OPENEDGE 字母数字序列函数
Posted
技术标签:
【中文标题】OPENEDGE 字母数字序列函数【英文标题】：OPENEDGE ALPHANUMERIC SEQUENCE FUNCTION 【发布时间】：2017-12-12 15:01:01 【问题描述】：我已经为字母数字编号 en PROGRESS OPENEDGE 创建了这个算法。我看到的问题是它完全是连续的，当序列增长时它会变得更慢。我想看看是否有办法重新排列函数，这样无论输入参数上给出哪个数字都会有效。
代码如下：
/* LOAN-ORDER-FUNCTION.i */
DEF VAR i-NUMBER-IN AS INT.
DEF VAR o-order AS CHAR.
DEF VAR cnt AS INTEGER.
DEF VAR NUMERAL AS INTEGER.
DEF VAR CODE-OUT AS CHAR FORMAT "X(5)".
DEF VAR LETTERs1 AS CHAR EXTENT 24
INITIAL ["A","B","D","E","F","G","H","I","J","K","L","M","N","O","P","R","S","T","U","V","W","X","Y","Z"].
DEF VAR LETTERs2 AS CHAR EXTENT 26
INITIAL ["A","B","C","D","E","F","G","H","I","J","K","L","M","N","O","P","Q","R","S","T","U","V","W","X","Y","Z"].
FUNCTION BIG-NUMBER RETURNS CHAR (INPUT COMPANY AS CHAR, INPUT NUMBER-IN AS INTEGER):
DEF VAR LETTER1 AS integer INITIAL 1 .
DEF VAR LETTER2 AS INTEGER INITIAL 1 .
DEF VAR LETTER3 AS INTEGER INITIAL 1 .
DEF VAR i AS integer INITIAL 1 NO-UNDO.
DEF VAR j AS integer INITIAL 1 NO-UNDO.
DEF VAR k AS integer INITIAL 1 NO-UNDO.
DEF VAR CODIGO AS CHAR.
DEF VAR in-letter2 AS INT NO-UNDO.
DEF VAR in-letter1 AS INT NO-UNDO.
CNT = 0.
IF NUMBER-IN < 100000 THEN
RETURN COMPANY + STRING(NUMBER-IN,"99999").
REPEAT LETTER1 = 1 TO 24:
DO i = 0 TO 9999:
CODIGO = COMPANY + LETTERS1[LETTER1] + string(i,"9999").
IF CNT + 100000 = NUMBER-IN THEN
RETURN CODIGO.
cnt = cnt + 1.
END.
DO i = 0 TO 999:
CODIGO = COMPANY + LETTERS1[LETTER1] + LETTERS1[LETTER2] + string(i,"999").
IF CNT + 100000 = NUMBER-IN THEN
RETURN CODIGO.
cnt = cnt + 1.
END.
DO letter2 = 1 TO 26:
DO letter3 = 1 TO 26:
DO i = 0 TO 99:
CODIGO = COMPANY + LETTERS1[LETTER1] + LETTERS2[LETTER2] + LETTERS2[LETTER3] + string(i,"99").
IF CNT + 100000 = NUMBER-IN THEN
RETURN CODIGO.
cnt = cnt + 1.
END.
END.
END.
ASSIGN letter2 = 1
letter3 = 1.
END.
END FUNCTION.
FUNCTION BIG-TO-NUMBER RETURNS INTEGER (INPUT codigo-in AS CHAR):
DEF VAR LETTER1 AS integer INITIAL 1 .
DEF VAR LETTER2 AS INTEGER INITIAL 1 .
DEF VAR LETTER3 AS INTEGER INITIAL 1 .
DEF VAR i AS integer INITIAL 1 NO-UNDO.
DEF VAR j AS integer INITIAL 1 NO-UNDO.
DEF VAR k AS integer INITIAL 1 NO-UNDO.
DEF VAR codigo AS CHAR.
CNT = 0.
IF codigo-in < "AA0000" THEN
RETURN integer(SUBSTRING(codigo-in, 2)).
REPEAT LETTER1 = 1 TO 24:
DO i = 0 TO 9999:
CODIGO = COMPANY + LETTERS1[LETTER1] + string(i,"9999").
