为什么AJAX单选按钮不起作用？

我有一对单选按钮，用来将它们的值存储到数据库中，以便在用户将来返回站点时保留它们的状态。问题是，无论用户选择按钮1(喜欢)还是按钮2(不喜欢)，值总是以“喜欢”的方式返回。有人能帮我弄明白为什么不返回“不喜欢”吗？

form.php:

<script language="javascript" type="text/javascript">
<!-- 
//Browser Support Code
function ajaxFunction(){
var ajaxRequest;  // The variable that makes Ajax possible!

try{
    // Opera 8.0+, Firefox, Safari
    ajaxRequest = new XMLHttpRequest();
} catch (e){
    // Internet Explorer Browsers
    try{
        ajaxRequest = new ActiveXObject("Msxml2.XMLHTTP");
    } catch (e) {
        try{
            ajaxRequest = new ActiveXObject("Microsoft.XMLHTTP");
        } catch (e){
            // Something went wrong
            alert("Your browser broke!");
            return false;
        }
    }
}
// Create a function that will receive data sent from the server
ajaxRequest.onreadystatechange = function(){
    if(ajaxRequest.readyState == 4){
        var ajaxDisplay = document.getElementById('ajaxDiv');
        ajaxDisplay.innerHTML = ajaxRequest.responseText;
    }
}
var entered = document.getElementById('entered').value;
var queryString = "?entered=" + entered;
ajaxRequest.open("GET", "check.php" + queryString, true);
ajaxRequest.send(null); 
}

//-->
</script>

<form name="myform" action="check.php" method="post">
        <fieldset>
        <legend>Posts</legend>

<div id="post_1" class="post">
    <b>Post #1</b><br>
    Content of post #1<br>
    <p><input type="radio" id="entered" name="like_1" value="like" onclick="ajaxFunction();" onchange="ajaxFunction();" /><label for="like1a">Like</label></p> <p><input type="radio" id="entered" name="like_1" value="dislike" onclick="ajaxFunction();" onchange="ajaxFunction();" /><label for="like1b"> Dislike</label></p>
</div>
 </fieldset>
</form>
<div id='ajaxDiv'>Your result will display here</div>

check.php:

<?php
// Retrieve data from Query String
$entered = $_GET['entered'];
// Escape User Input to help prevent SQL Injection
$entered = mysql_real_escape_string($entered);
echo $entered;
?>

$Enter只存储了“like”，无论选择哪个单选按钮，都无法更改存储的值，但完全起作用。我是不是遗漏了什么？

When you get an element by an ID, it only returns one element, and which one is arbitrary because there should only be one element per ID. Instead, read the checked attribute to figure out which radio button is checked.

var entered=null;
var arr=document.getElementsByName("like_1");
for(var i=0;i<arr.length;i++){
    if(arr[i].checked){
        entered=arr[i].value;
        break;
    }
}

// HTML file:

<script type="text/javascript" src="http://ajax.googleapis.com/ajax/libs/jquery/1.6.1/jquery.min.js"></script>
<script language="javascript" type="text/javascript">
$(document).ready(function() {
    $('.button').click(function() {
        var valueSelected = this.value;
        var buttonSelected = this.id.replace(valueSelected + '_','');
        //alert('Button Selected: ' + buttonSelected + "
Value Selected: " + valueSelected);
        $.ajax({
            type: "GET",
            url: '/test.php?entered=' + valueSelected + '&id=' + buttonSelected,
            data:  '',
            cache: false,
            success: function(result) {
                $('#ajaxDiv').html(result);
            },
            error: function (response, desc, exception) {
                // custom error
            }
        });

    });
});

</script>
<fieldset>
<legend>Posts</legend>
    <div>
        <h1>Post #1</h1>
        <div>Content of post #1</div>
        <input type="radio" id="like_1" value="like" name="action1" class="button" /> <label for="like_1">Like</label>
        <br/>
        <input type="radio" id="dislike_1" value="dislike" name="action1" class="button" /> <label for="dislike_1">Dislike</label>
    </div>
    <div>
        <h1>Post #2</h1>
        <div>Content of post #2</div>
        <input type="radio" id="like_2" value="like" name="action2" class="button" /> <label for="like_2">Like</label>
        <br/>
        <input type="radio" id="dislike_2" value="dislike" name="action2" class="button" /> <label for="dislike_2">Dislike</label>
    </div>
</fieldset>
</form>
<div id="ajaxDiv">Your result will display here</div>

PHP file:

<?php

echo 'ID selected: ' . $_GET['id'] . ' - Value selected: ' . $_GET['entered'];