Loops 如何在COBOL中对表进行排序？

我正在学习COBOL，并且我很难找出如何对这个表进行排序。我甚至不认为我正确地实现了这个表，所以任何关于如何改进代码的帮助都是非常好的。我迷茫了

IDENTIFICATION DIVISION.
PROGRAM-ID. STUDENT.

DATA DIVISION. 
WORKING-STORAGE SECTION.
01 CLASSROOM-TABLE.
    05 STUDENT OCCURS 5 TIMES. 
        10 STUDENT1.
            15 STUDENT1-N PIC A(25).
            15 STUDENT1-A PIC 99. 
        10 STUDENT2.
            15 STUDENT2-N PIC A(25).
            15 STUDENT2-A PIC 99. 
        10 STUDENT3.
            15 STUDENT3-N PIC A(25).
            15 STUDENT3-A PIC 99. 
        10 STUDENT4.
            15 STUDENT4-N PIC A(25).
            15 STUDENT4-A PIC 99. 
        10 TEMP-STUDENT.
            15 STUDENT-N PIC A(25).
FIND.

END-METHOD.
            15 STUDENT-A PIC 99. 
01 I PIC 9 VALUE 0. 
01 J PIC 9 VALUE 1.

PROCEDURE DIVISION. 
MAIN-PARA.
    MOVE "MICHAELA" TO STUDENT (1) (1:25).
    MOVE 21 TO STUDENT (1) (26:2).
    MOVE "KEVIN" TO STUDENT (2) (1:25).
    MOVE 25 TO STUDENT (2) (26:2).
    MOVE "KENNY" TO STUDENT (3) (1:25).
    MOVE 16 TO STUDENT (3) (26:2). 
    MOVE "ANDREA" TO STUDENT (4) (1:25).
    MOVE 18 TO STUDENT (4) (26:2). 

    PERFORM PUT-ORDER.
    PERFORM PRINT.
STOP RUN. 

PUT-ORDER.
PERFORM VARYING I FROM 1 BY 1 UNTIL I > 4
    ADD 1 TO J  
    IF STUDENT(I)(26:2) > STUDENT(J)(26:2)
        ADD 1 TO I
        DISPLAY "INSIDE SORT"
        MOVE STUDENT(I)(26:2) TO STUDENT(5)(26:2)
        MOVE STUDENT(J)(26:2) TO STUDENT(I)(26:2)
        MOVE STUDENT(5)(26:2) TO STUDENT(J)(26:2)
    END-IF
END-PERFORM.

DISPLAY "SORT IS DONE".

END-METHOD.

PRINT.
    DISPLAY STUDENT (1).
    DISPLAY STUDENT (2).
    DISPLAY STUDENT (3).
    DISPLAY STUDENT (4).
    END-METHOD.

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我知道代码正在进行排序，因为它同时打印内部排序和完成排序。但当它再次打印表格时，一切都是一样的。我认为我没有正确地遍历表，但我尝试过的所有其他方法都会给我带来错误

J需要一个内环，而我只需要到3

PERFORM VARYING I FROM 1 BY 1 UNTIL I > 3
    PERFORM VARYING J FROM I+1 BY 1 UNTIL J > 4
        IF ...
        END-IF
    END-PERFORM
END-PEFORM

删除“将1添加到I”

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我不知道是否需要进行其他更改。

这既是一个答案，也是编写Cobol的指南

01 CLASSROOM-TABLE.
   05 STUDENT OCCURS 5 TIMES. 
      10 STUDENT-Name         PIC A(25).
      10 STUDENT-Age          PIC 99. 

这个过程就变成了

MOVE "MICHAELA" TO STUDENT-Name (1).
MOVE 21         TO STUDENT-Age  (1).
MOVE "KEVIN"    TO STUDENT-Name (2).
MOVE 25         TO STUDENT-Age  (2).
  ...

你也可以这样做

01 CLASSROOM-TABLE.
   05 Student-Table.
      10 STUDENT OCCURS 5 TIMES. 
         15 STUDENT-Name      PIC A(25).
         15 STUDENT-Age       PIC 99. 

  05 redefines Student-Table.
     10 filler                Pic X(25)  value 'MICHAELA'
     10 filler                pic 99     value 21. 
     10 filler                Pic X(25)  value 'KEVIN'
     10 filler                pic 99     value 25. 
       ....

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您还需要定义一个临时学生

  05 Temp-Student            Pic x(27).

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Cobol有一个

e、 g

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如果要在代码中进行排序，最容易实现的排序过程是。你可以

但基本上正如rcgldr所说，您需要2个循环

perform varying i from 1 by 1 until i > 4
   Add 1           to i giving j
   perform  until j > 5
       if Student-Age(i) > Student-Age(j)
          Move Student(i)     to Temp-Student
          Move Student(j)     to Student(i)
          Move Temp-Student   to Student(j)
       end-if
       add 1                  to j
   end-perform
end-perform 

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你为什么要做学生（I）（26:2）？，你应该只做学生名字（I）我在第二次表演中，I+1有错误，有一个意想不到的文字。我试图通过创建另一个值为1的变量来解决这个问题，但是错误是意外的“+”，预期为。括号也不能解决这个问题。这有什么问题？另外，以重定义的方式编写表是否更好？您肯定需要i+1，在+周围有空格；但我怀疑你需要按照上面更新的方式来做；我还修复了移动中的第二次执行和一个索引

perform varying i from 1 by 1 until i > 4
   Add 1           to i giving j
   perform  until j > 5
       if Student-Age(i) > Student-Age(j)
          Move Student(i)     to Temp-Student
          Move Student(j)     to Student(i)
          Move Temp-Student   to Student(j)
       end-if
       add 1                  to j
   end-perform
end-perform