Add item to array of arrays based on id

Question

I have an array of arrays set up like so. There are a total of 10 arrays but I will just display the first 2. The second column has a unique id of between 1-10 (each only used once).

Array
(
   [0] => Array
        (
            [0] => User1
            [1] => 5
        )

    [1] => Array
        (
            [0] => User2
            [1] => 3
        )
)

I have another array of arrays:

Array
(
    [0] => Array
        (
            [0] => 3
            [1] => 10.00
        )

    [1] => Array
        (
            [0] => 5
            [1] => 47.00
        )
)

where the first column is the id and the second column is the value I want to add to the first array.

Each id (1-10) is only used once. How would I go about adding the second column from Array#2 to Array#1 matching the ID#?

Axel A. García · Accepted Answer · 2014-10-03 20:23:09Z

2

There are tons of ways to do this :) This is one of them, optimizing the second array for search and walking the first one:

Live example

<?
$first_array[0][] = 'User1';
$first_array[0][] = 5;
$first_array[1][] = 'User2';
$first_array[1][] = 3;

$secnd_array[0][] = 3;
$secnd_array[0][] = 10.00;
$secnd_array[1][] = 5;
$secnd_array[1][] = 47.00;

// Make the user_id the key of the array
foreach ($secnd_array as $sca) {
    $searchable_second_array[ $sca[0] ] = $sca[1];
}
// Modify the original array
array_walk($first_array, function(&$a) use ($searchable_second_array) {
    // Here we find the second element of the first array in the modified second array :p
    $a[] = $searchable_second_array[ $a[1] ];
});

// print_r($first_array);

edited Oct 3, 2014 at 20:23

answered Oct 3, 2014 at 20:12

Axel A. García

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Comments

helllomatt · Accepted Answer · 2014-10-03 20:08:31Z

1

Assuming that 0 will always be the key of the array and 1 will always be the value you'd like to add, a simple foreach loop is all you need.

Where $initial is the first array you provided and $add is the second:

<?php

$initial = array(array("User1", 5),
    array("User2", 3));

$add = array(
    array(0, 10.00),
    array(1, 47.00));

foreach ($add as $item) {
    if (isset($initial[$item[0]])) {
        $initial[$item[0]][] = $item[1];
    }
}

printf("<pre>%s</pre>", print_r($arr1[$item[0]], true));

answered Oct 3, 2014 at 20:08

helllomatt

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1 Comment

Axel A. García Over a year ago

Shouldn't be: printf("<pre>%s</pre>", print_r($initial[$item[0]], true));? And this only return one element

ThatMSG · Accepted Answer · 2014-10-03 20:17:30Z

I don't know if I got you right, but I've come up with a solution XD

<?php
$array_1 = array(
    0 => array(
      0 => 'ID1',
      1 => 5
    ),
    1 => array(
      0 => 'ID2',
      1 => 3
    )
);

$array_2 = array(
    0 => array(
      0 => 3,
      1 => 10.00
    ),
    1 => array(
      0 => 5,
      1 => 47.00
    )
);


foreach($array_1 as $key_1 => $arr_1){
  foreach($array_2 as $key_2 => $arr_2){
    if($arr_2[0] == $arr_1[1]){
      $array_1[$key_1][2] = $arr_2[1];
    }
  }
}

var_dump($array_1);

?>

Demo: https://eval.in/201648

The short version would look like this:

<?php
$array_1 = array(array('ID1',5),array('ID2',3));

$array_2 = array(array(3,10.00),array(5,47.00));


foreach($array_1 as $key => $arr_1){
  foreach($array_2 as$arr_2){
    if($arr_2[0] == $arr_1[1]){
      $array_1[$key][2] = $arr_2[1];
    }
  }
}

var_dump($array_1);

?>

Demo: https://eval.in/201649

Hope that helps :)

malifa · Accepted Answer · 2014-10-03 20:17:32Z

1

A quick and dirty way just to show you one of the more self-explaining ways to do it :)

$users = array(
    0 => array(
        0 => 'User1',
        1 => 123
    ),
    1 => array(
        0 => 'User2',
        1 => 456
    )
);

$items = array(
    0 => array(
        0 => 123,
        1 => 'Stuff 1'
    ),
    1 => array(
        0 => 456,
        1 => 'Stuff 2'
    )
);

foreach($items as $item){
    foreach($users as $key => $user){
        if($item[0] == $user[1])
            array_push($users[$key], $item[1]);
    }
}

answered Oct 3, 2014 at 20:17

malifa

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