这是我写的第一个php程序。我的Ajax请求看起来没问题,我的状态为200,但没有得到任何回应。为什么这个简单的PHP没有返回响应?
<?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;
$numOfEmployees = 0;
?>
字面上我的第一个PHP程序,所以我肯定这是愚蠢的。因为该行的
这就像你的接缝不会在您的文章发送一个“提交”值,$ _ POST [“提交”]是不是isset – MacBooc
谢谢你会建议?将它添加到ajax或将该条件更改为其他内容? – Zachscs
对我来说并不是真的最好练习你的问题,两者都应该工作 – MacBooc