5
작은 문제가 있습니다. POST_TITLE 및 POST_ID를 반환하는 실시간 검색을 원합니다. 제목은 사람들이 볼 수 있지만 내 주된 이유는 POST_ID를 사용하여 작업하도록하려는 것입니다. 누군가가 나를 도와 드릴까요, 나는 아래의 코드를 게시Ajax Live Search - 1 대신 2 필드 가져 오기
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<script>
//Gets the browser specific XmlHttpRequest Object
function getXmlHttpRequestObject() {
if (window.XMLHttpRequest) {
return new XMLHttpRequest();
} else if(window.ActiveXObject) {
return new ActiveXObject("Microsoft.XMLHTTP");
} else {
alert("Your Browser Sucks!\nIt's about time to upgrade don't you think?");
}
}
//Our XmlHttpRequest object to get the auto suggest
var searchReq = getXmlHttpRequestObject();
//Called from keyup on the search textbox.
//Starts the AJAX request.
function searchSuggest() {
if (searchReq.readyState == 4 || searchReq.readyState == 0) {
var str = escape(document.getElementById('txtSearch').value);
searchReq.open("GET", '/wp-content/themes/twentyten/livesearch.php?search=' + str, true);
searchReq.onreadystatechange = handleSearchSuggest;
searchReq.send(null);
}
}
//Called when the AJAX response is returned.
function handleSearchSuggest() {
if (searchReq.readyState == 4) {
var sx = document.getElementById('restaurantid')
sx.innerHTML = '';
var ss = document.getElementById('search_suggest')
ss.innerHTML = '';
var str = searchReq.responseText.split("\n");
for(i=0; i < str.length - 1; i++) {
//Build our element string. This is cleaner using the DOM, but
//IE doesn't support dynamically added attributes.
var suggest = '<div onmouseover="javascript:suggestOver(this);" ';
suggest += 'onmouseout="javascript:suggestOut(this);" ';
suggest += 'onclick="javascript:setSearch(this.innerHTML);" ';
suggest += 'class="suggest_link">' + str[i] + '</div>';
ss.innerHTML += suggest;
ss
}
}
}
//Mouse over function
function suggestOver(div_value) {
div_value.className = 'suggest_link_over';
}
//Mouse out function
function suggestOut(div_value) {
div_value.className = 'suggest_link';
}
//Click function
function setSearch(value) {
document.getElementById('txtSearch').value = value;
document.getElementById('restaurantid').value = value;
document.getElementById('search_suggest').innerHTML = '';
}
</script>
<form id="frmSearch" action="">
<input type="text" id="restaurantid" name="restaurantid" style="display: none;" />
<input type="text" id="txtSearch" name="txtSearch" alt="Search Criteria" onkeyup="searchSuggest();" autocomplete="off" />
<input type="submit" id="cmdSearch" name="cmdSearch" value="Search" alt="Run Search" />
<div id="search_suggest"></div>
</form>
</code>
livesearch.php - THE AJAX PAGE
<code>
<?php
$con = mysql_connect('x', 'x', 'x);
if (!$con)
{
die('Could not connect: ' . mysql_error());
}
mysql_select_db("xx", $con);
if (isset($_GET['search']) && $_GET['search'] != '') {
//Add slashes to any quotes to avoid SQL problems.
$search = addslashes($_GET['search']);
//Get every page title for the site.
$suggest_query = mysql_query('SELECT * FROM `mrr_posts` WHERE `post_title` LIKE \'%'.$search.'%\' AND `post_status` LIKE \'publish\' LIMIT 0, 30') or trigger_error("Query: $suggest_query\n<br />MySQL Error: " .mysql_error());
while ($suggest = mysql_fetch_array($suggest_query, MYSQL_ASSOC)) {
//while($suggest = db_fetch_array($suggest_query)) {
//Return each page title seperated by a newline.
echo $suggest['post_title'] . "\n";
}
}
mysql_close($con);
?>
+1 – yoda
+1 또한 경고를 보내 주셔서 감사합니다. 미소를 지어 주셔서 감사합니다. - 싱글 대답은 상황에 가장 적합한 것으로 보입니다. 행운을 빕니다! – Zydeco