Check variable type with if statement

Question

Hi I have a simple question regarding the array.

I am trying to use foreach loop to echo the vars. However, there are times that the variable wont' be an array I have created a if statement to check the variable type, but I am not sure if it's the best practice doing it.

Are there any better way to do what I need? Thanks a lot!

My codes

$test = $_GET['testVar'];

if(is_array($test)){
 foreach($test as $t){
   echo $t;
 }
}else{
   echo $test;
}

Nope. That's how you would do it.

Kermit
– Kermit

2013-03-06 23:14:03 +00:00
Commented Mar 6, 2013 at 23:14 — Kermit
– Kermit, Commented Mar 6, 2013 at 23:14

hek2mgl · Accepted Answer · 2013-03-06 23:19:17Z

2

is_array is the best way to check if a variable is an array. So your code is ok.

However here comes a generic solution that will work for all data types not just arrays with the functiongettype() You can refine the results if $type is 'Object' using the function get_class()

$type = gettype($var);

// get class name for objects if so desired
if($type === 'object') {
    $type = get_class($var);
}

edited Mar 6, 2013 at 23:19

answered Mar 6, 2013 at 23:14

hek2mgl

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5 Comments

Kermit Over a year ago

I think OP is only concerned if it's an array; in which case is_array is perfectly suitable.

hek2mgl Over a year ago

Ok maybe I misunderstood someting, however this is generic solution and should help in any case

Kermit Over a year ago

Just stick to gettype and add that is_array is acceptable as well.

user1467267 Over a year ago

get_class doesn't work indeed. This is not JavaScript where everything is an Object-type :P. If you check for any type; use gettype() for unity. If you only check if it's an array, then use is_array as it returns a clean TRUE/FALSE :)

hek2mgl Over a year ago

@AarolamaBluenk Thanks. Anwers to the easiest questions are most vulnerable for earning downvotes if not read 100% exactly ;) So I hope all are happy now :)

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