drqvr26084 2019-04-22 06:47 采纳率: 100%
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使用JQuery AJAX和php将数据提取到mysql数据库

Im trying to insert data into my database with AJAX but dont working. I can verify that im connected to the database but when i click it doesnt insert the data. thanks

with a click function i take the 2 parameter that i wanna insert in my database.

$( "#q_answer1" ).click(function () {
      var q_no = $("#q_no").val(); 
      var main_no = $("#total_no").val();

      $.ajax({
         url: "server.php",
         type: "post",
         async: false,
         data: {
            "done": 1,
            "username": q_no,
            "comment": main_no

         }, 
         success: function(){
            $("#q_no").val('');
            $("#total_no").val('');
         }
      });
  });

And here is the php file, first connect to the ddbb and insert the 2 values with the mysql_query.

<?php
include("dbh.php");
if (isset($_POST['done'])) {
   $q_no = mysql_escape_string($_POST['username']);
   $total_no = mysql_escape_string($_POST['comment']);

   mysql_query("INSERT INTO variables(id, names) VALUES('{$q_no}', '{$total_no}')");
   exit();
}
?>

html is like this:

<div id="total_no">1</div>
<div id="q_answer1" class="btn left_b">yes</div>

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4条回答 默认 最新

  • duan246558 2019-04-22 09:04
    关注

    I think you should use PDO, to connect to the database instead of the old driver, which PHP no longer supports. with PDO you can use prepared statements to prevent sql injections

    PDO tutorial

    filter_var() Constants

    dbh.php

    $servername = "localhost";
    $username = "user";
    $password = "pass";
    $dbname = 'db';
    
    try {
    
      $db = new PDO("mysql:host=$servername;dbname=$dbname", $username, $password);
        // set the PDO error mode to exception
        $db->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
      }
      catch(PDOException $e)
      {
        exit($e->getMessage());
      }
    
    ?>
    

    serve.php

    <?php
    
    include("dbh.php");
    if (isset($_POST['done'])) {
    
       $q_no = filter_var($_POST['username'], FILTER_SANITIZE_STRING);
       $total_no = filter_var($_POST['comment'], FILTER_SANITIZE_STRING);
    
    try {
    
      $stmt = $db->prepare("INSERT INTO variables(id, names) VALUES(?, ?)");
      $stmt->execute(array($q_no, $total_no));
    
      echo json_encode(["message" => "success"]); // sends success response to front-end 
    
    } catch (\Exception $e) {
      echo json_encode(["message" => $e->getMessage() ]); // sends error response to front-end
    }
    
    
    }
     ?>
    

    in your ajax check if the data was inserted or not.

    $("#q_answer1").click(function() {
      var q_no = $("#q_no").val();
      var main_no = $("#total_no").val();
    
      $.ajax({
        url: "file.php",
        type: "post",
        async: false,
        data: {
          "done": 1,
          "username": q_no,
          "comment": main_no
    
        },
        success: function(data) {
    
          const respose = JSON.parse(data);
    
          if (respose.message === 'success') { // data was inserted 
    
            $("#q_no").val('');
            $("#total_no").val('');
    
          }else {
            alert(respose.message); // some error has occured
          }
        }
      });
    });
    
    本回答被题主选为最佳回答 , 对您是否有帮助呢?
    评论
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