Incrementing the array value within an array using PHP

Question

So my end result will be this (the end result will have 48 entries):

$theArray=array(
$theArray1,
$theArray2,
$theArray3,
$theArray4,
$theArray5,
$theArray6
);

I have tried a few things but I think this is the closest, but I'm still not there yet, any help appreciated.

$i = 0;
while ($i <= 48){
  $theArray[]=${"theArray".$i.","}
  $i++
 }
$theArray[]=${"theArray".$i};

Why don't you have $theArray1, $theArray2, etc defined as array values in the first place, so no numbers or hacky conversions necessary? Where are these values coming from? — Mark Baker
– Mark Baker, Commented May 17, 2016 at 9:23
Are you looking for this, Is your desire output like: https://3v4l.org/5khJe — Murad Hasan
– Murad Hasan, Commented May 17, 2016 at 9:44
Yes, but rather the entries in the arrays rather than a string — c7borg
– c7borg, Commented May 17, 2016 at 11:02
Perhaps I should add $theArray1, $theArray2.. etc are arrays — c7borg
– c7borg, Commented May 17, 2016 at 11:24

Dhara Parmar · Accepted Answer · 2016-05-17 09:26:08Z

2

You missed ; at the end of line:

$i = 0;
while ($i <= 48){
  $theArray[]=${"theArray".$i.","}; // missed ; here
  $i++; // missed ; here
 }
$theArray[]=${"theArray".$i};

answered May 17, 2016 at 9:26

Dhara Parmar

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1 Comment

c7borg Over a year ago

Sorry that was in haste, but it still doesn't work how I intended it.

user3416687 · Accepted Answer · 2016-05-17 09:36:02Z

0

You have to use the function compact for create an array containing variables. Try to read the documentation http://php.net/manual/en/function.compact.php

And here are some example http://www.w3schools.com/php/func_array_compact.asp

answered May 17, 2016 at 9:36

user3416687

1 Comment

c7borg Over a year ago

Yes perhaps this is the answer, but I'm struggling to incorporate it into a loop, I'll keep trying though.

Sachin I · Accepted Answer · 2016-05-17 09:42:02Z

0

Keep it simple as below:

<?php
$i = 0;
$theArray = array();
while ($i <= 48){
  array_push($theArray, '$theArray'.$i);
  $i++;
 }
echo '$arr = array('.implode(',', $theArray).');';
?>

Just run this on your end. Cheers!

answered May 17, 2016 at 9:42

Sachin I

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1 Comment

c7borg Over a year ago

Thanks I think this is nearly it, but rather than it output each array as a string I need it to be the actual array. The above outputs the following: Array ( [0] => $theArray0 [1] => $theArray1 [2] => $theArray2 [3] => $theArray3 ) rather than the values within $theArray0 etc..

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