Check if variable is set in PHP

Question

So I'm trying to see if the user inputted anything:

$test = $_GET["restaurantName"];

if(isset($test))
{
    echo "you inputed something";
    echo "$test";
}

if(!isset($test))
{
    echo "you did not input anything";
    echo "$test";
}

die("The End");

For some reason even when I don't input anything it still passes the first if statement and says that something has been inputted even when I don't I looked up the documentation about isset() and I'm pretty sure that this is how you are supposed to use it.

Test the original variable i.e. if (isset($_GET["restaurantName"])) thats what you are interested in being set to something afterall — RiggsFolly
– RiggsFolly, Commented Apr 6, 2016 at 20:59
also i wouldn't make a habit of placing variables in a string (especially if it serves no purpose at all) echo $test; works fine — Jester
– Jester, Commented Apr 6, 2016 at 21:10

GrumpyCrouton · Accepted Answer · 2016-04-07 03:43:16Z

2

You should do it this way if you want to keep the same layout style.

if(isSet($_GET["restaurantName"])) {
     $test = $_GET["restaurantName"];
}

if(isset($test))
    {
        echo "you inputed something";
        echo "$test";
    } else { //!isset($test)
        echo "you did not input anything";
        echo "$test";
    }

Your problem is that you are setting the variable, even if the GET doesn't exist.

How I would do it personally, as it makes the code much shorter with the same outputs:

if(isSet($_GET["restaurantName"])) {
    $test = $_GET["restaurantName"];
    echo "Your input: ".$test;
} else {
    echo "No Input";
}

edited Apr 7, 2016 at 3:43

answered Apr 6, 2016 at 20:59

GrumpyCrouton

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2 Comments

RiggsFolly Over a year ago

Well you dont really need $test at all if you think about this logically

GrumpyCrouton Over a year ago

@RiggsFolly correct, but I like to keep shorter variable names. I wouldn't really want to type $_GET["restaurantName"] every time I needed that variable.

Jester · Accepted Answer · 2016-04-06 21:08:17Z

1

You are setting it: $test = $_GET["restaurantName"]; the isset checks whether a variable has been set, not whether the variable contained is null or empty, you could use !empty

you can also check isset($_GET["restaurantName"];) but beware even if you have the get variable in your url as ?restaurantName= than it's still set, it's just empty

Best thing to do would be to check if it's set and not an empty string:

if(isset($_GET["restaurantName"]) && $_GET["restaurantName"] != "")
{
    echo "you inputed something";
    echo $_GET["restaurantName"];
} else {
    echo "you did not input anything";
}

die("The End");

i also removed the second if, cause you can just use an else clause instead of checking twice.

Some links to read: http://php.net/manual/en/function.isset.php http://php.net/manual/en/function.empty.php

edited Apr 6, 2016 at 21:08

answered Apr 6, 2016 at 20:59

Jester

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3 Comments

Random Over a year ago

Beware "!empty" condition... value "0" is considered as "empty"

Jester Over a year ago

true, but in case of a restaurant name i don't mind not allowing 0 as a restaurant name ;) but you're right.

Jester Over a year ago

@Random It's a weak reason though from my side though :p i edited my answer :)

Don't Panic · Accepted Answer · 2016-04-06 21:46:20Z

If this $_GET['restaurantName'] comes from a submitted (GET) form input rather than a query string in a link (which may or may not be present), it will always be set. If the user did not enter anything, it will be set to an empty string. You can check it using empty instead of isset (empty includes a check for isset).

if (!empty($_GET['restaurantName'])) {
    echo "you input " . $_GET['restaurantName'];
} else {
    echo "you did not input anything";
}

It may be a good idea to also check the trimmed entry in case it is something like ' ', which you can do with

if (!empty($_GET['restaurantName']) && trim($_GET['restaurantName'])) { ...

But that is starting to get more into form validation, which is another topic in itself.

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