duandaijian1583 2010-04-25 22:10
浏览 47
已采纳

通过ajax在html / php中搜索表单

I've a search form wherein the database query has been coded in php and the html file calls this php file via ajax to display the results in the search form. The problem is, I would like the result to be displayed in the same form as search.html; yet while the ajax works, it goes to search.php to display the results.

search.html:

<!DOCTYPE html>
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8">
<script src="scripts/search_ajax.js" type="text/javascript"></script> 
</head>

<body>
<form id="submitForm" method="post">
<div class="wrapper">
<div class="field">
<input name="search" id="search" />
</div><br />
<input id="button1" type="submit" value="Submit" class="submit" onclick="run_query();" /><br />
</div>
<div id="searchContainer">
</div>
</form>
</body>
</html>

If I add action="search.php" to the form tag, it displays the result but on search.php. I'd like it to display on the same form [i.e search.html, and not search.php] if I just add the javascript function [as done above], it displays nothing on search.html.

search_ajax.js:

var xmlHttp


function run_query() {
xmlHttp=GetXmlHttpObject();
if (xmlHttp==null) {
alert ("This browser does not support HTTP Request");
return;
} // end if
var url="search.php";
xmlHttp.onreadystatechange=stateChanged;
xmlHttp.open("GET",url,true);
xmlHttp.send(null);
} //end function

function stateChanged(){
if (xmlHttp.readyState==4 || xmlHttp.readyState=="complete"){
document.getElementById("searchContainer").innerHTML=xmlHttp.responseText;
} //end if
} //end function

function GetXmlHttpObject() {
var xmlHttp=null;
try {
// For these browsers: Firefox, Opera 8.0+, Safari
xmlHttp=new XMLHttpRequest();
}catch (e){
//For Internet Explorer
try{
xmlHttp=new ActiveXObject("Msxml2.XMLHTTP");
} catch (e) {
xmlHttp=new ActiveXObject("Microsoft.XMLHTTP");
}
}
return xmlHttp;
} //end function

search.php:

<?php
include 'config.php';

$search_result = "";

$search_result = $_POST['search'];

$result = mysql_query("SELECT cQuotes, vAuthor, cArabic, vReference FROM thquotes WHERE cQuotes LIKE '%$search_result%' ORDER BY idQuotes DESC", $conn)
  or die ('Error: '.mysql_error());
  • 写回答

3条回答 默认 最新

      报告相同问题?

      相关推荐 更多相似问题

      悬赏问题

      • ¥30 VB6.0操作 webview2内核的浏览器如何精确实现网页弹窗处置
      • ¥15 pr导出的视频打不开,提示“缺少编解码器”怎么解决
      • ¥15 html里js获取php参数值不成功,帮改代码
      • ¥20 如何控制ant design的InputNumber组件 最多输入5位小数
      • ¥15 c语言学生基本信息管理系统
      • ¥100 火车头采集器采集求解
      • ¥88 关于#运行时间 时间重叠 和非重叠#的问题,如何解决?
      • ¥15 C语言,密切接触者追踪
      • ¥20 关于计算机网络问题,请附带讲解
      • ¥30 自动识别图像目标并判断