전에 작업하다 말은 디비연동 최근게시물 순서바꾸기 정보
전에 작업하다 말은 디비연동 최근게시물 순서바꾸기
본문
http://zom.kr/
테스트:admin /admin
간단하게 관리자만 바꿀수있게 해봤습니다.
이제 최근게시물 생성페이지만 만들어주면 끝날꺼같은데..
jquery-1.3.2.min.js 이랑 jquery-ui-1.7.1.custom.min.js 은 검색하면 널렸습니다.
디비만드시고
CREATE TABLE `records` (
`recordID` int(11) NOT NULL auto_increment,
`recordText` varchar(255) default NULL,
`record_skin` varchar(25) default NULL,
`record_board` varchar(25) default NULL,
`record_row` varchar(25) default NULL,
`record_no` varchar(25) default NULL,
`record_option` varchar(25) default NULL,
`recordListingID` int(11) default NULL,
PRIMARY KEY (`recordID`)
);
INSERT INTO `records` VALUES ('1', '최근글1','스킨이름', '게시판명','몇줄?','제목글자수','옵션 있으면넣고 없으면공백','1');
INSERT INTO `records` VALUES ('2', '최근글2','스킨이름', '게시판명','몇줄?','제목글자수','옵션 있으면넣고 없으면공백','2');
INSERT INTO `records` VALUES ('3', '최근글3','스킨이름', '게시판명','몇줄?','제목글자수','옵션 있으면넣고 없으면공백','3');
INSERT INTO `records` VALUES ('4', '최근글4','스킨이름', '게시판명','몇줄?','제목글자수','옵션 있으면넣고 없으면공백','4');
INSERT INTO `records` VALUES ('5', '최근글5','스킨이름', '게시판명','몇줄?','제목글자수','옵션 있으면넣고 없으면공백','5');
--index.php--
<?php
<?php
include_once("./_common.php");
include_once("$g4[path]/lib/latest.lib.php");
include_once("$g4[path]/_head.php");
?>
<script type="text/javascript" src="<?=$g4['path']?>/js/jquery-1.3.2.min.js"></script>
<script type="text/javascript" src="<?=$g4['path']?>/js/jquery-ui-1.7.1.custom.min.js"></script>
<style>
@font-face {font-family:맑은 고딕; src:url(<?=$g4['path']?>/gothic-kcy1019_.eot) };
left {
margin: 0;
}
#contentWrap {
width: 700px;
margin: 0 auto;
height: auto;
}
#contentLeft {
float: left;
width: 400px;
}
.contentLeft3 {
list-style: none;
margin:3px;
color:#000;
}
#contentRight {
float: right;
width: 260px;
padding:10px;
color:#FFFFFF;
}
#adm {
width: 260px;
padding:10px;
background-color:#6870ae;
color:#FFFFFF;
}
#adm a{
color:#FFFFFF;
}
</style>
<script type="text/javascript">
$(document).ready(function(){
$(function() {
$("#left").sortable({ opacity: 0.6, cursor: 'move', update: function() {
var order = $(this).sortable("serialize") + '&action=updateRecordsListings';
$.post("<?=$g4[admin_path]?>/index/updateDB.php", order, function(theResponse){
$("#contentRight").html(theResponse);
});
}
});
});
});
</script>
<?
$newsArray = latestskin
?>
<div id="adm"><a href="<?=$g4[path]?>/">돌아가기</a></div>
<div id="contentLeft">
<div id="left">
<?
$query = "SELECT * FROM records ORDER BY recordListingID ASC";
$result = mysql_query($query);
while($row = mysql_fetch_array($result, MYSQL_ASSOC))
{
?>
<div class="contentLeft3" id="leftArray_<?=$row['recordID']; ?>">
<?=$row['recordText']; ?> 위치 바꾸기
<?=latest($row['record_skin'], $row['record_board'], $row['record_row'], $row['record_no']);?>
</div>
<? } ?>
</div>
</div>
<div id="contentRight"></div>
<div id="left">213
</div>
<?
include_once("$g4[path]/_tail.php");
?>
--updataDB.php--
<?
include_once("./_common.php");
$action = $_POST['action'];
$updateleftArray = $_POST['leftArray'];
if ($action == "updateRecordsListings"){
$listingCounter = 1;
foreach ($updateleftArray as $recordIDValue) {
$query = "UPDATE records SET recordListingID = " . $listingCounter . " WHERE recordID = " . $recordIDValue;
mysql_query($query) or die('저장실패');
$listingCounter = $listingCounter + 1;
}
echo 'F5를 눌려서 페이지를 갱신하면 순서가 저장됩니다.';
}
?>
참쉽죠? ^^
0
댓글 1개
