dynamically creating array in php - php

Question

I am trying to create arrays dynamically and then populate them by constructing array Names using variable but I am getting the following warnings
Warning: in_array() expects parameter 2 to be array, null given
Warning: array_push() expects parameter 1 to be array, null given
For single array this method worked but for array of arrays this is not working. How should this be done?
<?php
for ($i = 1; $i <= 23; ++$i)
{
$word_list[$i] = array("1");
}
for ($i = 1; $i <= 23; ++$i)
{
$word = "abc";
$arrayName = "word_list[" . $i . "]";
if(!in_array($word, ${$arrayName}))
{
array_push($$arrayName , $word);
}
}
?>

Answer 1

Why are even trying to put array name in a variable and then de-reference that name? Why not just do this:
for ($i = 1; $i <= 23; ++$i)
{
$word = "abc";
$arrayName = "word_list[" . $i . "]";
if(!in_array($word, $word_list[$i]))
{
array_push($word_list[$i] , $word);
}
}

Answer 2

You get the first warning because your $arrayName variable is not actually an array, you made it into a string.
So instead of:
$arrayName = "word_list[" . $i . "]";
You should have this:
$arrayName = $word_list[$i];
You get your second warning because your first parameter is not an array.
So instead of:
array_push($$arrayName , $word);
You should have this:
array_push($arrayName , $word);
If you make these changes you will get an array that looks like this in the end:
$wordlist = array( array("1", "abc"), array("1", "abc"), ... ); // repeated 23 times

Answer 3

And in the for loop, you are accessing the array the wrong way
Here is your corrected code
for ($i = 1; $i <= 23; ++$i)
{
$word = "abc";
$arrayName = $word_list[$i];
if(!in_array($word, $arrayName))
{
array_push($arrayName , $word);
$word_list[$i] = $arrayName;
}
}