Assign variable within condition if true

@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...);

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.

Instead you can Use ternary operator like this

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