IF CODIGO = codigo-IN THEN
RETURN CNT + 100000.
cnt = cnt + 1.
END.
DO i = 0 TO 999:
CODIGO = COMPANY + LETTERS1[LETTER1] + LETTERS1[LETTER2] + string(i,"999").
IF CODIGO = codigo-IN THEN
RETURN CNT + 100000.
cnt = cnt + 1.
END.
DO letter2 = 1 TO 26:
DO letter3 = 1 TO 26:
DO i = 0 TO 99:
CODIGO = COMPANY + LETTERS1[LETTER1] + LETTERS2[LETTER2] + LETTERS2[LETTER3] + string(i,"99").
IF CODIGO = codigo-IN THEN
RETURN CNT + 100000.
cnt = cnt + 1.
END.
END.
END.
ASSIGN letter2 = 1
letter3 = 1.
END.
END FUNCTION.
提前感谢您的时间和精力，
雨果
hugoyamil@yahoo.com
波多黎各
【问题讨论】：
如果你想让它快速抛弃字母，只使用 int64。这将允许您使用内置序列。另外一个由多个实体（公司+序列）组成的字段是违反规范化的，很可能导致下游数据质量问题。 0。您的示例无法编译 - 1. 这将有助于添加一些预期的输入和输出值 2. 所以您真的在使用蛮力方法来获取数字？！？！？阅读 ASC 和 CHR 函数的作用。 好的，这是因为它作为包含文件运行，请在顶部添加 DEF VAR COMPANY 作为 CHAR 首字母“A”。 @tom Bascom @Stefan Drissen 向发票功能添加其他 var def var in-data 作为 int64 不可撤销。 def var iter as integer no-undo。最后让输入或生成数字重复：设置数据。 display big-number(company,in-data) big-to-number(big-number(company,in-data). end. 或者可能重复 iter = 123456 到 234567: display big-number(company,iter) big -to-number(big-number(company,iter)).end. 【参考方案1】：您可以尝试下面的代码来生成字母数字序列（短、快、简单），
DEFINE VARIABLE chText AS CHARACTER NO-UNDO.
DEFINE VARIABLE inLoop1 AS INTEGER NO-UNDO.
DEFINE VARIABLE inLoop2 AS INTEGER NO-UNDO.
DEFINE VARIABLE inLength AS INTEGER NO-UNDO.
DEFINE VARIABLE chLetter AS CHARACTER NO-UNDO.
DEFINE VARIABLE loNext AS LOGICAL NO-UNDO.
SET chtext FORMAT "X(15)".
inLength = LENGTH(chtext).
DO inLoop1 = 1 TO 1000:
MESSAGE chtext
VIEW-AS ALERT-BOX INFO BUTTONS OK.
DO inLoop2 = 1 TO inLength:
chLetter = SUBSTRING(chtext, (inLength + 1 - inLoop2), 1).
IF chLetter = "Z" THEN
ASSIGN chtext = SUBSTRING(chtext,1,(inLength - inLoop2)) + STRING(CHR(ASC(chLetter) - 25)) + SUBSTRING(chtext,(inLength + 2 - inLoop2))
loNext = TRUE.
ELSE IF chLetter = "9" THEN
ASSIGN chtext = SUBSTRING(chtext,1,(inLength - inLoop2)) + STRING(CHR(ASC(chLetter) - 9)) + SUBSTRING(chtext,(inLength + 2 - inLoop2))
loNext = TRUE.
ELSE
ASSIGN chtext = SUBSTRING(chtext,1,(inLength - inLoop2)) + STRING(CHR(ASC(chLetter) + 1)) + SUBSTRING(chtext,(inLength + 2 - inLoop2))
loNext = FALSE.
IF NOT loNext THEN
LEAVE.
END.
END.
将其放入您的函数中并传递序列字段的值。 我使用do循环创建了1000个进行测试；你需要删除它。
【讨论】：
以上是关于OPENEDGE 字母数字序列函数的主要内容，如果未能解决你的问题，请参考以下文章
怎样用Excel随机函数rand()生成随机数字和大写字母，能自动产生4位混合的
excel 如何自动生成32位的大写字母跟数字组合的随机序列，类似于验证码的那种随机的