我想基于索引合并两个关联数组

Question

我有两个索引为1到10的数组，然后必须根据索引合并它。因此，只需要增加一些员工的工资。我尝试使用，array_merge()但输出未合并。我无法获得正确的输出。输出只是employeeSalary数组。为什么会这样呢？$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")

Answer 1

尝试以下解决方案，使用数组的值$employee和$employee_salary问题中给出的值。

$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 2

我认为也将有一个简单的解决方案。因为您已经使用id键创建了employee_salary数组。所以会是这样。

Foreach($employee as $key=> $value)

{ 

     $employee[$key]['salary'] = $employee_salary[$value['employee_id']]['salary'];

 }

 Print_r($employee);die;

我认为这会帮助您。谢谢