PHP的if else问题，

Question

我写PHP函数用于获取周末不能得到正确的答案，这个“$数据[‘项目’]”不是空的，但它重定向到$end = date("Y-m-d", strtotime('next sunday'));

function weekendignSet(){ 
    if(!empty($data['projects'])){ 
     $end = $data['projects'][0]->week_end_day; 
     }else{ 
     if(!empty($_GET['week_ending'])){ 
      $end = $_GET['week_ending']; 
     }else{ 
      $end = date("Y-m-d", strtotime('next sunday')); 
     } 
     } 
     return $end; 
} 

什么是错误？谢谢

Answer 1

你需要传递$data运作然后如果（它会检查），否则你总是去在其他条件下尝试

function weekendignSet($data){ 

上呼吁： -

weekendignSet($data); 

Answer 2

你的功能需求一个名为“$ data”的输入参数

Answer 3

使$data成为一个全局变量，或者将其传递给函数：

global $data; 
$data = array(); 
function weekendignSet(){ 
    global $data; 
    if(!empty($data['projects'])){ 
     $end = $data['projects'][0]->week_end_day; 
     }else{ 
     if(!empty($_GET['week_ending'])){ 
      $end = $_GET['week_ending']; 
     }else{ 
      $end = date("Y-m-d", strtotime('next sunday')); 
     } 
     } 
     return $end; 
} 

或

$data = array(); 
function weekendignSet($data){ 
    if(!empty($data['projects'])){ 
     $end = $data['projects'][0]->week_end_day; 
     }else{ 
     if(!empty($_GET['week_ending'])){ 
      $end = $_GET['week_ending']; 
     }else{ 
      $end = date("Y-m-d", strtotime('next sunday')); 
     } 
     } 
     return $end; 
} 
weekendignSet($data);