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更新

我需要获得所有组件的 BOM，目前使用 Tc。CK86 ，但这并没有提供足够的信息，所以我想通过 SQL 来完成，我来自 oracle 背景，不知道如何在 DB2 R/3 中完成这项工作，我无权访问查询生成器或 quickview ，但我确实可以通过 SQL 进行读取访问，我目前正在尝试找出一种使用表获取此信息的方法：

  • MAST 材料到 BOM 链接
  • STKO BOM 表头
  • STPO BOM 项目

你们中有人对此有解决方案吗？

在 Oracle 中，我做了如下的事情：

SELECT DISTINCT LEVEL
    ,sys_connect_by_path(msil.segment1, ' @ ') AS "BOM TREE"
    ,msi.segment1
    ,lpad(' ', LEVEL, '') || msil.segment1 Cod_Component
    ,msil.item_type
    ,msil.description Desc_Component
    ,BIC.component_quantity
    ,msiL.primary_unit_of_measure
FROM mtl_system_items msi
    ,bom_bill_of_materials bom
    ,BOM_INVENTORY_COMPONENTS BIC
    ,MTL_SYSTEM_ITEMS MSIL
WHERE msi.organization_id = 332
    AND BOM.ASSEMBLY_ITEM_ID = MSI.INVENTORY_ITEM_ID
    AND BOM.ORGANIZATION_ID = MSI.ORGANIZATION_id
    AND bom.bill_sequence_id = bic.bill_sequence_id
    AND nvl(bic.disable_date, sysdate) >= SYSDATE
    AND BIC.component_ITEM_ID = MSIL.INVENTORY_ITEM_ID
    AND Bom.ORGANIZATION_ID = MSIL.ORGANIZATION_ID
    AND msil.inventory_item_status_code = 'Active'
    AND msi.inventory_item_status_code = 'Active' 
    connect BY prior bic.component_item_id = bom.assembly_item_id
    START WITH msi.segment1 = trim(:parte)
    ORDER BY 2

我正在尝试以下操作，试图保持简单，但无论我尝试什么，它都会在第 18 行给我一个错误，显然在 DB2 中，在我的 oracle 工作示例中，我需要将“连接方式”放在“开始”之后它首先有“连接”，不知道它是否有区别，但无论我怎么写，它都会给我一个错误：“错误 [42601] [IBM][DB2/AIX64] SQL0104N 一个意外的令牌”之前" 在 "ASQ19130' CONNECT BY" 之后发现。预期的标记可能包括："PRIOR"。"

这是我到目前为止得到的：

SELECT DISTINCT level
    ,sys_connect_by_path(msil.stlnr, ' @ ') AS "BOM TREE"
    ,msi.stlnr as parent
    --,lpad(' ', LEVEL, '') || MSIL.MATNR Cod_Component
  --,lpad(' ', LEVEL, '') || MSIL.MATNR as Cod_Component
  ,CAST(SPACE((LEVEL - 1) * 4) || '/' || MSIL.MATNR AS VARCHAR(40)) as Cod_Component
    ,BIC.menge as qty
  ,bic.stlnr as compnumb
    ,msiL.mein as uom
FROM 
   MAST msi
    ,STKO bom
    ,STPO BIC
    ,MAST MSIL
WHERE 
BOM.STLNR = MSI.STLNR
AND BIC.STLNR = MSIL.STLNR 
START WITH msi.MATNR = 'ASQ19130'
CONNECT BY PRIOR BIC.stlnr = bom.stlnr
order by 2
4

1 回答 1

1

这就是最终的结果，它对我有用。

（为修正错误而编辑）

如果有人需要这个：

WITH myquery (
    root,
    matnr,matd,
    bom_tree,
    lvl,
    parent_stlkn,
    stlkn,
    idnrk,
    meins, menge
) AS (
    SELECT
        m.matnr root,
        m.matnr,MAKT.MAKTX Matd,
        p.idnrk bom_tree,
        1 lvl,
        p.stlkn,
        p.stlkn,
        p.idnrk,
        p.meins, p.menge
    FROM
        mast m 
        JOIN stko k ON k.stlnr = m.stlnr AND K.STLAL=M.STLAL
        JOIN stpo p ON p.stlnr = k.stlnr,makt
        where m.stlal='01' /*and k.stlal='01'*/
        and m.matnr=MAKT.matnr
),x (
    root,
    matnr,matd,
    bom_tree,
    lvl,
    parent_stlkn,
    stlkn,
    idnrk,
    meins, menge
) AS (
    SELECT
        root,
        matnr,matd,
        bom_tree,
        lvl,
        parent_stlkn,
        stlkn,
        idnrk,
        meins, menge
    FROM
        myquery
    UNION ALL
    SELECT
        x1.root,
        x2.matnr,x2.matd,
        x1.bom_tree
         || ' @ '
         || x2.idnrk bom_tree,
        x1.lvl + 1 lvl,
        x1.parent_stlkn,
        x2.stlkn,
        x2.idnrk,
        x2.meins,x2.menge
    FROM
        myquery x1,
        myquery x2
    WHERE
        x2.matnr = x1.bom_tree
        
) SELECT 
    x.matnr,matd Description,
    marc.herkl country,
    eina.urzla country2,
    mbew.stprs comp_price,
    bom_tree,
    lvl,
    stlkn,
    idnrk comp,
    x.meins UOM,
    menge qpa
FROM
    x, marc, mbew, eina
WHERE
    root  = :p
    and mbew.matnr=marc.matnr
    and mbew.matnr=eina.matnr
    and marc.werks=1850
    and idnrk=mbew.matnr
    --and x.matnr=mbew.matnr
    and mbew.bwkey=1850
ORDER BY
    root,
    parent_stlkn,
    lvl,
    stlkn
于 2020-02-27T00:06:36.813 回答