使用SORT的大型机JCL记录转置 [英] Mainframes JCL Record transposing using SORT
问题描述
我想在使用SORT(snycsort或DFSORT)将记录转置为列之后进行.它应该可扩展到任意数量的记录.这可能吗?
I want to do following transposing of records into column using SORT (snycsort or DFSORT). It should be scalable to any number of records .Is this possible ?
DE001XYX A
CD100000 B
CD200000 C
DE001KKK A
CD100000 B
DE003ZZZ A
DE001XYX A
CD100000 B
CD200000 C
DE001KKK A
CD100000 B
........
转置
DE001XYX CD100000 CD200000
DE001KKK CD100000
DE003ZZZ
DE001XYX CD100000 CD200000
DE001KKK CD100000
.....
推荐答案
这是我最喜欢的一个问题，并渴望自己解决一段时间.经过研究和试用，我只是想出了一个解决方案.
This is one question which I Favorited and longed to solve it myself for quite some time. After some research & try outs, I've just come up with a solution.
更新: 下面提供了一种简易的方法.它通过1遍本身即可解决问题，而最初的方法需要3遍.
UPDATE: Improvised approach is given below. It solves the problem in 1 pass itself whereas the initial approach took 3 passes.
改进的方法: 不需要3次通过.
Improvised approach: No need of 3 passes.
***************************** Top of Data ******************************
//JOBNAME JOB ('ACCOUNT INFORMATION'),'TRANSPOSE',
// CLASS=2,MSGCLASS=H,NOTIFY=&SYSUID
//STEP1 EXEC PGM=SORT
//SORTIN DD *
DE001XYX A
CD100000 B
CD200000 C
DE001KKK A
CD100000 B
DE003ZZZ A
DE001XYX A
CD100000 B
CD200000 C
DE001KKK A
CD100000 B
//SORTOUT DD SYSOUT=*
//SYSOUT DD SYSOUT=*
//SYSIN DD *
SORT FIELDS=(30,1,CH,A)
INREC IFTHEN=(WHEN=GROUP,BEGIN=(12,1,CH,EQ,C'A'),
END=(12,1,CH,EQ,C'C'),PUSH=(14:ID=1)),
IFTHEN=(WHEN=(12,1,CH,EQ,C'A'),BUILD=(1:1,8,19Z,28:12,1,
30:14,1)),
IFTHEN=(WHEN=(12,1,CH,EQ,C'B'),
BUILD=(1:9Z,10:1,8,18:10Z,28:12,1,30:14,1)),
IFTHEN=(WHEN=(12,1,CH,EQ,C'C'),
BUILD=(1:18Z,19:1,8,28:12,1,30:14,1))
SUM FIELDS=(1,8,10,8,19,8),FORMAT=BI
OUTREC FIELDS=(1:1,8,9:X,10:10,8,18:X,19:19,8)
**************************** Bottom of Data ****************************
结果与OP在其问题中显示的输出相同.
Results in the same output which OP has shown in his/her question.
初始方法:
第1步:
我将WHEN=GROUP与BEGIN，END和PUSH参数一起使用.
Step 1:
I used WHEN=GROUP with BEGIN, END and PUSH parameters.
//JOBNAME JOB ('ACCOUNT NAME'),'TRANSPOSE',
// CLASS=2,MSGCLASS=H,NOTIFY=&SYSUID
//STEP1 EXEC PGM=SORT
//SORTIN DD *
DE001XYX A
CD100000 B
CD200000 C
DE001KKK A
CD100000 B
DE003ZZZ A
DE001XYX A
CD100000 B
CD200000 C
DE001KKK A
CD100000 B
//SORTOUT DD SYSOUT=*
//SYSOUT DD SYSOUT=*
//SYSIN DD *
SORT FIELDS=COPY
OUTREC IFTHEN=(WHEN=GROUP,BEGIN=(12,1,CH,EQ,C'A'),
END=(12,1,CH,EQ,C'C'),PUSH=(14:ID=1))
PUSH=(14:ID=1)表示1的位置14是对每个组递增的标识符.不管先前的组是否以C终止，每个A都指示一个新组.启动新组时，标识符+1.
PUSH=(14:ID=1) means position 14 for 1 is an identifier that increments for each group. Each A indicates a new group regardless of whether the previous group was terminated with C. The identifier is +1'd when a new group is started.
