我是jquery / ajax的新手,经过两天的搜索,我终于寻求帮助了。显然有些基本的我不理解。
这是.ajax代码:
$.ajax({
url: "../ajax/create_employee.php",
type: 'json',
async: false,
data: $serialFormData,
// callback handler that will be called on success
success: function(response, textStatus, jqXHR){
这是php文件(create_employee.php)的调用:
<?php
require ("../models/m_employee.php");
$myemployee = new m_employee();
$response = $myemployee->create_new_employee($_POST);
//echo " php file response = " . $response;
return $response;
?>
这是目标代码:
}
$q = "insert into employee(employee_num,employee_fname,employee_lname,employee_position,employee_start_time,employee_stop_time)";
$q .= "values ('$num','$fname','$lname','$position','$start_time','$stop_time')";
if(!$r = mysqli_query($dbc,$q) ) {
$data['error'] = "true";
$data['message'] = "insert failed";
$data['success'] = "false";
return json_encode($data);
}
返回到调用文件的数据是正确的,但它似乎没有进入“成功”功能。返回到.ajax调用的数据应该来自我的create_employee.php文件,还是仅报告文件执行的服务器?
答案 0 :(得分:0)
请尝试使用echo
:
<?php
require ("../models/m_employee.php");
$myemployee = new m_employee();
$response = $myemployee->create_new_employee($_POST);
//echo " php file response = " . $response;
echo $response;
?>
答案 1 :(得分:0)
这不是ajax相关的,而是一个mysql问题。我建议用测试数据自己运行php。您可能需要重新计算$ _POST变量。
关注维和部队关于使用回声而不是回归的建议。