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Поиск на Ajax22.02.11. |
HTML
<form id="searchForm" name="searchForm" method="post" action="javascript:insertTask();"> <div class="searchInput"> <input name="searchq" type="text" id="searchq" size="30" onkeyup="javascript:searchNameq()"/> <input type="button" name="submitSearch" id="submitSearch" value="Search" onclick="javascript:searchNameq()"/> </div> </form> <h3>Search Results</h3> <div id="msg">Type something into the input field</div> <div id="search-result"></div> JS
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 searchNameq() { searchq = encodeURI(document.getElementById('searchq').value); document.getElementById('msg').style.display = "block"; document.getElementById('msg').innerHTML = "Searching for <strong>" + searchq+""; // Set te random number to add to URL request nocache = Math.random(); http.open('get', 'in-search.php?name='+searchq+'&nocache = '+nocache); http.onreadystatechange = searchNameqReply; http.send(null); } function searchNameqReply() { if(http.readyState == 4){ var response = http.responseText; document.getElementById('search-result').innerHTML = response; } } php in-search.php
<?php
include('config.php'); $searchq = $_GET['name']; $getName = mysql_query('SELECT * FROM USER WHERE name LIKE "%'.addslashes($searchq).'%"'); while ($row = mysql_fetch_array($getName)) echo $row['name'] . '<br/>'; ?> Вернуться назад |