将数据从html到php并将其发布到数据库 [英] Bringing data from html to php and posting it onto database

问题描述

<img id="image" src="jj.png" onclick="randomizeImage();"/>
<h2 id="Score" name="Score1">0</h2>

<script>
    var counter=0;
    function randomizeImage() {
        counter++;
        var test= document.getElementById('Score');
        test.innerHTML= counter;
    }
</script>

<?php
    $con = mysql_connect("localhost","root");
    if (!$con)
    {
        die('Could not connect: ' . mysql_error());
    }

    mysql_select_db("database", $con);

    $user_Name = $_COOKIE["username"]; //i stored the username as a cookie and retrieved it here 
    echo $user_Name; //echo it so i can check if it is right

    $New= $_POST["Score"]; //**this is the part which is undefined**

    $sql = "UPDATE User SET Score = '$New' WHERE ID='$user_Name'";

    if (!mysql_query($sql,$con))
    {
        die('Error please try again ' . mysql_error());
    }

    mysql_close($con)

?>

我有一个图像，每当我点击它，它会调用一个函数，这个分数将反映在html侧，并增加，每当我点击图像。但现在我想把这个计数器数据到php，我可以将其上传到一个数据库，它匹配用户的用户名，他/她在以前的phpfile中输入。我不能带来分数的价值？它不断地说未定义的索引。

I have an image that whenever i click on it, it will call a function which will increase a counter by 1. This score will be reflected on the html side and increase whenever i click on the image. However now i want to bring this counter data into php where i can upload it onto a database where it matches the user's username which he/she entered in a previous phpfile. I cant bring the value of score over? It keeps saying undefined index.

我有两个名为ID和分数的列。 ID是我为用户的用户名分数，Score是来自计数器的分数。

I have two colums called ID and Score. ID is where i score the user's username and Score is the score from the counter. I want the Score to updated everytime the image is pressed with respect to the username.

数据库名称是：database
表名是：User

Database name is :database Table name is: User

有没有AJAX没有做到这一点？

Is there anyway to do this without AJAX?

推荐答案

这是如何发送 ajax 请求到服务器。请务必先添加 jquery 。

This is how you can send an ajax request to the server. Make sure to include jquery first.

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>

 <script>
    var counter=0;
    function randomizeImage() {
      counter++;
      var test= document.getElementById('Score');
      test.innerHTML= counter;

      $.ajax
      (
          "ajax.php",
          {
              method: "POST",
              data: { score: counter}
          }
      );
    }
  </script>

接下来你必须把你的PHP代码放到一个单独的文件中，我称之为ajax.php 。

Next you have to put your PHP code into a separate file, I call it "ajax.php".

<?php

    $score = filter_input(INPUT_POST, "score");
    if (isset($score) && $score != "") {

    $con = mysql_connect("localhost","root");
    if (!$con)
    {
       die('Could not connect: ' . mysql_error());
    }
    mysql_select_db("database", $con);
    $user_Name = $_COOKIE["username"]; //i stored the username as a cookie and retrieved it here 
    echo $user_Name; //echo it so i can check if it is right

   $sql = "UPDATE User SET Score = '$score' WHERE ID='$user_Name'";
   $result =mysql_query($sql) or die("could not update" . mysql_error);

  if (!mysql_query($sql,$con))
  {
    die('Error please try again ' . mysql_error());
  }
  mysql_close($con)
  }
 ?>

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