Assign variable within condition if true

Question

I know you can assign a variable in a condition like this:

if ($var = $foo)

However I don't need to do anything in the condition itself, so I'm often left with empty brackets. Can I simply assign $var if $foo is true in some other way without needing to do something else later?

Also can I assign $var to $foo if $foo is true but if $foo is false do something else? Like:

if ($var = !$foo) {
    if ($var = !$bar) {
        //Etc...
    }
}

Basically I want to have more fallbacks/defaults.

Solution 1

@chandresh_cool's suggestion is right but to allow multiple possiblities / fallbacks you would have to nest the ternary expressions:

$var = ($foo == true) ? $foo: 
       ($bar == true) ? $bar: 
       ($fuzz == true) ? $fuzz:
       $default;

Note: the first 3 lines end in colons not semi-colons.

However a simpler solution is to do the following:

$var = ($foo||$bar||$fuzz...);

Solution 2

Although this is a very old post. Fallback logic on falsify values can be coded like this.

$var = $foo ?: $bar ?: "default";

In this case when $foo is a falsified value (like false, empty string, etc.) it will fall back to $bar otherwise it uses $foo. If bar is a falsified value, it will fallback to the string default.

Keep in mind, that this works with falsified values, and not only true.

example:

$foo = "";
$bar = null;
$var = $foo ?: $bar ?: "default";  

$var will contain the text default because empty strings and null are considered "false" values.

[update]

In php 7 you can use the new null coalescing operator: ??, which also checks if the variable exists with isset(). This is usefull for when you are using a key in an array.

Example:

$array = [];
$bar = null;
$var = $array['foo'] ?? $bar ?? "default";

Before php 7 this would have given an Undefined index: foo notice. But with the null coalescing operator, that notice won't come up.

Solution 3

Instead you can Use ternary operator like this

$var = ($foo == true)?$foo:"put here what you want";

Solution 4

You can assign values like this:

$var = $foo;

Setting them within an if statement is also possible, PHP will evaluate the resulting $var which you just assigned.

I dont really get your question, but you could do something like this:

if(!($var = $foo)){
   //something else.
}

View more solutions

Comments

[Jared]

Your simpler solution is exactly what I'm looking for. It's short, sweet and does what I need it to do: set $var to the first positive value it can find.

[Jared]

It's still a little long-winded, but it works so I'll give you +1 for your answer.

[Jared]

Your if statement code looks like what I'm trying to do, but RobJ's answer was better. I'll still give you a +1 for it.

[Veda]

If you're going for the first option, please don't do the "== true", just do "$var = ($foo) ? $....."

[HPierce]

@jconder - I'm rolling back your edit to this answer. Leroy says "otherwise it uses $foo", but the edit you've made completely invalidates that. Leroy is intentionally using two ternary operators here (each with the first expression being empty), allowing for 3 possible outcomes: $foo, $bar, or "default", rather than using a single ternary operator which could only return 2 possible outcomes: $bar or "default".

[jconder]

Yep you're right about that, my bad, I missed the entire point of the post.

[iyrin]

This makes if expressions a lot cleaner. I'm using this to check for a custom post type in WordPress $is_entry = is_singular( 'myentry' ) ?: is_post_type_archive( 'myentry' ) ?: false;. Now I can use if ( $is_entry && $other_condition ) to determine if the current page belongs to my post type without the if expression getting too ugly.

[Leroy]

In php7 there is a new null coalescing operator: ??, which also checks if the variable exists like isset(). This is usefull for when you are using a key in an array. For example: $var = $array['key'] ?? $bar ?: "default". Now when the key in $array does not exists, it won't throw any errors. Because it wil also look if the key exists.