$( ".home .images .image-slide div" ).each(function() {
var count = $(this).children().length;
if (count > 1) {
$(this).find("img:first-child").addClass("active");
var self = $(this);
setInterval(function(){
if(self.find('.active').is(":last-child")){
self.find('.active').removeClass("active").fadeOut().parent().find("img:first-child").addClass("active").fadeIn();
} else {
self.find('img.active').removeClass("active").fadeOut().next().fadeIn().addClass("active");
}
}, 4000);
}
});
如何对输出进行排序,以使CREATE TABLE #sorttest (test int, test2 int)
INSERT INTO #sorttest
VALUES (1, 2), (5, 4), (4, 3), (NULL, 1),
(3, NULL), (2, 5)
SELECT *
FROM #sorttest
ORDER BY
CASE WHEN test IS NULL THEN 1 ELSE 0 END, test DESC
DROP TABLE #sorttest
对于两列都排在最后?
答案 0 :(得分:1)
DECLARE @sorttest TABLE ( test1 INT, test2 INT)
INSERT INTO @sorttest values
(1,2),(5,4),(4,3),(NULL,1),(3,null),(2,5)
DECLARE @temp1 TABLE ( ID1 INT IDENTITY(1,1) PRIMARY KEY, test3 INT )
DECLARE @temp2 TABLE ( ID2 INT IDENTITY(1,1) PRIMARY KEY, test4 INT )
INSERT INTO @temp1
SELECT test1 FROM @sorttest ORDER BY test1 DESC
INSERT INTO @temp2
SELECT test2 FROM @sorttest ORDER BY test2 DESC
SELECT test3, test4 FROM @temp1 t1
LEFT JOIN @temp2 t2 ON t1.ID1 = t2.ID2
我使用了另外两个临时表来操纵数据,以使用join语句获得所需的输出。
如果需要在一行中重新排列值,则比单个select语句需要更多的逻辑。
输出:
test3 test4
5 5
4 4
3 3
2 2
1 1
NULL NULL
答案 1 :(得分:0)
您可以这样做
SELECT * FROM #sorttest ORDER BY
CASE WHEN test IS NULL and test2 is null THEN 2 ELSE
case when test is null or test2 is null then 1 end
END ,test desc
但是非null列的顺序有点奇怪