Multiple if statements in PHP

Question

I know that this is not the way to do it but I'm having some trouble trying to set a variable using conditional statements.

For some context, $foo, $foo_1 and $foo_2 are checking to see if a form field is set, if it is, to set $bar equal to the calculation that accompanies each condition.

if(isset($foo)) {
    $bar = ($baz * 52) * ($qux);
} 

if(isset($foo_1)) {
    $bar = ($baz_1 * 12) * ($qux);
}

if(isset($foo_2)) {
    $bar = ($baz_2 * $qux);
}

This is because the existing code uses $bar in a calculation

$a = ($b / $bar)

and $a is passed on further down in a variety of cases.

As it is, only the first $foo condition is passing $bar properly, all other conditions $bar outputs as NULL.

EDIT: To clarify the code above (which was incorrect) and update my example below:

$foo = isset($form['first'] ? $form['first'] : '';
$foo_1 = isset($form['second'] ? $form['second'] : '';
$foo_2 = isset($form['third'] ? $form['third'] : '';
if(isset($foo)) {
    $bar = ($foo * 52) * $qux;
} 

if(isset($foo_1)) {
    $bar = ($foo_1 * 12) * $qux;
}

if(isset($foo_2)) {
    $bar = $foo_2 * $qux;
}

Revent, Oswald and RoyalBG's suggestions have pointed out that my code isn't setting variables properly, which is the case. $baz, $baz_1, $baz_2 were deprecated and non-existent and replaced with $foo, $foo_1, $foo_2 to accompany their conditionals for the desired inputs. Once I clean this up I will use a switch statement instead of what I've written above. All of the comments below have been very helpful Thanks for the help.

You are multiplying, no need to parenthesis. Also, keep in mind these if are not mutual excluding. — fedorqui
– fedorqui, Commented Mar 28, 2013 at 21:19
you haven't shown enough of your code for me to really help, but from what I can see, I would say it's nearly certain your code could be written much better. You might benefit from asking for some help with improving the code over at codereview.stackexchange.com — Spudley
– Spudley, Commented Mar 28, 2013 at 21:38

Oswald · Accepted Answer · 2013-03-28 21:24:42Z

1

$foo_1 and $foo_2 are never set. That's why if $foo is not set, $bar won't be set either.

answered Mar 28, 2013 at 21:24

Oswald

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Last Breath · Accepted Answer · 2013-03-28 21:26:59Z

0

Please clarify your code much more because i don't see where your $baz_1 variable states and where it is initialized ?

As the logic say you must multiply the $foo_1 by the $qux i don't know what you mean by the $baz_1,$baz_2 variables

Then if you are testing the $foo_1,$foo_2 you must make your calculation base on the tested variable not on a new variable that may it is not set

More explaining what if the $foo,$foo_1,$foo_2 are set but the $baz,$baz_1,$baz_2 not set i don't see any manipulation for the $baz variables please clarify the situation .

answered Mar 28, 2013 at 21:26

Last Breath

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Ivan Yonkov · Accepted Answer · 2013-03-28 21:32:02Z

0

If the code in your condition is not passed to the PHP engine, when you ask the variable to be set, then it's most likely not set... you could also try with switch/case block

answered Mar 28, 2013 at 21:32

Ivan Yonkov

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Revent · Accepted Answer · 2013-03-28 21:47:12Z

0

Oswald and Royal Bg are correct that your variables are likely never set. I would write the code this way (assuming you allow a zero value for the $baz and $qux variables):

$bar = 0;   // initialize the variable so it won't be null
if(isset($foo) && isset($baz) && isset($qux)) {
  $bar = ($baz * 52) * ($qux);
} 
else if(isset($foo_1) && isset($ba_1z) && isset($qux)) {
  $bar = ($baz_1 * 12) * ($qux);
}
else if(isset($foo_2) && isset($baz_2) && isset($qux)) {
  $bar = ($baz_2 * $qux);
}

You can then check the value of $bar using the empty() function, which returns true if the value is null, zero, or an empty string.

answered Mar 28, 2013 at 21:47

Revent

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