使用带有ajax的复选框来更新数据库

Question

So, I am try to update the fields in my database (id[int], status[boolean]) using a checkbox with Ajax.因此，我尝试使用带有 Ajax 的复选框来更新我的数据库中的字段（id[int]、status[boolean]）。 This should be done without refreshing the page and not using a submit button.这应该在不刷新页面和不使用提交按钮的情况下完成。 I have tried multiple tutorials and non seems to work for me.我尝试了多个教程，非似乎对我有用。 To name a few: Ajax updating database when a checkbox is checked , checkbox on change ajax call , update database with checkbox仅举几例：选中复选框时 Ajax 更新数据库、更改 ajax 调用复选框、使用复选框更新数据库

Please I really need help with this one been on this for days.请我真的需要帮助这个已经好几天了。 This is my attempt at it:这是我的尝试：

index.html索引.html

<!DOCTYPE html>
<html>
<head>
  <meta charset="UTF-8">
<script type="text/javascript" src="https://code.jquery.com/jquery-3.4.1.min.js"></script>
<script src="jquery.min.js"></script>

</head>
<body>

    <input type="checkbox" name="action1" id="action1" title="Action 1" value="1"/>
  
<script type="text/javascript">
 $("#action1").change(function () {
    var value = $(this).val();
    $.ajax({
        type: "POST",
        url: "update.php",
        async: true,
        data: {
            action1: value // as you are getting in php $_POST['action1'] 
        },
        success: function (msg) {
            alert('Success');
            if (msg != 'success') {
                alert('Fail');
            }
        }
    });
});
</script>


</body>
</html>

update.php更新.php

 <?php
$servername = "localhost";
$username = "root";
$password = "";
$database = "api";
// Create connection
$conn = new mysqli($servername, $username, $password, $database);

// Check connection
if ($conn->connect_error) {
  die("Connection failed: " . $conn->connect_error);
}
//echo "Connected successfully";

 if($_POST['action1']==1){  //as used variable name "value" in ajax post data
          $sql = "UPDATE toggle SET status = '1' WHERE id = '1'"; //query was incomplete and missing ";"
          
          echo 'success';
     }
     else{
          $sql = "UPDATE toggle SET status = '0' WHERE id = '1'"; // missing ";"
          
          echo 'success';
     }
     $result=$conn->query($sql);
?>

Answer 1

Your update file returns 'success' no matter what data is sent to it.无论发送什么数据，您的更新文件都会返回“成功”。 Try it this way:试试这个方法：

 if($_POST['action1']==1){  //as used variable name "value" in ajax post data
      $sql = "UPDATE toggle SET status = '1' WHERE id = '1'";
      
      echo 'success';
 }
 else{
      $sql = "UPDATE toggle SET status = '0' WHERE id = '1'";
      
      echo 'fail';
 }
 $result=$conn->query($sql);

?> ?>

使用带有ajax的复选框来更新数据库

问题描述

index.html索引.html

update.php更新.php

1 个解决方案

解决方案1
0 2021-10-18 20:28:19

使用带有ajax的复选框来更新数据库

问题描述

index.html索引.html

update.php更新.php

1 个解决方案

解决方案1 0 2021-10-18 20:28:19

解决方案1
0 2021-10-18 20:28:19