我已创建此算法用于字母数字编号en PROGRESS OPENEDGE。我看到的问题是它是完全顺序的，当序列增长时，它会变得更慢。我想看看是否有一种重新排列功能的方法，无论输入参数给出哪个数字都会有效。
这是代码：
/* 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
答案 0 :(得分：0)
您可以尝试下面的代码来生成字母数字序列（短，快速和简单），
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；您需要将其删除。