php - 如何使用 AngularJS 和 php 后端插入表格

Question

我有一个 View ，其中客户端的数据显示在表“clientttente”中，由 ng-repeat 显示，受到数组的影响，对于属于每个客户端的每一行，我都有一个“添加”按钮，此按钮必须将每个客户端添加到另一个表“客户端”，当我尝试此代码时，它会生成此错误:undefined property "code"，请问错误在哪里!

注册.html

<div class="padding" ng-controller="RegisterCtrl" ng-init="loadClient1()">

<a class="button button-info"  href="#/ajoutClient" style="background-color:#6EB4DC;float:right">Ajouter un client</a>
<table>
<tr>
          <th>Code.Client</th>
          <th>Nom</th>
          <th>Prenom</th>
          <td>Ajouter</td>
</tr>

<tr ng-repeat="x in names">
          <td ng-model="code">{{x.CodeClient}} </a> </td>
          <td ng-model="nom">{{x.NomClient}}       </td>
          <td ng-model="prenom">{{x.PrenomClient}}</td> >
          <td><a class="button button-info" ng-click="insertClient()" >Ajouter</a> </td>        
</tr>
</table>

应用程序.js

$scope.insertClient = function(){ 
      if(confirm("Êtes-vous sûr?"))  
           {   
           $http.post(  
                "http://localhost/deb/insertClient.php",  
                {  
                'code'   :$scope.code, 
                'nom'    :$scope.nom,
                'prenom' :$scope.prenom,
                }  

           ).success(function(data){  

           });  
         }
         else
         {
          return false;
         }
      }

插入客户端.php

 <?php  

 $connect = mysqli_connect("localhost", "root", "", "test");  

 $data = json_decode(file_get_contents("php://input"));  
 if(count($data) > 0)  
 {    $CodeClient=mysqli_real_escape_string($connect,$data->code);
      $NomClient = mysqli_real_escape_string($connect, $data->nom); 
      $PrenomClient = mysqli_real_escape_string($connect, $data->prenom); 

      $query = "INSERT INTO client (NomClient,PrenomClient) VALUES 
      ('$NomClient','$PrenomClient') WHERE   CodeClient='$CodeClient'"; 

      if(mysqli_query($connect, $query))  
      {  
           echo "The client has been successfully added";  
      }  
      else  
      {  
           echo 'Error';  
      }  
 }  
 ?>

Answer 1

没有 INSERT INTO ... WHERE 语法。你想要做的是 UPDATE ..SET ..WHERE:

$query = "UPDATE client  
  SET NomClient ='$NomClient' ,PrenomClient = '$PrenomClient'
  WHERE   CodeClient='$CodeClient'"; 

正如我在评论中建议的那样:

使用准备好的语句而不是 mysqli_real_escape_string 来防止 SQL 注入(inject)。