6

我只是在学习 COBOL;我正在编写一个简单地回显用户输入的程序。我已将变量定义为:

User-Input PIC X(30).

后来当我接受用户输入,然后显示用户输入“加上一些额外的文本”时,它有一堆空格来填充 30 个字符。有没有一种标准的方法(比如 Ruby 的 str.strip!)来删除多余的空格?

4

5 回答 5

6

人们希望有一种更优雅的方式来简单地修剪文本字符串,但这几乎是标准解决方案......修剪部分在 SHOW-TEXT 段落中完成。


      *************************************                    
      * TRIM A STRING... THE HARD WAY...                       
      *************************************                    
       IDENTIFICATION DIVISION.                                
       PROGRAM-ID. TESTX.                                      
       DATA DIVISION.                                          
       WORKING-STORAGE SECTION.                                
       01  USER-INPUT         PIC X(30).                       
       01  I                  PIC S9(4) BINARY.                
       PROCEDURE DIVISION.                                     
           MOVE SPACES TO USER-INPUT                           
           PERFORM SHOW-TEXT                                   

           MOVE '  A B C' TO USER-INPUT                        
           PERFORM SHOW-TEXT                                   

           MOVE 'USE ALL 30 CHARACTERS -------X' TO USER-INPUT 
           PERFORM SHOW-TEXT                                 
           GOBACK                                            
           .                                                 
       SHOW-TEXT.                                            
           PERFORM VARYING I FROM LENGTH OF USER-INPUT BY -1 
                     UNTIL I LESS THAN 1 OR USER-INPUT(I:1) NOT = ' '
           END-PERFORM                                       
           IF I > ZERO                                       
              DISPLAY USER-INPUT(1:I) '@ OTHER STUFF'        
           ELSE                                              
              DISPLAY '@ OTHER STUFF'                        
           END-IF                                            
           .                                                 

产生以下输出:


@ OTHER STUFF                              
  A B C@ OTHER STUFF                       
USE ALL 30 CHARACTERS -------X@ OTHER STUFF

请注意,PERFORM VARYING 语句依赖于 UNTIL 子句的从左到右评估,以避免在 USER-INPUT 仅包含空格的情况下越界下标。

于 2010-01-12T17:22:22.987 回答
3

使用 OpenCOBOL 1.1 或更高版本。

 Identification division.
 Program-id. 'trimtest'.
*> Compile:
*> cobc -x -free -ffunctions-all  TrimTest.cbl
*>
 Data division.
 Working-Storage Section.
1 myBigStr Pic X(32768) Value Spaces.

 Procedure Division.

Display "Enter Something? " With no advancing.
Accept myBigStr.
Display "[" Trim(myBigStr) "]".
Goback.

修剪功能也有选项;领先或落后。cobc -h 了解更多信息。

于 2012-09-06T21:48:56.713 回答
1

如果您在 OpenVMS 上工作,这是一个解决方案:

   01 WS-STRING-LENGTH                 PIC S9(04) COMP.

   CALL "STR$TRIM" USING BY DESCRIPTOR user_output,
                                       user_input,
                                       BY REFERENCE WS-STRING-LENGTH.
于 2010-01-12T16:34:56.420 回答
0

更通用的解决方案:

01 length pic 99.

perform varying length from 1 by 1 
  until length > 30 or user-input[length] = space
end-perform.
if length > 30
  display user-input 'plus some extra text'
else
  display user-input[1:length] 'plus some extra text'
end-if.

