Referencing values of an array inside of an array

[Ethan27]

If I wanted to only reference the one or more of the values of an array inside another array how would I do this…?
I have got the idea of referencing both I think and also the first one but the second Im not sure about. Would it just be a question of leaving the first square brackets with only speechmarks, no value inside it…

Here is some example code in case my wording is a bit confusing.

$array1 = array("cat","fox", "dog", array("horse",shoes","food"));

echo $array1[1][2];

If I wanted to just echo “food”, how would this be done…?

Thanks guys

[MrSoundless]

I don’t get your whole story but to answer your last question: echo $array1[3][2];

If that doesn’t help, try to explain what you are trying to do.

Good luck

[Ethan27]

Thanks for your reply

I want to reference the values of the array inside the array

Therefore in this case I only want to echo “food”, nothing else.

I noticed you put ‘3’ in the first reference. Is this because there is no ‘3’ value and this is a simple workaround…?

[MrSoundless]

An array starts counting by 0. So by using 3 in the first reference, you refer to the 4th element in the first array. Now since the 4th element in the array is an array you can add a second reference, since food is the 3th element in that array, you use 2 as the 2nd reference.

[Ethan27]

So since the third element is an array you write “3” then reference the value right…?

The value in the first set of brackets, to put in a basic way doesnt have anything to do with the values in the first array, excluding the fact you’re referncing the second array…?

Thanks man

[richtestani]

You should loop though them.

foreach($array as $val) {



if(is_array($val) {



print_r($val);



}

}

You could even loop back though it by doing this in a function:

function my_array($array) {



foreach($array as $val) {

if(is_array($val) {

my_array($array);

} else {

echo '
'.$val.'';

}

}





}my_array($array);

sorry for formatting, not sure how to get it pretty

[Author]

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