正在进行中的 GROUP BY 函数 OpenEdge答案

【问题标题】：GROUP BY Function in Progress OpenEdge正在进行中的 GROUP BY 函数 OpenEdge
【发布时间】：2014-06-26 14:22:48
【问题描述】：

我正在尝试找出与“GROUP BY”函数等效的进度。我正在尝试编写一个程序来列出多个位置的产品库存信息。现在，它正在列出两次产品信息；每个位置一次。我尝试了 BREAK BY 功能，但没有成功。我当前和所需的输出和代码如下：

电流输出：

期望的输出：

DEF INPUT PARAMETER ip-um AS CHARACTER NO-UNDO.

MESSAGE
  "ProdCode" + "^" +
  "ProdName" + "^" +
  "ProdUM" + "^" +
  "GrossPkgdWeight" + "^" +
  "QtyOH - LOC1" + "^" +
  "QtyOH - LOC2"
  SKIP.

FOR EACH product-um WHERE
     product-um.gross-pkgd-weight <= 0.0000
     NO-LOCK,
EACH product WHERE
     product.product-key = product-um.product-key AND
     product.can-be-sold = YES
     NO-LOCK,
EACH inventory WHERE
     inventory.product-key = product.product-key AND
     inventory.qoh > 0 AND
     inventory.level = 2
     NO-LOCK,
EACH um WHERE
     um.um-key = product-um.um-key AND
     um.um = ip-um
     NO-LOCK
BREAK BY product.product-code:

MESSAGE
     product.product-code + "^" +
     product.product-name + "^" +
     um.um-code + "^" +
     STRING(product-um.gross-pkgd-weight) + "^" +
     IF inventory.level-key-2 = '00000001' THEN STRING(inventory.qoh) ELSE "0" 
     + "^" + IF inventory.level-key-2 = '00000002' THEN STRING(inventory.qoh) ELSE "0" 
     SKIP.
END.

【问题讨论】：

标签： progress-4gl openedge

【解决方案1】：

因为您累积 Inventory.qoh 依赖于 inventory.level-key-2，所以 ACCUMULATE stmt 并不可行，因此手动编码累积将是最佳选择

DEFINE VARIABLE loc1 AS INTEGER NO-UNDO.
DEFINE VARIABLE loc2 AS INTEGER NO-UNDO.
FOR EACH  product-um NO-LOCK
    WHERE product-um.gross-pkgd-weight <= 0.0000
,
EACH  product NO-LOCK
WHERE product.product-key = product-um.product-key
  AND product.can-be-sold = YES
,
EACH  inventory NO-LOCK
WHERE inventory.product-key = product.product-key
  AND inventory.product-code = product.product-code
  AND inventory.qoh > 0
  AND inventory.level = 2
,
EACH  um NO-LOCK
WHERE um.um-key = product-um.um-key
  and um.um = ip-um
BREAK
   BY product.product-code:

  CASE (inventory.level-key-2):
    WHEN "00000001"
      THEN loc1 = loc1 + inventory.qoh.
    WHEN "00000002"
      THEN loc2 = loc2 + inventory.qoh.
   END CASE.
  IF LAST-OF(product.product-code)
  THEN DO:
    MESSAGE
      product.product-code + "^" +
      product.product-name + "^" +
      um.um-code + "^" +
      STRING(product-um.gross-pkgd-weight) + "^" +
      STRING(loc1) + "^" +
      STRING(loc2)
      SKIP.
    ASSIGN
      loc1 = 0
      loc2 = 0
    .
  END.
END.

【讨论】：

【解决方案2】：

BREAK BY 告诉编译器标记 FOR EACH 何时到达中断组的开始或结束。要检测这些更改，您需要使用以下函数之一来检测该更改：FIRST(table.field )、FIRST-OF(table.field)、LAST(table.field) 和 LAST-OF(table.field)。

一旦所需的函数返回 true，您就可以使用 ABL 提供的函数（如 ACCUMULATE、COUNT、TOTAL 等）来显示所需的结果。就我个人而言，我发现这些概念有点难以理解，所以我声明了一些局部变量并以这种方式进行总计。

【讨论】：