如何从html表单执行php查询？

Question

我知道php是一种服务器端语言，html是一种客户端语言，为此，我会使用javascript，但我不知道该怎么做。有人可以帮我写一个名为changeFunction（）的脚本吗？

这是我在index.php中的表单：

<form method="POST">
            <div class="row"style="margin-top: 5%;">
                <div class="col-3">

                </div>
                <div class="col-6">
                   <label for="inputUsername">Insert username</label>
                   <input type="text" class="form-control" name="inputUser" style="margin-bottom: 2%">
                   <div class="input-group mb-3">
                    <div class="input-group-prepend">
                      <label class="input-group-text" for="inputGroupSelect01">Convert account type to: </label>
                    </div>
                    <select class="custom-select" id="inputGroupSelect01" name="inputSelected">
                      <option selected>Choose...</option>
                      <option value="admin">Admin</option>
                      <option value="user">User</option>
                    </select>
                       <button type="button" class="btn btn-outline-primary" onclick="changeFunction()">Submit</button>
                  </div>
                </div>
                <div class="col-3">
                </div>
            </div>
        </form>

这是我在query.php中的查询：

 <?php     
 $var1= $_POST["inputGroupSelect01"];
 $var2= $_POST["inputUser"];
 require 'connection.php';
 $sql = "UPDATE user SET typeAcc =".$var1." WHERE username=".$var2;
 ?>

Answer 1

我希望这是你想要的解决方案。

首先在你的html表单<div id='resultDiv' ></div>中添加以下div

所以你的HTML表单代码是这样的

<form method="POST" >
        <div class="row"style="margin-top: 5%;">
            <div class="col-3">
                <h2>Testing Form</h2>
            </div>
            <div class="col-6">
               <label for="inputUsername">Insert username</label>
               <input type="text" class="form-control" name="inputUser" style="margin-bottom: 2%" id="inputUser">
               <div class="input-group mb-3">
                <div class="input-group-prepend">
                  <label class="input-group-text" for="inputGroupSelect01">Convert account type to: </label>
                </div>
                <select class="custom-select" id="inputGroupSelect01" name="inputSelected">
                  <option selected>Choose...</option>
                  <option value="admin">Admin</option>
                  <option value="user">User</option>
                </select>
                   <button type="button" class="btn btn-outline-primary" onclick="changeFunction()">Submit</button>
              </div>
              <!-- this need to be added -->
              <div id='resultDiv' ></div>
            </div>
            <div class="col-3">
            </div>
        </div>
    </form>

第二步将jQuery和后续脚本添加到您的代码中

两个jquery变量定义userName和role，后来作为data传递，如果服务器响应是SUCCESS然后执行成功条件，或者你可以通过设置ERROR与YOUR ERROR TEXT发送数据或特定错误消息

 <script src="https://code.jquery.com/jquery-3.3.1.min.js" 
     integrity="sha256-FgpCb/KJQlLNfOu91ta32o/NMZxltwRo8QtmkMRdAu8="   
      crossorigin="anonymous"></script> 

    <script type='text/javascript'>

           function changeFunction(){
          var userName = $('#inputUser').val();
          var role     =$('#inputGroupSelect01').val();

         $.ajax({
           method: "POST",
           url: "server.php",
           data: JSON.stringify( { "userId": userName, "userRole": role } ),
           dataType: "text",

           success: function (response){
               if(response=='SUCCESS'){
                  $('#resultDiv').append('<br/>Successful Updated'+response);

            }else{
                $('#resultDiv').append('<br/>Error'+ response);
            }
        }

        });
      }
  </script>

**第三个**您必须使用以下代码编写新的server.php文件

<?php     
    $var1= $_POST["userId"]; // change here userId as used in jquery block 
    $var2= $_POST["userRole"];// change here userRole as used in jquery  Block
     //require 'connection.php';
     $sql = "UPDATE user SET typeAcc =".$var1." WHERE username=".$var2;

    //suppose query is executed successfull then 
    //it will return no of rows updated


   $query_result = mysql_query($query);
   //SUCCESS_CONDITION is normally if number of row updated is 1 in return 
   ///of query execution so kindly define by your own way 

   $num_rows = mysql_num_rows($query_result);
   //if($num_rows==1 || define your own SUCCESS_CONDITION){
   if($num_rows ==1){ 
     $result ='SUCCESS';
   }else{
      $result ='ERROR';
   }

  echo $result;
?>

我希望它会有所帮助谢谢如果需要请求帮助

Answer 2

您可以使用PHP执行此操作，但需要检查$ _POST数据。

if(isset($_POST['inputGroupSelect01']) && isset($_POST['inputUser;]){
$var1= $_POST["inputGroupSelect01"];
 $var2= $_POST["inputUser"];
 require 'connection.php';
 $sql = "UPDATE user SET typeAcc =".$var1." WHERE username=".$var2;
}

顺便说一句，该代码完全容易受到攻击。您正在从用户那里获取2个输入并直接放置它们，并将它们放入SQL查询中。

如果我对var2的输入是'; DROP TABLE user;怎么办？