How to use in array ? Warning: in_array() expects parameter 2 to be array, string given in

Question

How can I deal with this error:

Warning: in_array() expects parameter 2 to be array, string given in C:\xampp\htdocs\php\index.php

My code is:

if (!isset($_GET['jenis'])) {
  $jenis =  "";
} else {
  $jenis = $_GET['jenis'];
}


<li><input type="checkbox" onclick="jeniss();" name="jenis[]" value="11" <?php if (in_array("11",$jenis))  { echo "checked"; } ?> > <a> 11 </a> </li>
<li><input type="checkbox" onclick="jeniss();" name="jenis[]" value="12" <?php if (in_array("12",$jenis))  { echo "checked"; } ?> > <a> 12 </a> </li>
<li><input type="checkbox" onclick="jeniss();" name="jenis[]" value="13" <?php if (in_array("13",$jenis))  { echo "checked"; } ?> > <a> 13 </a> </li>

Note: the error at HTML input type when the page has not posted anything yet.

Well, $jenis is a string, so it's kind of hard to do an array search on it. — Loek
– Loek, Commented Jul 10, 2018 at 9:22
You could probably fix it by just changing $jenis = ""; to $jenis = []; — CD001
– CD001, Commented Jul 10, 2018 at 9:36

Nageen · Accepted Answer · 2018-07-10 09:23:20Z

3

You have to check $_GET['jenis'] is string or array by using

var_dump($_GET['jenis']);

if it is string than you have to define as a array just like that

$jenis = array($_GET['jenis']);

and you can use in_array

answered Jul 10, 2018 at 9:23

Nageen

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Niranjan N Raju · Accepted Answer · 2018-07-10 09:23:38Z

1

As your error says, you are trying to pass string instead of array for in_array(). Please check this to have a better look at in_array()

According to your code

if (!isset($_GET['jenis'])) {
    $jenis =  array(); //<--- have empty array be default.
} else {
    $jenis = (array) $_GET['jenis']; //<---- change this line. You can typecast to array if you are getting only one value.
}

answered Jul 10, 2018 at 9:23

Niranjan N Raju

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Danyal Sandeelo · Accepted Answer · 2018-07-10 09:28:29Z

1

The second parameter needs to be an array and you are passing the string instead of array.

It is like this in_array("value_you_wanna_look",$array)

$_GET['jenis'] is a string value not an array.

You may check against the value of $jenis like this

<?php if ($jenis=="11") echo "checked"; ?> // and so on for other values

OR

Enter the value in array like this:

$jenis = array($_GET['jenis']);

edited Jul 10, 2018 at 9:28

answered Jul 10, 2018 at 9:22

Danyal Sandeelo

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Comments

Webber · Accepted Answer · 2018-07-10 09:37:06Z

0

In order to use in_array() you have to make sure that your second parameter is an array, you are providing a string as the second parameter.

Reason for this is that when $jenis is set by the form, it already is an array how you defined it, however before posting the form $jenis is not set and thus it will be set an empty string on line 2.

All you have to change is change line 2 to $jenis = array();.

Good practice is to make sure that a variable always has the same type or null. It will generally prevent these kind of mistakes (that might also become hard to test in complex cases).

answered Jul 10, 2018 at 9:37

Webber

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