【发布时间】:2016-05-16 19:21:24
【问题描述】:
在我的 AngularJS 应用程序中,输入表单的数据在点击提交按钮后不会存储在 MySQL 数据库中。然而,成功提交表单数据后的警报表明它正在工作。
另外,在点击提交按钮后,我希望应用程序继续到下一个视图 (#/setup-step2) - 但它仍停留在第 1 步。
html部分(#/setup-step1):
<form ng-submit="submitForm1()">
Job title: <input type="text" name="jobtitle" ng-model="formData1.jobtitle">
Vacancy ID: <input type="text" name="vacancyid" ng-model="formData1.vacancyid">
<button type="submit" class="animatedbutton"> Proceed </button>
</form>
controller.js:
var ctr = angular.module('myApp.controller', []);
ctr.controller
('Step1Controller', ['$scope', '$routeParams', '$http', function($scope, $routeParams, $http){
$scope.formData1 = {};
$scope.submitForm1 = function() {
$http({
method: 'POST',
url: 'php/setup-step1.php',
data: $.param($scope.formData1),
headers: { 'Content-Type': 'application/x-www-form-urlencoded'}
})
.success(function(data){
console.log(data);
alert("It worked");
})
.error(function(data) {
console.log(data);
alert("It didn't work");
})
}
}]);
文件夹/php中的setup-step1.php:
<?php
include_once('db.php');
// Check connection
if(mysqli_connect_errno())
{echo '<p>Connection to MySQL server failed: '.mysqli_connect_error().'</p>';}
else
{echo '<p>Connection to MySQL server successful.</p>';}
$_POST = json_decode(file_get_contents("php://input"), true);
if (empty($_POST['jobtitle']))
{$errors['jobtitle'] = 'Job title is required.';}
else {
// Escape user inputs for security
$jobtitle = $_POST['jobtitle'];
$vacancyid = $_POST['vacancyid'];
$data = "INSERT INTO campaign (Job_title, Job_id) VALUES ('$jobtitle','$vacancyid')";mysql_query("INSERT INTO campaign (Job_title, Job_id) VALUES ('$jobtitle', '$vacancyid')");}
if ($connect->query($data) === TRUE)
{$conn->close();}
else
{echo "Error: " . $sql . "<br>" . $conn->error;}
exit();
?>
【问题讨论】:
-
您在 chrome 控制台中看到什么了吗?服务器日志中有什么? mysql日志?
-
将控制台日志添加到问题中
标签: php jquery mysql angularjs