Is it possible to loop through all the items of two arrays using foreach?

Question

I have two arrays: $Forms and $formsShared .

<?php foreach ($Forms as $r): ?>
    $("#shareform<?=$r['Form']['id'];?>").hide();
    $(".Share<?=$r['Form']['id'];?>").click(function () {
        $("#shareform<?=$r['Form']['id'];?>").toggle("show");
    });
<?php endforeach; ?>

Currently, I have this hide and toggle function for each Form in the $Forms array. I want these functions to be enabled for the forms in the $formsShared array also.

If I add another for loop for $formsShared , like this:

<?php foreach ($formsShared as $r): ?>
    $("#shareform<?=$r['Form']['id'];?>").hide();
    $(".Share<?=$r['Form']['id'];?>").click(function () {
        $("#shareform<?=$r['Form']['id'];?>").toggle("show");
    });//.Share click
<?php endforeach; ?>

I achieve what I want, but it seems to be a repetition of the same code.

Is there any way in cakePHP to loop through two arrays in a single foreach loop?

Solution: array_merge() only accepts parameters of type array. So use typecasting to merge other types.

  <?php foreach (array_merge((array)$Forms,(array)$formsShared) as $r): ?>

    $("#shareform<?=$r['Form']['id'];?>").hide();
     $(".Share<?=$r['Form']['id'];?>").click(function () {
        $("#shareform<?=$r['Form']['id'];?>").toggle("show");
    });//.Share click

  <?php endforeach;?>

Answer 1

It sounds like you don't want to loop over two lists at the same time . You want to loop over two lists separately, executing the same code for each element in either list. So why not concatenate the lists:

foreach (array_merge($Forms, $FormsShared) as $r)
    // do stuff

Answer 2

Instead of repeating so much JavaScript, you could make use of jQuery's selectors . Instead of using unique class es on your form elements (or in addition to), use a generic class name like form-field . Then you can add a click event to all of them at the same time:

$('.form-field').click(function () {
    // Without knowing your HTML structure, I can't make this accurate, but maybe
    // this will work, assuming #shareform... is the parent <form> element:
    $(this).closest('form').toggle('show');
});

This JavaScript would only need to be output once and would apply to all elements with class="form-field" . Note that you can have multiple classes on an element too: class="Share123 form-field" gives the element class Share123 and class form-field .

Answer 3

You can do that using for() loop, though it looks pretty nasty:

<?php

$arr1 = array('first 1', 'first 2', 'first 3');
$arr2 = array('second 1', 'second 2', 'second 3');

for($c1 = 0, $c2 = 0; $c1 < count($arr1), $c2 < count($arr2); $c1++, $c2++ )
{
    echo $arr1[$c1] . '<br />';
    echo $arr2[$c2] . '<br />';
}

?>

That's just PHP example, but you know how to edit it ;)

Edit:

Karl is right, duplicating indexes is not so great, so another solution, though you have to use @ to avoid warnings:

<?php

$arr1 = array('first 1', 'first 2', 'first 3', 'first 4');
$arr2 = array('second 1', 'second 2', 'second 3');

for($c = 0; $c < max( count($arr1), count($arr2) ); $c++ )
{
    echo @$arr1[$c] . '<br />';
    echo @$arr2[$c] . '<br />';
}

?>