我想根據索引合並兩個關聯數組

Question

我有兩個索引為 1 到 10 的數組，然后我必須根據索引將其合並。 所以只有一些員工必須添加薪水。

我試過使用array_merge()但輸出沒有合並。 我無法獲得正確的輸出。 輸出只是employeeSalary數組。 為什么會這樣？

$employee = array
(
0=>
    array("employee_id"=>1, "firstName"=>"Zahir", "lastName"=>"Alam", "Age"=>25, "Company"=>"Switchme", "Role"=>"Developer", "Department"=>"Tech"
        ,"Head"=>
            array("Id"=>3 , "Name"=>"Sourasis Roy")
    )
,
1=>
    array("employee_id"=>2, "firstName"=>"Amith", "lastName"=>"Manniken", "Age"=>25, "Company"=>"Switchme", "Role"=>"Developer", "Department"=>"Tech"
        ,"Head"=>
            array("Id"=>3 , "Name"=>"Sourasis Roy")
    )
,
2=>
    array("employee_id"=>3, "firstName"=>"Sourasis", "lastName"=>"Roy", "Age"=>28, "Company"=>"Switchme", "Role"=>"CTO")
,
3=>
    array("employee_id"=>4, "firstName"=>"Aditya", "lastName"=>"Mishra", "Age"=>29, "Company"=>"Switchme", "Department"=>"Tech", "Role"=>"CEO")
,
4=>
    array("employee_id"=>5, "firstName"=>"Priti", "lastName"=>"Lata", "Age"=>24, "Company"=>"Switchme", "Role"=>"HR")
,
5=>
    array("employee_id"=>6, "firstName"=>"Sumita", "lastName"=>"Nath", "Age"=>24, "Company"=>"Switchme", "Role"=>"HLA Head", "Department"=>"Crm")
,
6=>
    array("employee_id"=>7, "firstName"=>"Tarini", "lastName"=>"Khanna", "Age"=>22, "Company"=>"Switchme", "Role"=>"Content Writer")
,
7=>
    array("employee_id"=>8, "firstName"=>"Abhisek", "lastName"=>"Soni", "Age"=>23, "Company"=>"Switchme", "Role"=>"HLA", "Department"=>"Crm","Head"=>array("Id"=>5 , "Name"=>"Sumita Nath")
    )
,
8=>
    array("employee_id"=>9, "firstName"=>"Ankit", "lastName"=>"Pump", "Age"=>23, "Company"=>"Switchme", "Role"=>"HLA", "Department"=>"Crm"
        ,"Head"=>
            array("Id"=>5 , "Name"=>"Sumita Nath")
    )
,
9=>
    array("employee_id"=>10, "firstName"=>"Pogo", "lastName"=>"Laal", "Age"=>23, "Company"=>"Switchme", "Role"=>"Designer")
,
10=>
    array("employee_id"=>11, "firstName"=>"Sabina", "lastName"=>"Sekh", "Age"=>28, "Company"=>"Switchme", "Role"=>"HLA Head", "Department"=>"Crm")
,
11=>
    array("employee_id"=>12, "firstName"=>"Sanjay", "lastName"=>"Poudal", "Age"=>24, "Company"=>"Switchme", "Role"=>"HLA Head", "Department"=>"Crm"
        ,"Head"=>
            array("Id"=>10 , "Name"=>"Sabina Sekh")
    )
,
);

$employee_salary = array
(
7=>
    array("employee_id"=>7, "salary"=>"55,000"
    )
,
2=>
    array("employee_id"=>2, "salary"=>"60,000"
    )
,
9=>
    array("employee_id"=>9, "salary"=>"50,000"
    )
,
10=>
    array("employee_id"=>10, "salary"=>"30,000"
    )
,
);

$ar= array();
for($j=0;$j<count($employee);$j++)
{
    error_reporting(0);
    if($employee[$j]=$employee_salary[$j])
    {
        $ar=$employee[$j]+$employee_salary[$j];
        print_r($ar);
        echo "<br>";
    }
    error_reporting(0);
}

Answer 1

正如克里斯·懷特所說，您忘記再添加一個 = 用於比較。並使用isset()檢查是否在 employee_salary 的第二個數組中設置了數組索引。 比如if($employee[$j]==isset($employee_salary[$j]))

