merge two array in one array - php

Question

I have two array, $result and $social_result. I have to merge both tables. social_icons_id of $result matches id of $social_result. If match then show link of $result array otherwise blank.
I have an array
Array
(
[0] => Array
(
[social_icons_id] => 14
[link] => www.instagram.com
[edittemplate_id] => 218
[name] => Email
[image] => email.png
)
[1] => Array
(
[social_icons_id] => 16
[link] => www.instagram.com
[edittemplate_id] => 218
[name] => Blogger
[image] => blogger.png
)
)
Another is:
Array
(
[0] => Array
(
[id] => 13
[name] => Address
[image] => address.png
)
[1] => Array
(
[id] => 14
[name] => Email
[image] => email.png
)
[2] => Array
(
[id] => 15
[name] => Fax
[image] => fax.png
)
[3] => Array
(
[id] => 16
[name] => Text
[image] => text.png
)
[4] => Array
(
[id] => 17
[name] => Website
[image] => Website.png
)
)
Now I have to merge both table in one table like:
Array
(
[0] =>
[1] => www.instagram.com
[2] =>
[3] =>
[4] =>
[5] => www.instagram.com
[6] =>
[7] =>
[8] =>
[9] =>
[10] =>
[11] =>
[12] =>
[13] =>
[14] =>
[15] =>
[16] =>
)
id of both tables matches and make one table.
I tried-
$result = $obj->select_social_ids($id); // for first table
$social_result = $obj->show_social_icons(); // for second table
for($j=0;$j<count($social_result);$j++)
{
if(in_array($social_result[$j]['id'], $result)) { // search value in the array
$link[] = $result[$j]['link'];
}
else
{
$link[] = '';
}
}
But not working.

Answer 1

Depending on where you're getting this information from (e.g. a database table), doing this operation in SQL may make more sense.
That said, given the data and code you've provided, I think your in_array() check is incorrect, as it will only check the top level of $result. The 'social_icon_id' value that you seem to want to compare to $social_results[$j]['id'] is contained in a nested array within $result.
You could do something like this:
<?php
$results = $obj->select_social_ids($id);
$results_ids = array_map(
function ($result) { return $result['id']; },
$results
);
$results = array_combine($results_ids, $results);
$social_results = $obj->show_social_icons();
foreach ($social_results as $social_result) {
$id = $social_result['id'];
if (isset($results[$id])) {
$link[] = $results[$id]['link'];
}
else
{
$link[] = '';
}
}

Answer 2

If I understand your question correctly, you want to loop thru $social_result and compare ID to those keys in $result, maybe something like this will work.
$link = array();
foreach($social_result as $social){
$key = array_search($social['id'], array_column($result, 'social_icons_id'));
if($key != ''){
$link[] = $result[$key]['link'];
}else{
$link[] = '';
}
}
I tested this code and it works to do what I beliueve you are trying to accomplish
$a = array('social_icons_id' => '14','link' => 'www.instagram14.com','edittemplate_id' => '218','name' => 'Email','image' => 'email.png');
$b = array('social_icons_id' => '16','link' => 'www.instagram16.com','edittemplate_id' => '218','name' => 'Blogger','image' => 'blogger.png');
$result = array($a,$b);
$social_result = array(array('id'=>'14','name'=>'address0','image'=>'adress.png'),array('id'=>'15','name'=>'address1','image'=>'adress.png'), array('id'=>'16','name'=>'address2','image'=>'adress.png'),array('id'=>'17','name'=>'address3','image'=>'adress.png'),array('id'=>'18','name'=>'address4','image'=>'adress.png'),array('id'=>'19','name'=>'address5','image'=>'adress.png'));
$link = array();
foreach($social_result as $social){
$key = array_search($social['id'], array_column($result, 'social_icons_id'));
echo "<p> k ".$key;
if($key != ''){
$link[] = $result[$key]['link'];
}else{
$link[] = '';
}
}
print_r($link);

Answer 3

Simply you can create a simple left join query to generate a flat array containing social image links.
select social_image.link
from social_icons
left join social_image
on social_icons.id = social_image.social_icons_id
order by social_icons.id
But be carefully with array size limitation on php, therefore that needs a proper result limitation.
select social_image.link
from social_icons
left join social_image
on social_icons.id = social_image.social_icons_id
order by social_icons.id
limit 1000
Hope this helps.