PHP Passing an Array to a Function

Question

How can I pass an array to a function?

Let's say I have this multi-dimensional array:

$TheArray = (
array("Value 1 0","Value 1 1","Value 1 2"),
array("Value 2 0","Value 2 1","Value 2 2"),
array("Value 3 0","Value 3 1","Value 3 2")
);

Instead of doing this...

for ($i=0; $i<=(count($TheArray)-1); $i++)
{
echo $TheArray[$i][0] . " " . $TheArray[$i][1] . " " . $TheArray[$i][2] . "<br />";
}

I want to do this...

function DoStuffWithTheArray($SubArr)
{
echo $SubArr[0] . " " . $SubArr[1] . " " . $SubArr[2] . "<br />";
}

for ($i=0; $i<=(count($TheArray)-1); $i++)
{
DoStuffWithTheArray($TheArray[$i]);
}

Hopefully, you can tell what I am trying to do, but I do not know how to get it to work. When I do try it the way I want, all the values are empty

What does all the values are empty mean? Do you not get any output? — Joseph Silber
– Joseph Silber, Commented Feb 1, 2013 at 2:38
Get in the habit of using foreach loops for array iteration whenever possible. In PHP, using an incremental for loop is actually not that common for basic iteration. — Michael Berkowski
– Michael Berkowski, Commented Feb 1, 2013 at 2:43
Try add echo 'function called<br/>'; into function DoStuffWithTheArray to check if you have called the function — Michael Law
– Michael Law, Commented Feb 1, 2013 at 2:44
This works exactly as you have it, except for the missing array( keyword. codepad.viper-7.com/Ysz6qT. If that is your issue (blank screen), turn on error reporting. error_reporting(E_ALL); ini_set('display_errors', 1); — Michael Berkowski
– Michael Berkowski, Commented Feb 1, 2013 at 2:48

Kerem · Accepted Answer · 2019-04-04 18:48:15Z

1

I think, you need to learn first how create an array;

// instead of this
$TheArray = (...
// you can create an array so
$TheArray = array(...

Second, change this style please and see: http://php.net/manual/en/control-structures.foreach.php

for ($i = 0; $i < count($TheArray); $i++)

And answer;

$array = array(
    array("Value 1 0", "Value 1 1", "Value 1 2"),
    array("Value 2 0", "Value 2 1", "Value 2 2"),
    array("Value 3 0", "Value 3 1", "Value 3 2")
);

function fn($a) {
    print "$a[0], $a[1], $a[1]\n";
}

foreach ($array as $a) fn($a);

Value 1 0, Value 1 1, Value 1 1
Value 2 0, Value 2 1, Value 2 1
Value 3 0, Value 3 1, Value 3 1

edited Apr 4, 2019 at 18:48

answered Feb 1, 2013 at 3:08

Kerem

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1 Comment

Soren Over a year ago

I do know how to create an array. It was a typo on my part. Thanks for your help.

Matt Runion · Accepted Answer · 2013-02-01 02:43:04Z

0

Just pass the array in:

function DoStuffWithTheArray($array) {
    $array[0]...$array[n];
}

That should work. Does it not?

answered Feb 1, 2013 at 2:43

Matt Runion

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2 Comments

Michael Berkowski Over a year ago

This is what the OP is doing with DoStuffWithTheArray($TheArray[$i]) inside the loop.

Matt Runion Over a year ago

The second example works for me. You do need to make the following correction: $TheArray = array(.... But when I past in the array and the second section of code, it works just fine on writecodeonline.com/php

जलजनक · Accepted Answer · 2013-02-01 03:22:33Z

0

If you want to alter the content of the array, then you need to pass it by reference:

Argument passing by Reference

answered Feb 1, 2013 at 3:22

जलजनक

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