$employee = array
(
0=>
    array("employee_id"=>1, "firstName"=>"Zahir", "lastName"=>"Alam", "Age"=>25, "Company"=>"Switchme", "Role"=>"Developer", "Department"=>"Tech"
        ,"Head"=>
            array("Id"=>3 , "Name"=>"Sourasis Roy")
    )
,
1=>
    array("employee_id"=>2, "firstName"=>"Amith", "lastName"=>"Manniken", "Age"=>25, "Company"=>"Switchme", "Role"=>"Developer", "Department"=>"Tech"
        ,"Head"=>
            array("Id"=>3 , "Name"=>"Sourasis Roy")
    )
,
2=>
    array("employee_id"=>3, "firstName"=>"Sourasis", "lastName"=>"Roy", "Age"=>28, "Company"=>"Switchme", "Role"=>"CTO")
,
3=>
    array("employee_id"=>4, "firstName"=>"Aditya", "lastName"=>"Mishra", "Age"=>29, "Company"=>"Switchme", "Department"=>"Tech", "Role"=>"CEO")
,
4=>
    array("employee_id"=>5, "firstName"=>"Priti", "lastName"=>"Lata", "Age"=>24, "Company"=>"Switchme", "Role"=>"HR")
,
5=>
    array("employee_id"=>6, "firstName"=>"Sumita", "lastName"=>"Nath", "Age"=>24, "Company"=>"Switchme", "Role"=>"HLA Head", "Department"=>"Crm")
,
6=>
    array("employee_id"=>7, "firstName"=>"Tarini", "lastName"=>"Khanna", "Age"=>22, "Company"=>"Switchme", "Role"=>"Content Writer")
,
7=>
    array("employee_id"=>8, "firstName"=>"Abhisek", "lastName"=>"Soni", "Age"=>23, "Company"=>"Switchme", "Role"=>"HLA", "Department"=>"Crm","Head"=>array("Id"=>5 , "Name"=>"Sumita Nath")
    )
,
8=>
    array("employee_id"=>9, "firstName"=>"Ankit", "lastName"=>"Pump", "Age"=>23, "Company"=>"Switchme", "Role"=>"HLA", "Department"=>"Crm"
        ,"Head"=>
            array("Id"=>5 , "Name"=>"Sumita Nath")
    )
,
9=>
    array("employee_id"=>10, "firstName"=>"Pogo", "lastName"=>"Laal", "Age"=>23, "Company"=>"Switchme", "Role"=>"Designer")
,
10=>
    array("employee_id"=>11, "firstName"=>"Sabina", "lastName"=>"Sekh", "Age"=>28, "Company"=>"Switchme", "Role"=>"HLA Head", "Department"=>"Crm")
,
11=>
    array("employee_id"=>12, "firstName"=>"Sanjay", "lastName"=>"Poudal", "Age"=>24, "Company"=>"Switchme", "Role"=>"HLA Head", "Department"=>"Crm"
        ,"Head"=>
            array("Id"=>10 , "Name"=>"Sabina Sekh")
    )
,
);

$employee_salary = array
(
7=>
    array("employee_id"=>7, "salary"=>"55,000"
    )
,
2=>
    array("employee_id"=>2, "salary"=>"60,000"
    )
,
9=>
    array("employee_id"=>9, "salary"=>"50,000"
    )
,
10=>
    array("employee_id"=>10, "salary"=>"30,000"
    )
,
);

$ar= array();
for($j=0;$j<count($employee);$j++)
{

    error_reporting(0);
    if($employee[$j]==isset($employee_salary[$j]))
    {
        $ar=$employee[$j]+$employee_salary[$j];

        print_r($ar);
        echo "<br>";
    }
    error_reporting(0);
}

工作示例：http: //phpfiddle.org/main/code/dbi8-4krn

Answer 2

嘗試以下解決方案，使用問題中給出的數組$employee和$employee_salary的值。

$ar= array();

foreach($employee as $j => $v){
    $id = $v['employee_id'];
    foreach($employee_salary as $i => $u){
     $id_s = $u['employee_id'] ;

     if($id == $id_s){
       $ar = $employee[$j] + $employee_salary[$i];
       print_r($ar);
       echo "<br>";
     }
    }
}

Answer 3

我認為也會有一個簡單的解決方案。 因為您已經使用 id 鍵創建了 employee_salary 數組。 所以就會這樣。

Foreach($employee as $key=> $value)
{ 
     $employee[$key]['salary'] = $employee_salary[$value['employee_id']]['salary'];

 }
 Print_r($employee);die;

我想這會幫助你。 謝謝

Answer 4

那應該做

    $ar= array();
    for($j=0;$j<count($employee);$j++)
    {
        $ar[$j]=$employee[$j];
        if(isset($employee_salary[$employee[$j]["employee_id"]]) && $employee_salary[$employee[$j]["employee_id"]]['employee_id'] == $employee[$j]["employee_id"] ){
            $ar[$j]=array_merge($employee[$j],$employee_salary[$employee[$j]["employee_id"]]);
        }
    }

注意數組鍵。