SQL中怎么實(shí)現(xiàn)嵌套查詢

SQL中怎么實(shí)現(xiàn)嵌套查詢，相信很多沒(méi)有經(jīng)驗(yàn)的人對(duì)此束手無(wú)策，為此本文總結(jié)了問(wèn)題出現(xiàn)的原因和解決方法，通過(guò)這篇文章希望你能解決這個(gè)問(wèn)題。

一個(gè)優(yōu)秀的系統(tǒng),會(huì)集成優(yōu)秀的程序和優(yōu)秀的數(shù)據(jù)庫(kù)設(shè)計(jì).
要做到這點(diǎn)得有足夠的經(jīng)驗(yàn).
這是我寫的一個(gè)結(jié)合UINON的嵌套查詢.
將五個(gè)方面的報(bào)表放到一個(gè)臨時(shí)表里,再?gòu)呐R時(shí)表里,將所要的數(shù)據(jù)查詢出來(lái). 

復(fù)制代碼 代碼如下:

$sql="SELECT type , sum( yjsl ) as yjsl , sum( yysl ) as yysl, sum( jyrs ) as jyrs, sum( jycs ) as jycs
FROM (
SELECT c.mc AS
TYPE , count( d.lsh ) AS yjsl, 0 AS yysl, 0 AS jyrs, 0 AS jycs
FROM sys_dzxxb AS b, sys_jcb AS c, sys_dzyjb AS d
WHERE b.bm = c.lsh
AND d.dzlsh = b.lsh
GROUP BY c.mc
UNION SELECT c.mc AS
TYPE , 0 AS yjsl, count( e.lsh ) AS yysl, 0 AS jyrs, 0 AS jycs
FROM sys_dzxxb AS b, sys_jcb AS c, sys_dzyy AS e
WHERE b.bm = c.lsh
AND e.dzlsh = b.lsh
GROUP BY c.mc
UNION SELECT c.mc AS
TYPE , 0 AS yjsl, 0 AS yysl, count( DISTINCT e.dzlsh ) AS jyrs, 0 AS jycs
FROM sys_dzxxb AS b, sys_jcb AS c, sys_ltxxb AS e
WHERE b.bm = c.lsh
AND e.dzlsh = b.lsh
GROUP BY c.mc
UNION SELECT c.mc AS
TYPE , 0 AS yjsl, 0 AS yysl, 0 AS jyrs, count( DISTINCT e.lsh ) AS jycs
FROM sys_dzxxb AS b, sys_jcb AS c, sys_ltxxb AS e
WHERE b.bm = c.lsh
AND e.dzlsh = b.lsh
GROUP BY c.mc
) AS temptable
GROUP BY TYPE ";

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