使用連接的分層查詢通過

Question

以下是與我的真實場景類似的示例上下文。

產品：XYZ QTY:1

需要原料B,0.002和半成品項目A,0.001。生產我需要的原料J,0.1和半成品K,0.9。有產品K我以前我需要原材料G 0.004 enter code here和T 0.005。

我需要得到所需的全部原材料的累計產量10產品數量XYZ的結果。

Answer 1

試試這個：

SELECT component AS material, 10 * quantity AS quantity 
    FROM (SELECT component, quantity, 
       CASE WHEN CONNECT_BY_ISLEAF = 1 THEN 'Raw' ELSE 'Semi-Finished' END AS type 
    FROM bill_of_materials 
    START WITH item = 'XYZ' CONNECT BY PRIOR component = item) 
WHERE type = 'Raw' 

上SQL Fiddle的例子給出了：

J |  1 
G |  0.04 
T |  0.05 
B |  0.02 

Answer 2

正如@KenGeis在評論中提到，爲Oracle 11g中，您可以使用遞歸查詢：

with t (p, i , q) as (
    select product, ingredient, qty from test where product = 'XYZ' 
    union all 
    select product, ingredient, qty*q from test, t where product = i) 
select i, sum(q) qty from t 
    where not exists (select 1 from test where product = i) group by i; 

如果由於某些原因您需要connect by版本，這裏是我的嘗試：

with t1 as (
    select ingredient i, sys_connect_by_path(ingredient, '/') path1, 
     sys_connect_by_path(qty, '/') path2 
    from test where connect_by_isleaf = 1 connect by prior ingredient = product 
    start with product = 'XYZ'), 
t2 as (
    select i, path1, path2, trim(regexp_substr(path2, '[^/]+', 1, lines.column_value)) val 
    from t1, 
     table (cast (multiset(
     select level from dual connect by level < regexp_count(t1.path2, '/')+1 
     ) as sys.ODCINumberList)) lines) 
select i, sum(s) qty 
    from (select i, exp(sum(ln(val))) s from t2 group by i, path1) group by i 

^{SQL Fiddle demo for both queries}

子查詢t1生成所需的原材料，並在列PATH2名單 - 這是我們需要乘以因素。 t2未轉義這些值，如果 是兩個使用相同原材料的半成品，則最終查詢執行乘法和組結果。 對於乘法我用從this SO問題的答案。