merge two array in one array

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.

have you considered array_merge?

Mike Volmar
– Mike Volmar

2020-05-11 12:18:07 +00:00
Commented May 11, 2020 at 12:18 — Mike Volmar
– Mike Volmar, Commented May 11, 2020 at 12:18
How to to that?

Vinod
– Vinod

2020-05-11 12:20:42 +00:00
Commented May 11, 2020 at 12:20 — Vinod
– Vinod, Commented May 11, 2020 at 12:20
php.net/manual/en/function.array-merge.php

Mike Volmar
– Mike Volmar

2020-05-11 12:21:52 +00:00
Commented May 11, 2020 at 12:21 — Mike Volmar
– Mike Volmar, Commented May 11, 2020 at 12:21
I don't think array_merge should work in my condition.

Vinod
– Vinod

2020-05-11 12:22:09 +00:00
Commented May 11, 2020 at 12:22 — Vinod
– Vinod, Commented May 11, 2020 at 12:22
Logic here is unclear to me.

u_mulder
– u_mulder

2020-05-11 12:24:26 +00:00
Commented May 11, 2020 at 12:24 — u_mulder
– u_mulder, Commented May 11, 2020 at 12:24

Matthew Turland · Accepted Answer · 2020-05-11 12:23:30Z

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[] = '';
    }
}

answered May 11, 2020 at 12:23

Matthew Turland

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3 Comments

Vinod Over a year ago

I have to use social_icons_id from first array and match this, with id of second array. If match then print something otherwise blank. But sequence should be the second array

Matthew Turland Over a year ago

I read your original post, and I can't see a reason why the code I've provided wouldn't do what you're asking. What is different between the output you're wanting to see and the output of the code I've given you? You'll have to tell me more than that it's "not working" for me to help you further.

Vinod Over a year ago

$link[] = $result[$j]['link']; Here $j is not found in first array. So only second row of first array is not printing.

Mike Volmar · Accepted Answer · 2020-05-11 13:05:53Z

0

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);

edited May 11, 2020 at 13:05

answered May 11, 2020 at 12:30

Mike Volmar

2,1031 gold badge25 silver badges39 bronze badges

2 Comments

Vinod Over a year ago

I have to use social_icons_id from first array and match this, with id of second array. If match then print something otherwise blank. But sequence should be the second array

Vinod Over a year ago

I am stucked whole day. Can we connect with teamviewer please?

OO7 · Accepted Answer · 2020-05-11 13:22:31Z

0

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.

answered May 11, 2020 at 13:22

OO7

6904 silver badges10 bronze badges

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