使用ajax保存单击时单选按钮的值

Question

Im creating a exam using php. 我使用php创建考试。

My problem is when the page is refresh the checked in the radio button dissappear.. i using ajax to save it in database everytime the radio button is click 我的问题是，刷新页面后，单选按钮中的选中项消失了。.每当单击单选按钮时，我就使用ajax将其保存在数据库中

<?php
session_start();
include './connect.php';
?>
<html>
<head>

</head>
<body>
<div id="online"></div>
<?php
$result = mysql_query("SELECT * FROM questions");
$count = mysql_num_rows($result);
while($rows = mysql_fetch_array($result))
{
?>
<script type="text/javascript" src="jquery.js"></script>
<script>
function saveans(){

var aa = $('input:radio[name=a1]:checked').val();
alert(aa);
 $.ajax({
      type: "POST",
      url:  "ans.php",
      data: "stdID=1&q=2&ans="+aa,
     })



}

</script>

<script language="javascript" type="text/javascript">
$(document).ready(function() {

   $("#online").load("./tmr.php");
   var refreshId = setInterval(function() {
   //alert("aa");
      $("#online").load('./tmr.php');
      $.ajax({
                       type: "POST",
                       url: "./update.php",
                       data: "stdID=1"
     })
   }, 1000);
   $.ajaxSetup({ cache: false });
});

</script>


<p><?php echo $rows["questions"]; ?></p>
<input type="radio" id="a<?php echo $rows['questionNum']; ?>" name="a<?php echo $rows['questionNum']; ?>" value="A" onClick="saveans(this);"/>A. <?php echo $rows['choi1']; ?><br/>
<input type="radio" id="a<?php echo $rows['questionNum']; ?>" name="a<?php echo $rows['questionNum']; ?>" value="B" onClick="saveans(this);"/>B. <?php echo $rows['choi2']; ?><br/>
<input type="radio" id="a<?php echo $rows['questionNum']; ?>" name="a<?php echo $rows['questionNum']; ?>" value="C" onClick="saveans(this);"/>C. <?php echo $rows['choi3']; ?><br/>
<input type="radio" id="a<?php echo $rows['questionNum']; ?>" name="a<?php echo $rows['questionNum']; ?>" value="D" onClick="saveans(this);"/>D. <?php echo $rows['choi4']; ?><br/>
<?php
}
?>
</body>
</html>

and the php code is

<?php
include './connect.php';$command = mysql_query("UPDATE ans_1 SET ans".$_POST['q']."='".$_POST['ans']."' WHERE std_id='".$_POST['stdID']."'") or die(mysql_error());
?>

the problem is i what to change the data: "stdID=1&q=2&ans="+aa.. the name of my radio button is a1,a2,a3 for multiple radio buttons.. i want to save it on my database depends the name Is it possible to insert database value in a ajax variable? 问题是我该如何更改数据：“ stdID = 1＆q = 2＆ans =” + aa ..对于多个单选按钮，我的单选按钮的名称是a1，a2，a3。我想将其保存在数据库中取决于name是否可以在ajax变量中插入数据库值？ and the std value is come from session 并且std值来自会话

Answer 1

You define your function as : 您将函数定义为：

function saveans(){
                 ^---no arguments

But invoke it in your onclick handers as: 但是在您的onclick处理程序中将其调用为：

onClick="saveans(this);"
                 ^^^^---passing in "this"

You want: 你要：

function saveans(obj) {
    answer = obj.value;
    ... do ajax ...
}

There's no need to look up by ID (and a hard-coded id at that), since this and obj will already be the exact same DOM element you're trying to look up by ID anyways. 无需按ID查找（以及在那里的硬编码ID），因为无论如何， this obj和obj都已经是您要尝试按ID查找的完全相同的DOM元素。

使用ajax保存单击时单选按钮的值

问题描述

1 个解决方案

解决方案1
1 2014-02-15 02:06:36

使用ajax保存单击时单选按钮的值

问题描述

1 个解决方案

解决方案1 1 2014-02-15 02:06:36

解决方案1
1 2014-02-15 02:06:36