注意:由于我们只允许一个字符作为ID，所以当ID计数器达到10时，位置14会显示一个"0".您也可以为该地址分配1个以上的字节ID.
Note: Since we are only allowing one character for the ID, when the ID counter gets to 10, a '0' will appear in position 14. You may also assign more than 1 byte for the ID.
第1步的输出:
********************************* TOP OF DATA **********************************
DE001XYX A 1
CD100000 B 1
CD200000 C 1
DE001KKK A 2
CD100000 B 2
DE003ZZZ A 3
DE001XYX A 4
CD100000 B 4
CD200000 C 4
DE001KKK A 5
CD100000 B 5
******************************** BOTTOM OF DATA ********************************
步骤2:获取步骤1的输出.分别根据三个键A，B和C修改DE，CD1和CD2记录的位置.
Step 2: Take the output from Step 1. Modify the position of DE, CD1 and CD2 records based on the three keys A, B and C respectively.
//JOBNAME JOB ('ACCOUNT NAME'),'TRANSPOSE',
// CLASS=2,MSGCLASS=H,NOTIFY=&SYSUID
//STEP1 EXEC PGM=SORT
//SORTIN DD *
DE001XYX A 1
CD100000 B 1
CD200000 C 1
DE001KKK A 2
CD100000 B 2
DE003ZZZ A 3
DE001XYX A 4
CD100000 B 4
CD200000 C 4
DE001KKK A 5
CD100000 B 5
//SORTOUT DD SYSOUT=*
//SYSOUT DD SYSOUT=*
//SYSIN DD *
SORT FIELDS=COPY
OUTREC IFTHEN=(WHEN=(12,1,CH,EQ,C'A'),BUILD=(1:1,8,19Z,28:12,1,
30:14,1)),
IFTHEN=(WHEN=(12,1,CH,EQ,C'B'),
BUILD=(1:9Z,10:1,8,18:10Z,28:12,1,30:14,1)),
IFTHEN=(WHEN=(12,1,CH,EQ,C'C'),
BUILD=(1:18Z,19:1,8,28:12,1,30:14,1))
/*
注意:在步骤3中，将二进制零插入其中作为占位符，以便可以用数据填充它们.
Note: Binary zeros are inserted in between, as place holders so that they can be filled in with data, in Step 3.
步骤2的输出
********************************* TOP OF DATA **********************************
DE001XYX A 1
CD100000 B 1
CD200000 C 1
DE001KKK A 2
CD100000 B 2
DE003ZZZ A 3
DE001XYX A 4
CD100000 B 4
CD200000 C 4
DE001KKK A 5
CD100000 B 5
******************************** BOTTOM OF DATA ********************************
在十六进制模式为ON的情况下，您将能够看到二进制零(X'00).
With Hex mode ON, you will be able to see the binary zeros (X'00).
***************************** Top of Data ******************************
DE001XYX A 1
CCFFFEEE0000000000000000000C4F444444444444444444444444444444444444444444
450017870000000000000000000101000000000000000000000000000000000000000000
-----------------------------------------------------------------------
CD100000 B 1
000000000CCFFFFFF0000000000C4F444444444444444444444444444444444444444444
000000000341000000000000000201000000000000000000000000000000000000000000
-----------------------------------------------------------------------
CD200000 C 1
000000000000000000CCFFFFFF4C4F444444444444444444444444444444444444444444
000000000000000000342000000301000000000000000000000000000000000000000000
-----------------------------------------------------------------------
DE001KKK A 2
CCFFFDDD0000000000000000000C4F444444444444444444444444444444444444444444
450012220000000000000000000102000000000000000000000000000000000000000000
-----------------------------------------------------------------------
CD100000 B 2
000000000CCFFFFFF0000000000C4F444444444444444444444444444444444444444444
000000000341000000000000000202000000000000000000000000000000000000000000
-----------------------------------------------------------------------
DE003ZZZ A 3
CCFFFEEE0000000000000000000C4F444444444444444444444444444444444444444444
450039990000000000000000000103000000000000000000000000000000000000000000
-----------------------------------------------------------------------
DE001XYX A 4
CCFFFEEE0000000000000000000C4F444444444444444444444444444444444444444444
450017870000000000000000000104000000000000000000000000000000000000000000
-----------------------------------------------------------------------
CD100000 B 4
000000000CCFFFFFF0000000000C4F444444444444444444444444444444444444444444
000000000341000000000000000204000000000000000000000000000000000000000000
-----------------------------------------------------------------------
CD200000 C 4
000000000000000000CCFFFFFF4C4F444444444444444444444444444444444444444444
000000000000000000342000000304000000000000000000000000000000000000000000
DE001KKK A 5
CCFFFDDD0000000000000000000C4F444444444444444444444444444444444444444444
450012220000000000000000000105000000000000000000000000000000000000000000
-----------------------------------------------------------------------
CD100000 B 5
000000000CCFFFFFF0000000000C4F444444444444444444444444444444444444444444
000000000341000000000000000205000000000000000000000000000000000000000000
-----------------------------------------------------------------------
**************************** Bottom of Data ****************************
第3步:将第2步的输出作为输入，使用SUM FIELDS组合组的DE，CD1和CD2记录，并在第30位的ID字段上使用SORT FIELDS .
