referencing php variables inside of an array

Question

I have a small problem that I cannot seem to wrap my head around and hoping someone here might be able to find the problem. I have an variable named $data which is an array of string values and when I use the following code everything works fine.

$data = array(1,2,3,4,5);

$users = implode("," , $data);

echo ($users);

This will produce something like 1,2,3,4,5 which is what I expected, but if I try to follow the same logic on the following piece of code then the result is an empty string.

$data = array(1,2,3,4,5);

$users = implode("," , $data);

$MyArray = array(
    $user_ids => array($users)
)

My question then is how do I need to reference the $users variable in this array so it will produce the results I need. (ie 1,2,3,4,5 ...)

Thx, Amy

prodigitalson · Accepted Answer · 2011-08-14 06:32:00Z

2

Not sure exactly what you want the key to be but right now you are using a variable as a key:

$users = implode("," , $data);
$user_ids = 'users_id_keyname';

$MyArray = array(
  $user_ids => $users 
);

echo $MyArray['users_id_keyname']; // outputs 1,2,3,4,5
echo $MyArray[$user_ids]; // outputs 1,2,3,4,5

You probably want to just use the string as the key:

$users = implode("," , $data);

$MyArray = array(
  'user_ids' => $users 
);

echo $MyArray['user_ids']; // outputs 1,2,3,4,5

answered Aug 14, 2011 at 6:32

prodigitalson

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14 Comments

Amy Over a year ago

Sorry, I don't understand what you mean by Key and all that but I tried your 2nd example and I received a parse error saying: Parse error: syntax error, unexpected T_VARIABLE, expecting '(' in C:\wamp\www\forum\practicepm.php on line 58

prodigitalson Over a year ago

the you have an error somewhere thats not directly part of the code i gave you. You do need to understand what a key is when dealing with arrays, though... its kind of essential.

Amy Over a year ago

I updated the code in the original post. I forgot to add the word array in front of ($users) so the old code just said user_ids => ($users) but it now says user_ids => array($users) that may clear up some of the confusion sorry about that.

prodigitalson Over a year ago

@Amy: if you do it that you will actually have the following if you were to var_dump $MyArray: Array('user_ids' => Array( 0 => "1,2,3,4,5")) is that what you actually want?

Amy Over a year ago

As long as it works I don't care but all I know is I need the word array in front of the ($users) part or I get tons of error messages about missing this and that.

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bozdoz · Accepted Answer · 2011-08-14 07:05:19Z

0

You don't need to make your data array into a string and convert it back into an array. You could just insert the array into another array and reference it that way. For example:

$data = array(1,2,3,4,5);

$MyArray[] = $data;

To recall this data:

foreach($MyArray as $array){
echo implode(',',$array);
}

edited Aug 14, 2011 at 7:05

answered Aug 14, 2011 at 6:50

bozdoz

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3 Comments

Amy Over a year ago

I need the word array infront of the word users because if I omit it then I get alot of errors.

bozdoz Over a year ago

I rewrote my answer, and tested this, and it works. Hope it works for you. Any questions, let me know.

Amy Over a year ago

You all decide who wins the pot, this is getting to confusing to me. I already wasted 2 days on this I think I will let someone else work on this project who knows what they are doing.

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