image'; 
         $var2='no class'; 
         $msg=var3($var1,$var2);
         };

echo $msg;

edited Mar 9, 2011 at 12:18

answered Mar 7, 2011 at 16:21

The Bndr

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2 Comments

smottt Over a year ago

You have a typo in the function var3().. You are using $1 and $2 instead of $v1 and $v2.

Bogdan Over a year ago

This is the way i ended up. It was obvious but the answer eluded me :) Thank you all! for helping

powtac · Accepted Answer · 2011-03-07 16:17:50Z

0

use this if condition:

if (isset($_POST['number']) AND strlen($_POST['number'])) {
...

answered Mar 7, 2011 at 16:17

powtac

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Comments

Czechnology · Accepted Answer · 2011-03-07 16:18:32Z

0

Use something like this:

if(empty($_POST['number'])) {
  $var1='no image'; 
  $var2='no class'; 
} else {
  $var1='image';
  $var2='class';
}

echo sprintf('This %s has this class assigned:%s', $var1, $var2);

sprintf is fun! (and useful)

answered Mar 7, 2011 at 16:18

Czechnology

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Comments

krtek · Accepted Answer · 2011-03-07 16:19:35Z

0

You can use sprintf to achieve what you want :

$var1='image';
$var2='class';
$format ='This %s has this class assigned: %s';

if(! isset($_POST['number']) || ! $_POST['number']) {
    $var1='no image'; 
    $var2='no class'; 
};

$msg = sprintf($format, $var1, $var2);

I took the liberty to improve a little your condition too.

answered Mar 7, 2011 at 16:19

krtek

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Comments

Senthil Prabu · Accepted Answer · 2011-03-07 16:21:04Z

0

here the solution for your problelem

$msg='';

$var1='image';

$var2='class';

$var3='This '.$var1.' has this class assigned:'.$var2;

if($_POST['number']) {

$msg=$var3;

} else {

$var1='no image'; 

$var2='no class'; 

$var3='This '.$var1.' has this class assigned:'.$var2;

$msg=$var3;

};

echo $msg;

answered Mar 7, 2011 at 16:21

Senthil Prabu

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