我如何从 1 个变量中获得最长的单词答案

【问题标题】：How do i get the longest word from 1 variable我如何从 1 个变量中获得最长的单词
【发布时间】：2020-02-26 10:11:55
【问题描述】：

将变量 cWord 定义为 Character no-undo。

cWord = "Web 开发工具"。

需要的输出

发展

当它只有 1 个变量时，我如何从中获得最长的单词，

顺便说一句，这是一个progress4gl代码

【问题讨论】：

到目前为止你的尝试是什么？它是如何失败的？

标签： openedge progress-4gl

【解决方案1】：

DEFINE VARIABLE cWord          AS CHARACTER   NO-UNDO.
DEFINE VARIABLE iWord          AS INTEGER     NO-UNDO.
DEFINE VARIABLE iLongest       AS INTEGER     NO-UNDO.
DEFINE VARIABLE iLength        AS INTEGER     NO-UNDO.
DEFINE VARIABLE iLongestLength AS INTEGER     NO-UNDO.
DEFINE VARIABLE iEntries       AS INTEGER     NO-UNDO.

ASSIGN cWord = "Web Development Tool"

       iEntries = NUM-ENTRIES (cWord, " ").

DO iWord = 1 TO iEntries:

    ASSIGN iLength = LENGTH (ENTRY (iWord, cWord, " ")) . 

    IF iLength > iLongestLength THEN
    DO:
        ASSIGN iLongest       = iWord
               iLongestLength = iLength .        
    END.
END.

MESSAGE ENTRY (iLongest, cWord, " ") 
    VIEW-AS ALERT-BOX INFORMATION BUTTONS OK.

【讨论】：

【解决方案2】：

仅仅因为我最喜欢的锤子是临时表；-)

def var cword as longchar no-undo init "Web Development Tool".

define temp-table tt no-undo
   field cc as char
   .

temp-table tt:read-json( 
    "longchar", 
    '~{"tt":[~{"cc":"' + replace( cword, ' ', '"},~{"cc":"' ) + '"}]}'
).

for each tt by length( cc ) descending:
    message tt.cc.
    leave.
end.

https://abldojo.services.progress.com:443/#/?shareId=5e56f4a84b1a0f40c34b8c3c

【讨论】：

【解决方案3】：

如果你有两个长度相同的单词，这将首先返回。

DEF VAR iCount      AS INT  NO-UNDO.
DEF VAR cLongest    AS CHAR NO-UNDO.
DEF VAR cString     AS CHAR NO-UNDO INIT 'Web Development Tool'.

DO  iCount = 1 TO NUM-ENTRIES(cString,' '):
    cLongest = (IF LENGTH(ENTRY(iCount,cString,' ')) > LENGTH(cLongest) THEN ENTRY(iCount,cString,' ') ELSE cLongest).
END.

MESSAGE cLongest
    VIEW-AS ALERT-BOX INFO BUTTONS OK.

【讨论】：