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SQL Tips

發(fā)布時(shí)間:2023/11/27 生活经验 29 豆豆
生活随笔 收集整理的這篇文章主要介紹了 SQL Tips 小編覺(jué)得挺不錯(cuò)的,現(xiàn)在分享給大家,幫大家做個(gè)參考.

出自:http://blog.csdn.net/etmonitor/

一.怎樣刪除一個(gè)表中某個(gè)字段重復(fù)的列呀,舉個(gè)例子
表[table1]
id??? name
1?aa
2?bb
3?cc
1?aa
2?bb
3?cc
我想最后的表是這樣的
id??? name
1?aa
2?bb
3?cc
回答:
將記錄存到臨時(shí)表#t中,重復(fù)的記錄只存一條,然后將臨時(shí)表#t中的記錄再存回原表中,注意“select distinct id,class,name”要包含你需要的所有字段,否則有些字段就被刪掉了。
在查詢管理器里執(zhí)行下面代碼:
-----------------------------
SELECT DISTINCT id,, name
INTO #t
FROM table1 DELETE table1
????????? INSERT
??????? INTO table1
????????????????? SELECT *
??????????????? FROM #t
------------------------------
二.找出既會(huì)VB又會(huì)PHP的人
表是這樣的:
ID?員工?技能
1?1?VB
2?1?PHP
3?1?ASP
4?2?PHP
5?3?ASP
6?4?VB
7?4?ASP
要從這張表中找出既會(huì)VB又會(huì)PHP的人,SQL該怎么寫(xiě)啊?
回答:
---------------------------------------------------------------------------------------------
SELECT 員工 FROM [Table] WHERE 員工 IN(SELECT 員工 FROM [Table] WHERE 技能='VB' ) AND 技能='PHP'
----------------------------------------------------------------------------------------------
三.數(shù)據(jù)庫(kù)合并問(wèn)題
access里的兩個(gè)表,想讓兩個(gè)表的內(nèi)容合并

表[a]結(jié)構(gòu)如下:
[id]??????? 編號(hào)??????? 自動(dòng)編號(hào)
[name]????? 名稱??????? 文本
[price]???? 價(jià)格??????? 數(shù)字
[guige]???? 規(guī)格??????? 文本
[changjia]? 生產(chǎn)廠家??? 文本
[baozhuang] 包裝??????? 文本
[danwei]??? 單位??????? 文本
共有900條記錄,除了id和name字段,其他均可以為空

表[b]結(jié)構(gòu)如下:
[id]??????? 編號(hào)??????? 自動(dòng)編號(hào)
[name]????? 名稱??????? 文本
[price]???? 價(jià)格??????? 數(shù)字
[changjia]? 生產(chǎn)廠家??? 文本
[danwei]??? 單位??????? 文本
[xingzhi]?? 性質(zhì)??????? 文本
共有800條記錄,除了id和name字段,比表[a]少幾個(gè)字段,但還多一個(gè)[xingzhi]的字符安其它均可以為空

現(xiàn)在想生成一個(gè)新表[c],結(jié)構(gòu)如下,而且內(nèi)容是兩個(gè)表的內(nèi)容之和,
[id]??????? 編號(hào)??????? 自動(dòng)編號(hào)
[name]????? 名稱??????? 文本
[price]???? 價(jià)格??????? 數(shù)字
[guige]???? 規(guī)格??????? 文本
[changjia]? 生產(chǎn)廠家??? 文本
[baozhuang] 包裝??????? 文本
[danwei]??? 單位??????? 文本
[xingzhi]?? 性質(zhì)??????? 文本

用sql語(yǔ)句也可以,手工操作也好,xml也好,別管怎么著吧,怎么實(shí)現(xiàn)呀,哥們要郁悶壞了,真要讓我們?cè)佥斎?00條記錄,我就掛了,
回答:
1.這樣
-----------------------------
insert into c(id,name,.....)
select id,name,.....
from a
insert into c(id,name,.....)
select max(id)+1,name,.....
from b
------------------------------
2.更正:
如果直接在查詢分析器里執(zhí)行:
-------------------------------
insert into c(name,.....)
select name,.....
from a
insert into c(name,.....)
select name,.....
from b
--------------------------------
3.用union方法
---------------------------------
insert into [c] ([id] ,編號(hào),自動(dòng)編號(hào))
select [id],編號(hào),自動(dòng)編號(hào) from [a]
union
select [id],編號(hào),自動(dòng)編號(hào) from [b]
-----------------------------------
4.asp的解決辦法
------------------------------------------------------------
<% '循環(huán)檢測(cè)a表
Set rs = Server.CreateObect("ADODB.RECORDSET")
????? rs.open "select * from a order by id",conn,1,1
????? Do while not rs.eof
????????? Call actAdd(rs("name"))??? '調(diào)用像b表添加內(nèi)容的函數(shù)!
????? rs.MoveNext
????? Loop
rs.Close
Set rs = Nothing

