Question

我是Angular JS的新手，我想将数据从表单插入到MySQL数据库表中（在我的情况下表名是employee）但是当我点击提交时它不插入。我在浏览器中编译了insert.php文件，这给了我这个错误 ＆＃34;试图获取非对象＆＃34; 的属性。我正在以JSON格式发送数据，然后我在index.php文件中进行了解码，但当我尝试通过$data对象访问解码数据时，它给出了我提到的错误。

myApp.js代码

var app = angular.module('myApp',[])

.controller("myController",function($scope,$http){
    $scope.insertData = function(){
        $http.post("insert.php",{'name': $scope.name,'fname': $scope.fname,'dept': $scope.dept})
        .success(function(data,status,headers,config){
           console.log("Inserted Successfuly!"); 
        });
    }
});

insert.php code

<?php
$data = json_decode(file_get_contents("php://input"));
$name = $data->name;
$fname = $data->fname;
$dept = $data->dept;

$host = 'localhost';
$dbname = 'company';
$dbusername = 'root';
$dbpass = '';

$connection = mysqli_connect($host,$dbusername,$dbpass,$dbname);

if (mysqli_connect_errno()){
  echo "Failed to connect to MySQL: " . mysqli_connect_error();
}else{
    $query = "INSERT INTO employe ('id','ename','fname','dept') VALUES('','$name','$fname','$dept')";
    $query_run = mysqli_query($connection,$query);
}
?>

index.html代码

<html ng-app="myApp">

    <head>
        <title> Using Angular with PHP! </title>
        <!--<script src="https://ajax.googleapis.com/ajax/libs/angularjs/1.5.0-rc.1/angular.min.js"></script>-->
        <style type="text/css" src="css/bootstrap.min.css"></style>
    </head>

    <body>   
        <div class="container">
            <div class="row">
                <div class="col-lg-12" ng-controller="myController">
                    <form>
                        Your Name:   <input type="text" ng-model="name" /><br><br>
                        Father Name: <input type="text" ng-model="fname" /><br><br>
                        Department : <input type="text" ng-model="dept" /><br><br>
                        <input class="btn btn-success" type="button" value="Submit" ng-click="insertData()" />
                    </form>
                </div>
            </div>
        </div> 


       <script type="text/javascript" src="js/angular/angular.min.js"></script>
       <script type="text/javascript" src="js/bootstrap/bootstrap.min.js"></script>
       <script type="text/javascript" src="js/app.js"></script>
    </body>
</html>

Answer 1

你会笑，但只要添加这些``围绕你的表名就可以了。

  $query = "INSERT INTO `employe`(`id`, `ename`, `fname`, `dept`) VALUES('','$name','$fname','$dept')";

为什么我要尝试在AngularJs中获取非对象错误的属性？

1 个答案: