爲什麼這個簡單的PHP沒有返回響應？

Question

這是我寫的第一個php程序。我的Ajax請求看起來沒問題，我的狀態爲200，但沒有得到任何迴應。

<?php 

class Employee{ 
    private $fn; 
    private $ln; 
    private $dpt; 
    private $ID; 
} 

function newEmployee(){ 
    $employee = new Employee(); 
    $fn = $_POST['firstname']; 
    $ln = $_POST['lastname']; 
    $dpt = $_POST['department']; 
    $id = sprintf('%08d', $GLOBALS['$ID']); 
    $GLOBALS['$ID'] = $GLOBALS['$ID'] + 1; 

    echo "First Name: $employee\nLast Name: $ln\nDepartment: $dpt\nID: $id"; 

    $employee -> fn = $_POST['firstname']; 
    $employee -> ln = $_POST['lastname']; 
    $employee -> dpt = $_POST['department']; 

    $GLOBALS['$employeeArray'][]= $employee; 

    $GLOBALS['$numOfEmployees'] = $GLOBALS['$numOfEmployees'] + 1; 
    $numemployees = $GLOBALS['$numOfEmployees']; 

    echo "First Name: $employee\nLast Name: $ln\nDepartment: $dpt\nID: $id\nNumber of Employees: $numemployees"; 


} 

if(isset($_POST['submit'])) 
{ 
    newEmployee(); 
} 

$employeeArray = array(); 
$ID = 0; 
$numOfEmployees = 0; 

?> 

字面上我的第一個PHP程序，所以我肯定這是愚蠢的。因爲該行的

Answer 1

您的代碼遊：

echo "First Name: $employee\nLast Name: $ln\nDepartment: $dpt\nID: $id"; 

發生了什麼事時，你想輸出你$employee = new Employee();爲字符串時，它的一個對象。所以PHP打破了這裏，不想繼續通過你的其他代碼。

也許你想這樣嗎？

echo "First Name: $fn\nLast Name: $ln\nDepartment: $dpt\nID: $id";  

您也想在函數的底部，以取代$employee其他呼叫你的第二個迴音。

正如Macbooc正確指出的那樣，您沒有向您的表單發送$_POST['submit']，也許可以改變它？

if(isset($_POST['submit']))  

- >

if(isset($_POST))