Sub actAdd(txt)
Dim ts, sql
sql = "insert into b(name) values('"& txt &"')"
Set ts = Conn.Execute(sql)
????? ts.Close
Set ts = Nothing
end Sub
%>
------------------------------------------------------------------
5.asp的解決辦法
-----------------------------------------------------------------------------------
<%
dim arr_temp1,arr_temp2,arr_data
set rs=conn.execute("select id,name,price,guige,changjia,baozhuang,danwei from a")
arr_temp1=rs.getrows
rs.close
set rs=nothing

set rs=conn.execute("select id,name,price,guige,changjia,danwei,xingzhi from b")
arr_temp2=rs.getrows
rs.close
set rs=nothing

rem 開(kāi)始處理
redim arr_data(ubound(arr_temp1,2)+ubound(arr_temp2,2),7)
rem 把兩個(gè)數(shù)組的內(nèi)容復(fù)制進(jìn)來(lái)
這一部分自己寫(xiě)了做兩個(gè)循環(huán)
然后再存進(jìn)數(shù)據(jù)庫(kù)
%>
---------------------------------------------------------------------------------------
最后轉(zhuǎn)一些經(jīng)典的SQL語(yǔ)句:
1.蛙蛙推薦:一些精妙的SQL語(yǔ)句
-----------------------------------------------------------------------------------------------------------------------
說(shuō)明:復(fù)制表(只復(fù)制結(jié)構(gòu),源表名:a 新表名:b)

SQL: select * into b from a where 1<>1

?

說(shuō)明:拷貝表(拷貝數(shù)據(jù),源表名:a 目標(biāo)表名:b)

SQL: insert into b(a, b, c) select d,e,f from b;

?

說(shuō)明:顯示文章、提交人和最后回復(fù)時(shí)間

SQL: select a.title,a.username,b.adddate from table a,(select max(adddate) adddate from table where table.title=a.title) b

?

說(shuō)明:外連接查詢(表名1:a 表名2:b)

SQL: select a.a, a.b, a.c, b.c, b.d, b.f from a LEFT OUT JOIN b ON a.a = b.c

?

說(shuō)明:日程安排提前五分鐘提醒

SQL:? select * from 日程安排 where datediff('minute',f開(kāi)始時(shí)間,getdate())>5

?

?

說(shuō)明:兩張關(guān)聯(lián)表,刪除主表中已經(jīng)在副表中沒(méi)有的信息

SQL:?

delete from info where not exists ( select * from infobz where info.infid=infobz.infid )

?

說(shuō)明:--

SQL:?

SELECT A.NUM, A.NAME, B.UPD_DATE, B.PREV_UPD_DATE

? FROM TABLE1,

??? (SELECT X.NUM, X.UPD_DATE, Y.UPD_DATE PREV_UPD_DATE

??????? FROM (SELECT NUM, UPD_DATE, INBOUND_QTY, STOCK_ONHAND

??????????????? FROM TABLE2

????????????? WHERE TO_CHAR(UPD_DATE,'YYYY/MM') = TO_CHAR(SYSDATE, 'YYYY/MM')) X,

??????????? (SELECT NUM, UPD_DATE, STOCK_ONHAND

??????????????? FROM TABLE2

????????????? WHERE TO_CHAR(UPD_DATE,'YYYY/MM') =

??????????????????? TO_CHAR(TO_DATE(TO_CHAR(SYSDATE, 'YYYY/MM') &brvbar;&brvbar; '/01','YYYY/MM/DD') - 1, 'YYYY/MM') ) Y,

??????? WHERE X.NUM = Y.NUM (+)

????????? AND X.INBOUND_QTY + NVL(Y.STOCK_ONHAND,0) <> X.STOCK_ONHAND ) B

WHERE A.NUM = B.NUM

?

說(shuō)明:--

SQL:?

select * from studentinfo where not exists(select * from student where studentinfo.id=student.id) and 系名稱='"&strdepartmentname&"' and 專業(yè)名稱='"&strprofessionname&"' order by 性別,生源地,高考總成績(jī)

?

