sum(x) over( partition by y ORDER BY z ) 分析
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sum(x) over( partition by y ORDER BY z ) 分析
小編覺得挺不錯(cuò)的,現(xiàn)在分享給大家,幫大家做個(gè)參考.
參考的博文出處:http://www.cnblogs.com/luhe/p/4155612.html,對(duì)博文進(jìn)行了修改新增,修改了錯(cuò)誤的地方
之前用過row_number(),rank()等排序與over( partition by?... ORDER BY ...),這兩個(gè)比較好理解:?先分組,然后在組內(nèi)排名。
今天突然碰到sum(...) over( partition by?... ORDER BY?... ),居然搞不清除怎么執(zhí)行的,所以查了些資料,做了下實(shí)操。
1. 從最簡(jiǎn)單的開始
sum(...) over( ),對(duì)所有行求和
sum(...) over( order by ...?),和 =? 第一行 到 與當(dāng)前行同序號(hào)行的最后一行的所有值求和,文字不太好理解,請(qǐng)看下圖的算法解析。
with aa as ( SELECT 1 a,1 b, 3 c FROM dual union SELECT 2 a,2 b, 3 c FROM dual union SELECT 3 a,3 b, 3 c FROM dual union SELECT 4 a,4 b, 3 c FROM dual union SELECT 5 a,5 b, 3 c FROM dual union SELECT 6 a,5 b, 3 c FROM dual union SELECT 7 a,2 b, 3 c FROM dual union SELECT 8 a,2 b, 8 c FROM dual union SELECT 9 a,3 b, 3 c FROM dual ) SELECT a,b,c, sum(c) over(order by b) sum1,--有排序,求和當(dāng)前行所在順序號(hào)的C列所有值 sum(c) over() sum2--無排序,求和 C列所有值 from aa
補(bǔ)充 by 松門一枝花:
WITH aa AS( SELECT 1 a,1 b, 3 c FROM dualUNIONSELECT 2 a,2 b, 3 c FROM dualUNIONSELECT 3 a,3 b, 3 c FROM dualUNIONSELECT 4 a,4 b, 3 c FROM dualUNIONSELECT 5 a,5 b, 3 c FROM dualUNIONSELECT 6 a,5 b, 3 c FROM dualUNIONSELECT 7 a,2 b, 3 c FROM dualUNIONSELECT 8 a,2 b, 8 c FROM dualUNIONSELECT 9 a,3 b, 3 c FROM dual) SELECT a,b,c,SUM(c) over(order by a) sum1,--有排序,求和當(dāng)前行所在順序號(hào)的C列所有值--【博主新增的】SUM(c) over(order by b) sum2,--有排序,求和當(dāng)前行所在順序號(hào)的C列所有值SUM(c) over() sum3 FROM aa order by a; --無排序,求和 C列所有值
2. 與 partition by 結(jié)合
sum(...) over( partition by...?),同組內(nèi)所行求和
sum(...) over( partition by...?order by ... ),同第1點(diǎn)中的排序求和原理,只是范圍限制在組內(nèi)
with aa as ( SELECT 1 a,1 b, 3 c FROM dual union SELECT 2 a,2 b, 3 c FROM dual union SELECT 3 a,3 b, 3 c FROM dual union SELECT 4 a,4 b, 3 c FROM dual union SELECT 5 a,5 b, 3 c FROM dual union SELECT 6 a,5 b, 3 c FROM dual union SELECT 7 a,2 b, 3 c FROM dual union SELECT 7 a,2 b, 8 c FROM dual union SELECT 9 a,3 b, 3 c FROM dual ) SELECT a,b,c,sum(c) over( partition by b ) partition_sum, sum(c) over( partition by b order by a desc) partition_order_sumFROM aa;
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