这是我写的第一个php程序。我的Ajax请求看起来没问题,我已经获得200状态,但没有得到任何响应。
<?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;
?>
答案 0 :(得分:1)
您的代码因此行而中断:
echo "First Name: $employee\nLast Name: $ln\nDepartment: $dpt\nID: $id";
正在发生的事情是,当您的$employee = new Employee();
作为对象时,您正试图将其作为字符串输出。所以PHP在这里打破了,不想继续浏览你的其余代码。
也许你想这样吗?
echo "First Name: $fn\nLast Name: $ln\nDepartment: $dpt\nID: $id";
您还希望在功能底部的第二个回音中替换$employee
的另一个调用。
正如Macbooc正确指出的那样,您没有向表单发送$_POST['submit']
,可能会改变它吗?
if(isset($_POST['submit']))
- &GT;
if(isset($_POST))