Step 3: With output from Step 2 as input, use SUM FIELDS to combine DE, CD1 and CD2 records of the group, with SORT FIELDS on the ID field in 30th Position.
//JOBNAME JOB ('ACCOUNT NAME'),'TRANSPOSE',
// CLASS=2,MSGCLASS=H,NOTIFY=&SYSUID
//STEP1 EXEC PGM=SORT
//SORTIN DD *
DE001XYX A 1
CD100000 B 1
CD200000 C 1
DE001KKK A 2
CD100000 B 2
DE003ZZZ A 3
DE001XYX A 4
CD100000 B 4
CD200000 C 4
DE001KKK A 5
CD100000 B 5
//SORTOUT DD SYSOUT=*
//SYSOUT DD SYSOUT=*
//SYSIN DD *
SORT FIELDS=(30,1,CH,A)
SUM FIELDS=(1,8,10,8,19,8),FORMAT=BI
OUTREC FIELDS=(1:1,8,9:X,10:10,8,18:X,19:19,8)
/*
使用OUTREC FIELDS=(1:1,8,9:X,10:10,8,18:X,19:19,8)，将DE，CD1和CD2字段之间的二进制零替换为空格.
With OUTREC FIELDS=(1:1,8,9:X,10:10,8,18:X,19:19,8), replace binary zeros in between DE, CD1 and CD2 fields with spaces.
第3步的输出结果:开始了！
***************************** Top of Data ******************************
DE001XYX CD100000 CD200000
DE001KKK CD100000
DE003ZZZ
DE001XYX CD100000 CD200000
DE001KKK CD100000
**************************** Bottom of Data ****************************
有关具有二进制零的SUM FIELDS的更多详细信息:
More details on SUM FIELDS with Binary zeros:
二进制加法以两个位模式进行操作.让我们考虑以下记录(十六进制模式为开，因为我想显示如何对十六进制值执行二进制加法运算).
Binary addition operates on two bit patterns. Let's consider the following records (Hex mode is ON as I would like to show how Binary addition is being performed on the Hex values).
****** ********
000001 DE001XYX
CCFFFEEE
45001787
---------------
000002
00000000
00000000
第一个字节的十六进制值X'C4'与第二个记录的第一个字节中的二进制零相加.
Hex value of the first byte, X'C4' is being added with the binary zeros in the 1st byte of 2nd record.
C4
+00
--
C4
X'C4'在EBCDIC中表示字母'D'
X'C4' denotes Letter 'D' in EBCDIC
如果我们将第一个字节中的Binary 0更改为Binary 1(X'01')，结果将有所不同.
If we change the Binary zero in the first byte to Binary one (X'01'), results will vary.
C4
+01
--
C5
X'C5'在EBCDIC中表示字母'E'.
X'C5' denotes Letter 'E' in EBCDIC.
使用二进制零，我们可以对字母数字数据项(EBCDIC字符a-z，A-Z，0-9)执行SUM FIELDS，以将它们分组为一条记录.
With Binary zeros we can perform SUM FIELDS on Alphanumeric data items (EBCDIC characters a-z, A-Z, 0-9) to group them onto one single record.
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