CODEWITHSUNDEEP
phparraysdeclaration
John AJ
John AJ

Reputation: 21

"Warning: array_push(): First argument should be an array" while trying to push a value into a string-type value

$info = array(
        "First_Names" => "John",
        "Last_Names" => "Smith",
        "Gender" => "Male",
    );

array_push($info["First_Names"], "$fname");

print_r ($info);

I started learning PHP through a high school class. I'm not exactly knowledgeable nor do I pay attention much, but I'm completely stuck on this;

What I'm trying to do is get the variable $fname which is defined by the user (Jack, James, Shelly, etc) to be pushed into the array First_Names which is inside the array of $info. I'm not sure where it's going wrong, but PHP is not declaring $info as an array.

I think, it states:

Warning: array_push() [function.array-push]: First argument should be an array in /home/a4938424/public_html/process.php on line 22

If I print out the array it will show up the default names and gender,and if I echo out the $fname variable it shows up properly.)

Upvotes: 0

Views: 104

Answers (4)

szymonm
szymonm

Reputation: 1051

You need to declare an array inside the array. Here is an example:

<?php
$fnames = array("John","Adam");
$lnames = array("Smith","Kowalski");
$gender = array("Male","Male");

$info = array(
            "First_Names" => $fnames,
            "Last_Names" => $lnames,
            "Gender" => $gender
        );

$new_name = "New John";

array_push($info["First_Names"],$new_name);

var_dump($info);
?>

Cheers :)

Upvotes: 0

Crackertastic
Crackertastic

Reputation: 4913

$info['First_Names'] isn't an array. If you are trying to push the name into $info you will need to create an array with your info and then push it (creating a multidimensional array in the process). 

$a = array("First_Names" => $fname);
array_push($info, $a);

If you are trying to overwrite the information in ['First_Name'] then simply access it by it's key.

$info['First_Name'] = $fname;

Upvotes: 0

Barmar
Barmar

Reputation: 782158

If you want the elements of your array to be arrays, you have to declare them as such:

$info = array(
        "First_Names" => array("John"),
        "Last_Names" => array("Smith"),
        "Gender" => array("Male"),
    );

However, IMHO this is poor design. Instead of an associative array whose values are indexed arrays, invert it to an indexed array whose elements are associative arrays.

$info = array(array('First_Name' => 'John',
                    'Last_Name' => 'Smith',
                    'Gender' => 'Male')
             );

This allows you to treat each person as a single element, instead of having to loop over all the sub-arrays in parallel. To add another person to the array, you do:

$info[] = array('First_Name' => $fname, 'Last_Name' => $lname, 'Gender' => $gender);

Upvotes: 1

Niet the Dark Absol
Niet the Dark Absol

Reputation: 324780

$info['First_Names'] is not an array, it is a string ("John"). If you define it as an array, it might work ;)

Upvotes: 2

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