Question

我刚刚学习COBOL;我正在写一个简单回应用户输入的程序。我已将变量定义为：

User-Input PIC X(30).

稍后当我接受用户输入，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。是否有标准方法（如Ruby的str.strip！）来删除多余的空格？

Answer 1

人们希望以更优雅的方式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语句依赖于左侧正确评估UNTIL条款以避免越界在USER-INPUT仅包含的情况下订阅空白。

Answer 2

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;领先或尾随。 cobc -h formore info。

Answer 3

如果您使用OpenVMS，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.

Answer 4

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

Answer 5

有三种方法可以做到这一点。

使用COBOL函数确定字符串的“长度”。这是几个功能的混合。这是我首选的方法，但需要声明额外的变量。
编写自己的函数以获取“长度”。
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，例如三个空格或一个低值字符。

此示例代码演示了所有这三个。

   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：

 -- METHOD ONE --            
RALPH WIGGAM LIKES LEARNDING.
 -- METHOD TWO --            
RALPH WIGGAM LIKES LEARNDING.
 -- METHOD THREE, NAIVE --   
RALPH LIKES.                 
 -- METHOD THREE, OK --      
RALPH WIGGAM LIKES LEARNDING.

从cobol输入中删除空格的标准方法？

5 个答案: