Tell me more ×
Stack Overflow is a question and answer site for professional and enthusiast programmers. It's 100% free, no registration required.
up vote 2 down vote favorite

I combined two arrays to create the following array, named $group_wages_array:

Array ( [1] => 500 [4] => 44 [6] => 80 [3] => 11.25 )

I am trying to test if the array key matches X, set a variable as it's value. Here's what I have:

NOTE: This whole thing is executed in a while loop, so the value of $thegroup['group_id'] will change. I've set it's value as "6" for this example.

$thegroup['group_id'] = "6" // This particular group (for simplicity)

if (array_key_exists($thegroup['group_id'], $group_wages_array)) {

    $this_wages = // Need this to be 80... how do I do it?

}

So, how do I get $this_wages to equal the key value?

share|improve this question
That should work. If not, try casting $thegroup['group_id'] to an int. – BoltClock♦ Oct 12 '11 at 21:38
You can use anything you want as an array key, as long as PHP can typecast it to a string or an integer. This includes embedding one array reference inside another: $outer[$inner[1]], so $group_wages_array[$thegroup['group_id']]. – Marc B Oct 12 '11 at 21:46
+1 for going into detail about what your variables are and how they're being used. I wish more people would do that instead of just dumping code without sufficient context. – Herbert Oct 12 '11 at 22:43

2 Answers

up vote 5 down vote accepted

You just use the key from the array to get it:

$thegroup['group_id'] = "6" // This particular group (for simplicity)

if (array_key_exists($thegroup['group_id'], $group_wages_array)) {
    $this_wages = $group_wages_array[$thegroup['group_id']];
}

Also, the array keys are not 0,1,2,etc because you explicitly set them in Array ( [1] => 500 [4] => 44 [6] => 80 [3] => 11.25 )

share|improve this answer
OK thanks for clarifying. It's working per your example. – Oseer Oct 12 '11 at 21:44
up vote 1 down vote

You are trying to do:

$group_wages_array[6];

And

$thegroup['group_id'] = 6;

You can substitute this in as the key.

if (array_key_exists($thegroup['group_id'], $group_wages_array)) {
    $this_wages = $group_wages_array[$thegroup['group_id']];   
}
share|improve this answer
OK great, it worked when I used double brackets $this_wages = $combined_group_wages[$thegroup['group_id']]; – Oseer Oct 12 '11 at 21:44
You are correct, that was a syntax error, it should be correct now. Thanks for pointing that out. – Gazler Oct 12 '11 at 21:46

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.