未经测试,目前我手头没有编译器

于 2010-01-12T16:58:45.017 回答
0

您可以通过三种方式做到这一点。

  1. 使用 COBOL 函数来确定字符串的“长度”。这是几个功能的混合。这是我的首选方法,但需要声明额外的变量。
  2. 编写自己的函数来获取“长度”。
  3. 使用“终止”字符串的知识。您必须知道哪些关键字符表示字符串结尾,例如三个空格或低值字符。

此示例代码演示了所有三个。

   IDENTIFICATION DIVISION.
   PROGRAM-ID. TESTPROG.
   DATA DIVISION.
   WORKING-STORAGE SECTION.
   01 ONE-A     PIC X(20) VALUE 'RALPH WIGGAM'.
   01 ONE-A-TLY PIC 9(02) VALUE ZERO.
   01 ONE-A-LEN PIC 9(02) VALUE ZERO.
   01 ONE-B     PIC X(20) VALUE 'LIKES LEARNDING'.
   01 ONE-B-TLY PIC 9(02) VALUE ZERO.
   01 ONE-B-LEN PIC 9(02) VALUE ZERO.
   01 TWO-A     PIC X(20) VALUE 'RALPH WIGGAM'.
   01 TWO-A-LEN PIC 9(02) VALUE ZERO.
   01 TWO-B     PIC X(20) VALUE 'LIKES LEARNDING'.
   01 TWO-B-LEN PIC 9(02) VALUE ZERO.
   01 THREE-A   PIC X(20) VALUE 'RALPH WIGGAM'.
   01 THREE-B   PIC X(20) VALUE 'LIKES LEARNDING'.
   01 THREE-C   PIC X(80) VALUE SPACES.
   PROCEDURE DIVISION.

       DISPLAY ' -- METHOD ONE -- '
       INSPECT FUNCTION REVERSE(ONE-A)
        TALLYING ONE-A-TLY FOR LEADING SPACES.
       SUBTRACT ONE-A-TLY FROM LENGTH OF ONE-A GIVING ONE-A-LEN.
       INSPECT FUNCTION REVERSE(ONE-B)
        TALLYING ONE-B-TLY FOR LEADING SPACES.
       SUBTRACT ONE-B-TLY FROM LENGTH OF ONE-A GIVING ONE-B-LEN.
       DISPLAY ONE-A(1:ONE-A-LEN)
               ' ' ONE-B(1:ONE-B-LEN)
               '.'.

       DISPLAY ' -- METHOD TWO -- '
       PERFORM VARYING TWO-A-LEN FROM LENGTH OF TWO-A BY -1
        UNTIL TWO-A-LEN < 1 OR TWO-A(TWO-A-LEN:1) > SPACE
       END-PERFORM.
       PERFORM VARYING TWO-B-LEN FROM LENGTH OF TWO-B BY -1
        UNTIL TWO-B-LEN < 1 OR TWO-B(TWO-B-LEN:1) > SPACE
       END-PERFORM.
       DISPLAY TWO-A(1:TWO-A-LEN)
               ' ' TWO-B(1:TWO-B-LEN)
               '.'.

       DISPLAY ' -- METHOD THREE, NAIVE -- '
  *    DELIMITING BY JUST ANY SPACES ISN'T GOOD ENOUGH.
       STRING THREE-A DELIMITED BY SPACES
              ' ' DELIMITED BY SIZE
              THREE-B DELIMITED BY SPACES
              '.' DELIMITED BY SIZE
              INTO THREE-C.
       DISPLAY THREE-C.

       DISPLAY ' -- METHOD THREE, OK -- '
       STRING THREE-A DELIMITED BY '  '
              ' ' DELIMITED BY SIZE
              THREE-B DELIMITED BY '  '
              '.' DELIMITED BY SIZE
              INTO THREE-C.
       DISPLAY THREE-C.
   EXIT-PROG.
       STOP RUN.

输出如下所示:

 -- METHOD ONE --            
RALPH WIGGAM LIKES LEARNDING.
 -- METHOD TWO --            
RALPH WIGGAM LIKES LEARNDING.
 -- METHOD THREE, NAIVE --   
RALPH LIKES.                 
 -- METHOD THREE, OK --      
RALPH WIGGAM LIKES LEARNDING.
于 2012-04-06T21:47:47.390 回答