說(shuō)明:

從數(shù)據(jù)庫(kù)中去一年的各單位電話費(fèi)統(tǒng)計(jì)(電話費(fèi)定額賀電化肥清單兩個(gè)表來(lái)源)

SQL:

SELECT a.userper, a.tel, a.standfee, TO_CHAR(a.telfeedate, 'yyyy') AS telyear,

????? SUM(decode(TO_CHAR(a.telfeedate, 'mm'), '01', a.factration)) AS JAN,

????? SUM(decode(TO_CHAR(a.telfeedate, 'mm'), '02', a.factration)) AS FRI,

????? SUM(decode(TO_CHAR(a.telfeedate, 'mm'), '03', a.factration)) AS MAR,

????? SUM(decode(TO_CHAR(a.telfeedate, 'mm'), '04', a.factration)) AS APR,

????? SUM(decode(TO_CHAR(a.telfeedate, 'mm'), '05', a.factration)) AS MAY,

????? SUM(decode(TO_CHAR(a.telfeedate, 'mm'), '06', a.factration)) AS JUE,

????? SUM(decode(TO_CHAR(a.telfeedate, 'mm'), '07', a.factration)) AS JUL,

????? SUM(decode(TO_CHAR(a.telfeedate, 'mm'), '08', a.factration)) AS AGU,

????? SUM(decode(TO_CHAR(a.telfeedate, 'mm'), '09', a.factration)) AS SEP,

????? SUM(decode(TO_CHAR(a.telfeedate, 'mm'), '10', a.factration)) AS OCT,

????? SUM(decode(TO_CHAR(a.telfeedate, 'mm'), '11', a.factration)) AS NOV,

????? SUM(decode(TO_CHAR(a.telfeedate, 'mm'), '12', a.factration)) AS DEC

FROM (SELECT a.userper, a.tel, a.standfee, b.telfeedate, b.factration

??????? FROM TELFEESTAND a, TELFEE b

??????? WHERE a.tel = b.telfax) a

GROUP BY a.userper, a.tel, a.standfee, TO_CHAR(a.telfeedate, 'yyyy')

?

說(shuō)明:四表聯(lián)查問(wèn)題:

SQL: select * from a left inner join b on a.a=b.b right inner join c on a.a=c.c? inner join d on a.a=d.d where .....

?

說(shuō)明:得到表中最小的未使用的ID號(hào)

SQL:

SELECT (CASE WHEN EXISTS(SELECT * FROM Handle b WHERE b.HandleID = 1) THEN MIN(HandleID) + 1 ELSE 1 END) as HandleID

?FROM? Handle

WHERE NOT HandleID IN (SELECT a.HandleID - 1 FROM Handle a)
-----------------------------------------------------------------------------------------------------------------------
2.刪除重復(fù)數(shù)據(jù)
-----------------------------------------------------------------------------------------------------------------------
一、具有主鍵的情況
a.具有唯一性的字段id(為唯一主鍵)
delete table
where id not in
(
select max(id) from table group by col1,col2,col3...
)
group by 子句后跟的字段就是你用來(lái)判斷重復(fù)的條件,如只有col1,
那么只要col1字段內(nèi)容相同即表示記錄相同。

b.具有聯(lián)合主鍵
假設(shè)col1+','+col2+','...col5 為聯(lián)合主鍵
select * from? table where col1+','+col2+','...col5 in (
? select max(col1+','+col2+','...col5) from table
where having count(*)>1
group by col1,col2,col3,col4
)
group by 子句后跟的字段就是你用來(lái)判斷重復(fù)的條件,
如只有col1,那么只要col1字段內(nèi)容相同即表示記錄相同。

c:判斷所有的字段
? select * into #aa from table group by id1,id2,....
? delete table
? insert into table
? select * from #aa

二、沒(méi)有主鍵的情況

a:用臨時(shí)表實(shí)現(xiàn)
select identity(int,1,1) as id,* into #temp from ta
delete #temp
where id not in
(
? select max(id) from # group by col1,col2,col3...
)
delete table ta
inset into ta(...)
?? select ..... from #temp

b:用改變表結(jié)構(gòu)(加一個(gè)唯一字段)來(lái)實(shí)現(xiàn)
alter table 表 add? newfield int identity(1,1)
delete 表
where newfield not in
(
select min(newfield) from 表 group by 除newfield外的所有字段
)

alter table 表 drop column newfield

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