Question

我提出了3个问题,要求重新排列我的桌子。

SELECT @i:=0;
UPDATE imagedata SET imagedata.rank = 0;
UPDATE imagedata SET imagedata.rank = (SELECT @i:=@i+1) WHERE imagedata.kategorieID = 1;

我从上至下执行。

问题: 这样做的越少?

thx for all the feedback.. but i have a differnet Idear now: Somehow I have to "connect" the "kategorieID" with the "id(Primary key)" collumn Therefore I need to store both information into the "rank" collumn in the format like:

cxxx

c=Category Number(0 to 4) xxx = id (1 to ...n)..unique!

考试:产出:

   rank
 +------+
 + 1001 +
 + 1002 +
 + 1003 +
 + ...  +
 + 1018 +
 + ...  +
 + 2001 +
 + 2002 +
 + 1019 +
 + 1020 +

到目前为止,我也这样做:

UPDATE imagedata SET imagedata.rank = (SELECT CONCAT(kategorieID,"",LPAD(id,3, 0 )) ) WHERE id=88;

一旦“等级”填满了数据,它就给我放弃2 先令。

Therefore I have to: 1) get "rank" collumn of id_1 and id_2 2) get the substring ( xxx ) of that string exampl: ("004" .. "012") 3) exchange substrings of id_1 "rank" and id_2 "rank"

SELECT @ix1:=SUBSTRING(rank, -3) FROM imagedata WHERE id=88;
SELECT @ix2:=SUBSTRING(rank, -3) FROM imagedata WHERE id=83;

。我知道如何做到这一点,但真心知道如何交流?

Answer 1

我只想修改第二点,以更新第三行的WON T,如:

SELECT @i:=0;
UPDATE imagedata SET imagedata.rank = 0 WHERE imagedata.kategorieID <> 1;
UPDATE imagedata SET imagedata.rank = (SELECT @i:=@i+1) WHERE imagedata.kategorieID = 1;

http://www.un.org。 Per Lamak s comment, if the kategorieID 页: 1

SELECT @i:=0;
UPDATE imagedata SET imagedata.rank = 0 WHERE imagedata.kategorieID <> 1 OR imagedata.kategorieID IS NULL;
UPDATE imagedata SET imagedata.rank = (SELECT @i:=@i+1) WHERE imagedata.kategorieID = 1;

此外,这一解决办法可能比你更慢,特别是如果在<代码>kategorie上有一个指数。我之所以这样说,是因为尽管你正在更新较少的浏览量,但现在必须首先在<条码>上过滤。或许,你应首先测试这一业绩,看看它是否真正更快。

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