Get a string from an array inside an array

Question

I have a function that returns an array from the database from which I have to display some data into the view.

If I do

print "<pre>";
print_r($names);
exit();

on the variable that stores the data it returns this:

Array
(
[0] => Array
        (
            [nume] => Eugen
        )
[1] => Array
        (
            [nume] => Roxanescu
        )
[2] => Array
        (
            [nume] => Georgiana
        )
[3] => Array
        (
            [nume] => Andrei
        )
)

I can't change the function I talk about in any way, I need to store each name in a different variable. At the end it should look like this:

name1 = "Eugen"
name2 = "Roxanescu"
name3 = "Georgiana"
name4 = "Andrei"

Thank you very much!

So this array is a result of a database query correct? how are you making that query? can you provide that code? also, can you provide the structure of your db table? — CodeGodie
– CodeGodie, Commented Aug 12, 2015 at 14:29
Why would you want to store the names in different variables? I think you are trying work harder then you should. consider iterating through your results array to access the names. — CodeGodie
– CodeGodie, Commented Aug 12, 2015 at 14:31
The query looks like this: "SELECT nume FROM rummy_users LEFT JOIN rummy_games ON rummy_users.id_user=rummy_games.id_user WHERE rummy_games.id_joc=".$id_joc.". — Vlad Eugen Nitu
– Vlad Eugen Nitu, Commented Aug 12, 2015 at 14:34
I only have to display 4 each time so it is a lot simpler for me as I am a beginner. Any tips are welcome but I'm not sure I'm able to make an iteration yet. — Vlad Eugen Nitu
– Vlad Eugen Nitu, Commented Aug 12, 2015 at 14:35

bicho · Accepted Answer · 2015-08-12 14:36:12Z

1

You could assign a variable the data of your array;

<?php
    $name1=$names[0];
    $name2=$names[1];
    $name3=$names[2];
    $name4=$names[3];

    echo $name1." ".$name2." ".$name3." ".$name4;
?>

answered Aug 12, 2015 at 14:36

bicho

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1 Comment

Vlad Eugen Nitu Over a year ago

Haha thanks! I have never thought of this way but now that I see it I realize I should've known it. Thanks!

acupofjose · Accepted Answer · 2015-08-12 14:36:43Z

1

Note entirely sure why you want to do this... But here's a function that should do what you're looking for.

This answer uses dynamic variable assignment, creating a new variable with a string ( $"name1", $"name2", etc ).

for($i = 0; $i < count($names); $i++) {
  $var_name = "name".$i;
  $$var_name = $names[$i]['nume'];
}

var_dump(get_defined_vars());

$name1, $name2, $name3 ... < count($names)

answered Aug 12, 2015 at 14:36

acupofjose

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2 Comments

Vlad Eugen Nitu Over a year ago

The purpose is an application to keep the score record while you play rummy (if you know the game, something like whist or rentz or something simmilar). You have to select the players from a dropdown list and in this application there are always 4 players so I have no reason to complicate the code. Still, thank you very much!

Lorenzo Marcon Over a year ago

@VladEugenNitu, this is the right answer for what you've asked and the right way to code. Learn how to make code reusable and parametric as soon as possible, you won't regret it!

Inceddy · Accepted Answer · 2015-08-12 14:35:45Z

0

$result = array();
array_walk($names, function ($value, $key) use (&$result) {
    $result[ 'name' . $key ] = $value['name'];
});
var_dump($result);

answered Aug 12, 2015 at 14:35

Inceddy

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1 Comment

Henrique Barcelos Over a year ago

Even this is a simple question, please add some contextual comments in order to help others that have beginner skills.

alanmm · Accepted Answer · 2015-08-12 14:37:10Z

0

I suggest you use an array and reference to each element of it as a variable, if it's really necessary.

$result = array();
foreach ($names as $n) {
    array_push($result, $n['nume']);
    echo "name".array_search($n['nume'], $result)." = ".$n['nume'];
}

answered Aug 12, 2015 at 14:37

alanmm

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