dynamically creating array in 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);
  }
}


?>

try echo $arrayName and echo $$arrayName and see what you get. I bet it's something useless and null like the error message says. — Scott M.
– Scott M., Commented Mar 12, 2012 at 19:43

Aleks G · Accepted Answer · 2012-03-12 19:45:19Z

6

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);
  }
}

answered Mar 12, 2012 at 19:45

Aleks G

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3 Comments

Kaii Over a year ago

i guess the OP constructed this example for us and in reality, he does not have the name of the array (except for the string).

Aleks G Over a year ago

@Kaii possibly. I just hate to see an over-engineered solution. When a simple solution exists, why not just use it?

mickmackusa Over a year ago

Why declare $arrayName at all?

Kaii · Accepted Answer · 2012-03-12 20:09:44Z

3

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

edited Mar 12, 2012 at 20:09

Kaii

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answered Mar 12, 2012 at 19:58

fvuong

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Comments

Starx · Accepted Answer · 2012-03-12 19:59:21Z

1

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;
  }

}

edited Mar 12, 2012 at 19:59

answered Mar 12, 2012 at 19:43

Starx

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1 Comment

Starx Over a year ago

@vaichidrewar, Misunderstood the question, BUT NOW fixed with tested solution

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