Php合并相同的名称值

Question

我把这个作为输出

Department  Employee Name   Employee Code   
Sales   Ganeshmurthy    264 
Sales   AZIM ahmed KHAN 2012    
Sales   DHANABAL    231 04-03-2019  
Service RAVIKIRAN   2003    04-03
Manufacturing   RAJAVARMAN  2073    04-03-2019  09:20   1h 10m
Manufacturing   MEGALA  332 04-03-2019  08:14   0h 4m
Manufacturing   SABARI  333 04-03-2019  08:13   0h 3m
Manufacturing   Sasi    1001    04-03-2019  08:16   0h 6m
Manufacturing   FRANCIES XAVIER 136 04-03-2019  08:13   0h 3m
Manufacturing   MANIKANDAN.V    151 04-03-2019  08:15   0h 5m

我得到这个td格式，我需要合并的副本为一个？我有一个由多个部门组成的数组，我需要将相同的部门名称合并到一个数组中。例如，如果有department1、department1、department1重复n次。我需要将它合并到一个数组中，作为department1。我怎样才能做到这一点?我需要在表- td中显示它，这样各部门就不会重复了。

[0] => Array
        (
            [department_id] => 15
            [department_name] => R&D
            [name] => SUBRAMANIAN
            [code] => 101
            [clockin] => 
            [late] => 
            [date] => 1970-01-01
        )

    [1] => Array
        (
            [department_id] => 3
            [department_name] => Manufacturing
            [name] => RAVI
            [code] => 103
            [clockin] => 
            [late] => 
            [date] => 1970-01-01
        )

    [2] => Array
        (
            [department_id] => 3
            [department_name] => Manufacturing
            [name] => AMULRAJAN.A
            [code] => 104
            [clockin] => 
            [late] => 
            [date] => 1970-01-01
        )

    [3] => Array
        (
            [department_id] => 3
            [department_name] => Manufacturing
            [name] => KANAKARAJ.R
            [code] => 105
            [clockin] => 
            [late] => 
            [date] => 1970-01-01
        )

    [4] => Array
        (
            [department_id] => 3
            [department_name] => Manufacturing
            [name] => ELAYARAJA
            [code] => 106
            [clockin] => 
            [late] => 
            [date] => 1970-01-01
        )

预期结果如下所示

[1] => Array
    (

        [department_name] => Manufacturing

    )

[2] => Array
    (

        [department_name] => R & D

    )

以下是我曾经尝试过的：

foreach($attendance as $dr){
           $alldep = $dr->department_name;
       echo $alldep;
}

请注意，我以数组格式获取所有数据，而不是作为对象

Answer 1

你应该像这样试试

$uniqArr= array();
foreach ($attendance as $dr) {
    $uniqArr[] = $dr->department_name;;
}
$uniqueRes = array_unique($uniqArr);

Answer 2

接受一个新的数组$departments

在给定数组上循环。

将department附加到$departments。

使用department作为密钥。

因为PHP数组总是有唯一的键，所以您只会得到唯一的部门。

工作代码：

<?php
$arr = [
[
'department_id' => 15 ,'department_name' => 'R&D' ,'name' => 'SUBRAMANIAN' ,'code' => 101
 ,'clockin' => '' ,'late' => '' ,'date' => '1970-01-01'
],
[
 'department_id' => 3 ,'department_name' => 'Manufacturing' ,'name' => 'RAVI' ,'code' => 103
 ,'clockin' => '' ,'late' => '' ,'date' => '1970-01-01'
],
[
 'department_id' => 3 ,'department_name' => 'Manufacturing' ,'name' => 'AMULRAJAN.A' ,'code' => 104 
 ,'clockin' => '' ,'late' =>'' ,'date' => '1970-01-01'
],
[
 'department_id' => 3 ,'department_name' => 'Manufacturing' ,'name' => 'KANAKARAJ.R' ,'code' => 105
 ,'clockin' => '' ,'late' => '' ,'date' => '1970-01-01'
],
[
 'department_id' => 3 ,'department_name' => 'Manufacturing' ,'name' => 'ELAYARAJA' ,'code' => 106
 ,'clockin' => '' ,'late' => ''
]];

$departments = '';
if (! empty($arr)) {
    foreach ($arr as $elem) {
 $departments[$elem['department_name']] = $elem['department_name'];
    }
}
echo '<pre>';print_r($departments);echo '</pre>';

输出：

Array
(
    [R&D] => R&D
    [Manufacturing] => Manufacturing
)

Answer 3

为所需的输出尝试以下代码

$arr = array_unique(array_column($attendance, 'department_name'));
foreach($arr AS $a)
{
    $newarr[]['department_name'] = $a;
}
echo "<pre>"; print_r($newarr);