Javascript jQuery脚本未在数据库Mysql中写入值

Javascript jQuery脚本未在数据库Mysql中写入值,javascript,jquery,angularjs,html,Javascript,Jquery,Angularjs,Html,我的数据库有一个名为utilizatori的表,其中包含id、username、nume\u prenume、parola、locatie和sambata列 通过表单,我试图在数据库中添加新条目,但当我提交表单时,它没有添加任何条目。 可以找到一个演示。 更新和删除操作工作正常 HTML格式: <script src="js/jQuery/jquery.min.js"></script> <!-- Include AngularJS library --> &

我的数据库有一个名为utilizatori的表,其中包含id、username、nume\u prenume、parola、locatie和sambata列

通过表单,我试图在数据库中添加新条目,但当我提交表单时,它没有添加任何条目。 可以找到一个演示。 更新和删除操作工作正常

HTML格式:

<script src="js/jQuery/jquery.min.js"></script>
<!-- Include AngularJS library -->
<script src="lib/angular/angular.min.js"></script>
<!-- Include Bootstrap Javascript -->
<script src="js/bootstrap.min.js"></script>
<form class="form-horizontal alert alert-warning" name="empList" id="empForm" ng-submit="insertInfo(empInfo);" hidden>
   <h3 class="text-center">Insert Employee Details Into Database</h3>
   <div class="form-group">
      <label for="Username">Nume utilizator:</label>
      <input type="text" name="username" class="form-control" placeholder="Enter Employee Name" ng-model="empInfo.username" value="" autofocus required />
   </div>
   <div class="form-group">
      <p class="text-danger" ng-show="empList.username.$invalid && empList.username.$dirty">Name field is Empty!</p>
   </div>
   <div class="form-group">
      <label for="Nume_prenume">Nume angajat:</label>
      <input type="text" name="nume_prenume" class="form-control" placeholder="Enter Employee Email Address" ng-model="empInfo.nume_prenume" autofocus required />
   </div>
   <div class="form-group">
      <p class="text-danger" ng-show="empList.nume_prenume.$invalid && empList.nume_prenume.$dirty">Invalid Email!</p>
   </div>
   <div class="form-group">
      <label for="Parola">Parola cont:</label>
      <input type="text" name="parola" class="form-control" placeholder="Enter Employee Name" ng-model="empInfo.parola" value="" autofocus required />
   </div>
   <div class="form-group">
      <p class="text-danger" ng-show="empList.parola.$invalid && empList.parola.$dirty">Name field is Empty!</p>
   </div>
   <div class="form-group">
      <label for="Rol_user">Rol:</label>
      <input type="text" name="rol_user" class="form-control" placeholder="Enter Employee Address" ng-model="empInfo.rol_user" autofocus required />
   </div>
   <div class="form-group">
      <p class="text-danger" ng-show="empList.rol_user.$invalid && empList.rol_user.$dirty">Address field is Empty!</p>
   </div>
   <div class="form-group">
      <button class="btn btn-warning" ng-disabled="empList.$invalid">Add Into Database</button>
   </div>
</form>
<script src="js/angular-script.js"></script>
插入信息PHP脚本:

<?php 
    // Including database connections
    require_once 'database_connections.php';
    // Fetching and decoding the inserted data
    $data = json_decode(file_get_contents("php://input")); 
    // Escaping special characters from submitting data & storing in new variables.
    $username = mysqli_real_escape_string($con, $data->username);
    $nume_prenume = mysqli_real_escape_string($con, $data->nume_prenume);
    $parola = mysqli_real_escape_string($con, $data->parola);
    $rol_user = mysqli_real_escape_string($con, $data->rol_user);

    // mysqli insert query
    $query = "INSERT into utilizatori (username,nume_prenume,parola,rol_user) VALUES ('$username','$nume_prenume','$parola','$rol_user')";
    // Inserting data into database
    mysqli_query($con, $query);
    echo true;
?>
和database_connections.php脚本:

<?php
$con = mysqli_connect("localhost", "user", "pasword", "database");
?>

我发现了问题。我试着写4列,但有6列是自动递增的

如果我重写此查询以写入所有6列,或者从数据库中删除其他2列,则可以工作

// mysqli insert query
    $query = "INSERT into utilizatori (username,nume_prenume,parola,rol_user) VALUES ('$username','$nume_prenume','$parola','$rol_user')";

id是自动递增主键。我的服务器是WAMP.update和delete在那个演示中对我不起作用,相反,我可以在控制台中看到错误消息。$http.post确实发送JSON吗?在那个演示中,更新和删除很可能被禁用。这是一个演示,但是如果你把它安装在你的服务器上,它就可以工作了。为什么要投否决票?我不明白。
// mysqli insert query
    $query = "INSERT into utilizatori (username,nume_prenume,parola,rol_user) VALUES ('$username','$nume_prenume','$parola','$rol_user')";