需要通過

Question

SELECT WK_NBR,region_cd,plant_cd, POS_GAP_1DAY 
    FROM(SELECT wk_nbr,region_cd,plant_cd, 
     round((CASE WHEN SUM(gap_1day)>0 THEN SUM(gap_1day) ELSE 0 END ),2) AS POS_GAP_1DAY 
     FROM 
     table a 
     group by wk_nbr,region_cd,plant_cd) 

解決集團在上面的查詢中，我面臨的問題是POS_GAP_1DAY應該總結只有在列，但上面的查詢正數不會得到正確的答案，因爲我查SUM(gap_1day)>0。

問題是我只需要分組wk_nbr,region_cd,plant_cd。

請建議

Answer 1

進行當場，所以我不知道，如果它的工作原理，但...

SELECT wk_nbr,region_cd,plant_cd, 
     round(SUM(CASE WHEN GAP_1DAY>0 then GAP_1DAY else 0 end),2) AS POS_GAP_1DAY 
    FROM 
    table a 
    group by wk_nbr,region_cd,plant_cd 

你想要的東西沒有做？

Answer 2

嘗試改變

round((CASE WHEN SUM(gap_1day)>0 THEN SUM(gap_1day) ELSE 0 END ),2) AS POS_GAP_1DAY 

到

ROUND(SUM(CASE WHEN SIGN(gap_1day) > 0 THEN gap_1day ELSE 0 END), 2) AS POS_GAP_1DAY