PHP can't get output from array

Question

I posted a question last night that turned out to not be the problem. Digging around, I have discovered that the below code is giving me a headche. I had this working, but now for some reason I get no output. When I var_dump the function that gives me the $finishmins value, it outputs everything correctly until the point where it has to search the array (as below). After this it shows NULL. I was originally using strpos to find out if it started with a zero, then stripped said zero to match to the array, but when it stopped working, I tried the below approach to reduce code.

The point of the code is to convert minutes in time to minutes in decimal notation. I.e. 1 minute = 02, thus 12:01 = 12.02.

$finishmins = '01';
$finishmins = $minarray[$finishmins];

$minarray = array(
00 => '00',
01 => '02',
02 => '03',
03 => '05',
04 => '07',
05 => '08',
06 => '10',
07 => '12',
08 => '13',
09 => '15',
10 => '17',
'18',
// Array continues to 59 => '98'
);

echo $finishmins;

I have pasted the complete code here: http://codepad.org/EUW3n7AB and still can't seem to find the problem.

Just saw this question, and I'm wondering if this is part of my problem: stackoverflow.com/questions/4750781/… — BigJazzz
– BigJazzz, Commented Jun 23, 2014 at 1:30
@Dr.Molle what would be my course of action? Revert back to stripping out the zeros using strpos and repost the code? — BigJazzz
– BigJazzz, Commented Jun 23, 2014 at 1:32

Ja͢ck · Accepted Answer · 2014-06-23 01:39:00Z

1

There are two issues here:

Array indices behave differently between strings and numbers,
Variable scope of $minarray.

$arr[01] and $arr['01'] are not the same thing, so you should be more explicit; in your case you can just leave the array numerically indexed, i.e.:

$minarray = array('00', '02', '03', '05', ...);

Then, you use an (int) cast on the given minutes:

$finishmins = $minarray[(int)$finishmins];

You can solve the second issue by passing the array as a function argument:

function finishtime($minarray, $finish)

Then calling it like so:

echo finishtime($minarray, '12:01');

answered Jun 23, 2014 at 1:39

Ja͢ck

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5 Comments

BigJazzz Over a year ago

So the reason I'm having issues is because I haven't passed the array into the function? Most of what you have said makes sense, just not sure if that's what you're meaning in regards to the array.

Ja͢ck Over a year ago

@BigJazzz Both issues I've mentioned cause your code to not give the expected result of "12.02".

BigJazzz Over a year ago

Sorry, poor choice of words. That probably should have said "one of the reasons".

Ja͢ck Over a year ago

Yes indeed; this is because $minarray isn't defined inside the scope of finishtime().

BigJazzz Over a year ago

Ok, I have it working here: codepad.org/QRqSvw2K but when I go back to my code it's still outputting 12.00. I have been var_dumping pretty much every step of the way, so I'll try building it from scratch again and see what happens. Cheers for this!

Lock · Accepted Answer · 2014-06-23 01:32:33Z

0

You have to use the keyword global in your function when referring to the $minarray variable:

function finishtime($finish) {
    global $minarray;
    $finishx = explode(':', $finish);
    $finishhours = $finishx[0];
    $finishmins = $finishx[1];
    $finishmins;
    var_dump($finishmins);
    $finishmins = $minarray[$finishmins];
    var_dump($finishmins);

    $finishtime = $finishhours . '.' . $finishmins;

    return $finishtime;
}

answered Jun 23, 2014 at 1:32

Lock

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1 Comment

BigJazzz Over a year ago

Thanks Lock. Was certainly part of the problem.

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