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我在让搜索功能使用 ajax 在用户输入名称时显示结果时遇到了一些麻烦。这是有问题的网站（您可以单击查看全部以查看可搜索的内容） http://ra-yon.com/beta/Test_sites/HFE/admin/contact.php正在 执行的查询 http://ra-yon .com/beta/Test_sites/HFE/include/queryc.php?query=ra-yon&clause=email

有问题的代码

<html>
    <head>

<style type="text/css">
body
{
background:black;
color:black;
width:100%;

}
#center
{
width:90%;
height:110%;
background:white;
margin:0 auto;
text-align:center;
color:black;
}
#insert{
background:navy; color:white; font-family:impact; font-size:18px;width:170px; height:170px; border-radius:50%;
margin:0 auto;
float: left; margin-left: 100px;
top:200px;
position: relative;
}
#insert:hover

{
background:white;
color:navy;
}
label
{width:150px;}
td
{
    border:solid 2px black;
    max-width: 250px;
    text-align: center;
}
</style>
<script language="javascript" type="text/javascript">
<!-- 
//Browser Support Code
function ajaxFunction(x,y){

    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){
            document.getElementById('table').innerHTML = ajaxRequest.responseText;
        }
    }
    ajaxRequest.open("GET", "../include/queryc.php?query=" + x + "&clause="+ y, true);
    ajaxRequest.send(null);

}

//-->
</script>
<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.7.1/jquery.min.js"></script></head>

<body>
   <a href="index.php"> <h4 style="float:right; color:blue; position: relative; left:-100px;" id="goback"> Go back? </h4></a>
    <div id="center">
<div id="queryform1" >
    <a href="#" onclick="$('#table2').css('display','block');">View All?</a>
    <span style="color:black;">search</span> <input type='text' onKeyUp="ajaxFunction(this.value,sel.options[sel.selectedIndex].value); " name='searchname' id="searchname" />

<span style="color:black;">search by</span> <select onChange="ajaxFunction(this.value,sel.options[sel.selectedIndex].value);" name="searchclause" id="searchclause">
        <option ></option>
    <option value="name">Name</option>
    <option value="email">Email Address</option>
    <option value="subject">Subject</option>
    <option value="date">Date</option>

</select>
<a href="../include/downloadcontact.php">Download file?</a>

<div id="table2" style="display: none; margin: 0 auto; position: relative; top:40px; max-width:700px;">
    <div id="viewall">
        <?php
        include '../include/include.php';
        $sql = 'select * from contactus' ;



//print_r($sql);
$result=mysql_query($sql);

        ?>
        <table>
<tbody>
<th>Name</th><th>Email</th><th>Subject</th><th style="width:200px;">Message</th><th>Date</th>
<tr>
<?php
while ($client = mysql_fetch_array($result, MYSQL_ASSOC)){
echo "
<tr>

<td >".$client[name]."</td>
<td >".$client[email]."</td>
<td >".$client[subject]."</td>
<td >".$client[message]."</td>
<td >".$client[date]."</td>




</tr>";
} ?>
</tbody></table>
</div></div><div id="table">TestTestTest</div>


</div></div>

</body>
</html>

现在有点乱，我打算等我用完后清理一下。非常感谢大家！

4

1 回答 1

1

你为什么不也用 jQuery 来提出你的 AJAX 请求呢？jQuery 提供了一个简单的 $.ajax() 函数，可以很容易地发送 AJAX 请求。http://api.jquery.com/jQuery.ajax/

同样使用 jQuery 函数，您的代码也不会那么混乱。- 凯文

编辑：带有回调的示例 jQuery 函数。

$.ajax({
        url: 'test',
        type: 'POST',
        data: 'data=foo',
        success: function(text){
            $('.content').html(text);
        }
    });

请求的响应文本在变量text中。

于 2013-03-17T22:26:28.057 回答