PHP function not seeing value of second parameter

Question

For some odd reason my PHP is not seeing the value of the second parameter.

My code:

PHP function:

public function getVacs($key, $id = null, $deleted = null, $deleted_key = null) {

        if(!$id) {
            $data = $this->_db->getAll('vacatures', $key);
        } elseif(!empty($deleted)) {
            $data = $this->_db->getAll('vacatures', $key, $id, $deleted, $deleted_key);
        } else {
            $data = $this->_db->getAll('vacatures', $key, $id);
        }

        if($data->count()) {
            $this->_data = $data->results();
            $this->_count = $data->count();
            return true;
        }
    }

Calling the function:

} elseif(isset($_POST['all'])) {

        $vacs = $v->getVacs('delete', '0');

        echo json_encode($v->data());
        exit();
    }

The problem is, the function does not see the value of $id.

It's running the first if while it should be running the else.

Ray · Accepted Answer · 2016-07-07 14:01:22Z

2

In php the string "0" evaluates to false.

This means your check if(!$id) will evaluate to true and by your logic id won't be set in $data.

If the string "0" is legitimate option, then check for null explicitly instead:

if(is_null($id)){

This will

edited Jul 7, 2016 at 14:01

answered Jul 7, 2016 at 13:52

Ray

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3 Comments

Chris Over a year ago

Jup, I used the following which worked, if(!is_null($id)). Thanks!

Ray Over a year ago

@Chris yes if(!is_null($id)) will evaluate to true when $id == '0' is that what you want? I thought you didn't want to go into the first block when the value of $id` is anything except null?

Chris Over a year ago

I moved the else to be the first if. So now it will first check if $id is not null

aynber · Accepted Answer · 2016-07-07 13:58:05Z

1

It is seeing the value of $id, but your if statement is set up wrong. 0 will evaluate to false on a check like that. So you really need to make sure that it's not null:

 if($id != null) {

If you want the first if to run only if there is NOT a valid id, then you need to check if it's empty (i.e. not null, 0, false, or an empty string)

if(empty($id)) {

edited Jul 7, 2016 at 13:58

answered Jul 7, 2016 at 13:52

aynber

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4 Comments

Chris Over a year ago

Still runs the first if.

aynber Over a year ago

I updated my answer with what you might be looking for.

Chris Over a year ago

I've tried if(!empty($id)) but it still doesn't run the else

aynber Over a year ago

Whoops, that should be if(empty($id)). I'll fix that.

RockTheShow · Accepted Answer · 2016-07-07 14:11:47Z

0

Using the strict comparison operator might be a good idea in these cases (and I would say in most cases):

0 == null; // evaluates to true
0 === null; // evaluates to false

Useful with strpos also (returns 0 means searched term at position 0 of haystack string, returns false means searched term not found).

answered Jul 7, 2016 at 14:11

RockTheShow

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