我有兩個索引為1到10的數組，然后必須根據索引合并它。因此，只需要增加一些員工的工資。我嘗試使用，array_merge()但輸出未合并。我無法獲得正確的輸出。輸出只是employeeSalary數組。為什么會這樣呢？$employee = array(0=> array("employee_id"=>1, "firstName"=>"Zahir", "lastName"=>"Alam", "Age"=>25, "Company"=>"Switchme", "Role"=>"Developer", "Department"=>"Tech" ,"Head"=> array("Id"=>3 , "Name"=>"Sourasis Roy") ),1=> array("employee_id"=>2, "firstName"=>"Amith", "lastName"=>"Manniken", "Age"=>25, "Company"=>"Switchme", "Role"=>"Developer", "Department"=>"Tech" ,"Head"=> array("Id"=>3 , "Name"=>"Sourasis Roy") ),2=> array("employee_id"=>3, "firstName"=>"Sourasis", "lastName"=>"Roy", "Age"=>28, "Company"=>"Switchme", "Role"=>"CTO"),3=> array("employee_id"=>4, "firstName"=>"Aditya", "lastName"=>"Mishra", "Age"=>29, "Company"=>"Switchme", "Department"=>"Tech", "Role"=>"CEO"),4=> array("employee_id"=>5, "firstName"=>"Priti", "lastName"=>"Lata", "Age"=>24, "Company"=>"Switchme", "Role"=>"HR"),5=> array("employee_id"=>6, "firstName"=>"Sumita", "lastName"=>"Nath", "Age"=>24, "Company"=>"Switchme", "Role"=>"HLA Head", "Department"=>"Crm"),6=> array("employee_id"=>7, "firstName"=>"Tarini", "lastName"=>"Khanna", "Age"=>22, "Company"=>"Switchme", "Role"=>"Content Writer"),7=> array("employee_id"=>8, "firstName"=>"Abhisek", "lastName"=>"Soni", "Age"=>23, "Company"=>"Switchme", "Role"=>"HLA", "Department"=>"Crm","Head"=>array("Id"=>5 , "Name"=>"Sumita Nath") ),8=> array("employee_id"=>9, "firstName"=>"Ankit", "lastName"=>"Pump", "Age"=>23, "Company"=>"Switchme", "Role"=>"HLA", "Department"=>"Crm" ,"Head"=> array("Id"=>5 , "Name"=>"Sumita Nath") ),9=> array("employee_id"=>10, "firstName"=>"Pogo", "lastName"=>"Laal", "Age"=>23, "Company"=>"Switchme", "Role"=>"Designer"),10=> array("employee_id"=>11, "firstName"=>"Sabina", "lastName"=>"Sekh", "Age"=>28, "Company"=>"Switchme", "Role"=>"HLA Head", "Department"=>"Crm")
3 回答
翻過高山走不出你
TA貢獻1875條經驗 獲得超3個贊
嘗試以下解決方案，使用數組的值$employee和$employee_salary問題中給出的值。
$ar= array();
foreach($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($ar);
echo "<br>";
}
}
}
largeQ
TA貢獻2039條經驗 獲得超8個贊
我認為也將有一個簡單的解決方案。因為您已經使用id鍵創建了employee_salary數組。所以會是這樣。
Foreach($employee as $key=> $value)
{
$employee[$key]['salary'] = $employee_salary[$value['employee_id']]['salary'];
}
Print_r($employee);die;
我認為這會幫助您。謝謝
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