Question

我是ajax的新手。来自wrook博客的基础研究我使用下面的代码使用ajax插入我的数据库但是除了“等一下”之外，代码不打印任何东西。

JsFiddle link Ajax代码如下： -

/* ---------------------------- */
/* XMLHTTPRequest Enable */
/* ---------------------------- */

function createObject() {
var request_type;
var browser = navigator.appName;
if (browser == "Microsoft Internet Explorer") {
    request_type = new ActiveXObject("Microsoft.XMLHTTP");
}
else {
    request_type = new XMLHttpRequest();
}
return request_type;
}

var http = createObject();

function insertReply() {
if (http.readyState == 4) {
    var response = http.responseText;
    // else if login is ok show a message: "Site added+ site URL".
    document.getElementById('insert_response').innerHTML = 'User added:' + response;
}
}

/* -------------------------- */
/* INSERT */
/* -------------------------- */
/* Required: var nocache is a random number to add to request. This value solve an Internet          Explorer cache issue */
var nocache = 0;

 function insert() {
// Optional: Show a waiting message in the layer with ID login_response
document.getElementById('insert_response').innerHTML = "Just a second...";
// Required: verify that all fileds is not empty. Use encodeURI() to solve some issues about   character encoding.

var username = encodeURI(document.getElementById('Username').value);
var name = encodeURI(document.getElementById('Name').value);
var password = encodeURI(document.getElementById('Password').value);
document.getElementById('insert_response').innerHTML = 'h ' + username + ' ' + name;
// Set te random number to add to URL request
nocache = Math.random();

// Pass the login variables like URL variable
http.open('get', 'insert.php?name=' + name + '&username=' + username + '&password=' + password);
http.onreadystatechange = insertReply;
http.send(null);
}

插入db的php代码是:: -

  if(isset($_GET['name']) && isset($_GET['username'])&& isset($_GET['password'])){

 $name= $_GET['name'];
 $username= $_GET['username'];
  $password= $_GET['password'];
  $sql = "INSERT INTO `vidyasims`.`user_accounts` (`UserID`, `UserName`, `Name`, `Password`)   VALUES (NULL, '. $username. ','. $name. ', '. $password. ');";
 $insertSite= mysql_query($sql) or die(mysql_error());

  //If is set URL variables and insert is ok, show the site name -->
 echo $username;
} else { 
 echo 'Error! Please fill all fileds!';

欢迎支持。提前谢谢。

注意 - 上面的代码会向db！添加一个新用户。

Answer 1

您缺少html表单输入的ID，并且无法使用document.getElementById（）快速修复其值：

<td><input type="text" id="Name" name='Name' value='' ><br></td>
                 <td><input type="text" id="Username" name='UserName' value=''></td>
                 <td><input type="password" id="Password" name='Password'></td>

Answer 2

为什么你不使用这样的jQuery Ajax，我认为这是你工作的简单方法：

JQuery代码

 <script language="javascript">
  $(document).ready(function(){
  $("#myHtmlButtonID").bind("click",function(){
       $('#insert_response').html("Just a second...");
       var username = $('#Username').val();
       var name = $('#Name').val();
       var password = $('#Password').val();
       /* add some validation if needed here */
       $.ajax({
                 url: "yourPHPpage.php",
                 data: {
                       name: name,
                       username:username,
                       password: password
                       },
                type: "get",
                success: function(txt){
                      $('#insert_response').html('User added:' + response);
                      },
                error: function(xhr){
                      $('#insert_response').html('Error: - '+xhr.status+ ' '+ xhr.statusText);  
                      }
               });
     });
});
</script>

ajax代码定义错误或不响应

2 个答案: