CODEWITHSUNDEEP
phphtmlarrays
newbie41
newbie41

Reputation: 45

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

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.

Upvotes: 1

Views: 5692

Answers (4)

Webber
Webber

Reputation: 5514

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).

Upvotes: 0

Danyal Sandeelo
Danyal Sandeelo

Reputation: 12391

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']);

Upvotes: 1

Niranjan N Raju
Niranjan N Raju

Reputation: 11987

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.
}

Upvotes: 1

Nageen
Nageen

Reputation: 1759

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

Upvotes: 3

Related Questions