我是COBOL的新手，我一直试图从表的输出文本文件中读取记录信息。
我可以接受大多数非comp数据类型，这就是我坚持使用的“ COMP”类型。
我一直试图整天弄清楚这个问题，因此我会尽可能地阅读。
以下日期字段是我无法转换为日期字符串的字段：
05 VALDATE PIC 9(6) COMP
05 PAYDATE PIC 9(6) COMP
05 SYSDATE PIC 9(6) COMP
据我了解，以上所有这些类型在文件中将每个为4个字节。
它们应该是代表YYMMDD的日期，但是数据似乎并不那么小。我研究了EBCDIC并反转了byte[]数据，并使用BitConverter.ToUNIT32()并更改了用于读取文件的编码，但是没有运气。
我读到，计算成整数的日期存储为从1601年1月1日开始的天数，因此为什么下面的代码试图将值添加到1601。（http://www.techtricky.com/cobol-date-functions-list-add-find-duration/）
我的问题是文本文件中的数据不正确，或者我缺少获取类似于YYMMDD的日期的步骤。
以上3个的数据如下：
[ 32] [237] [ 44] [ 4] | 00100000 11101101 00101100 00000100
[ 33] [ 14] [ 32] [237] | 00100001 00001110 00100000 11101101
[131] [ 48] [ 48] [ 48] | 10000011 00110000 00110000 00110000
我是如何打开文件的，我将编码更改为ascii，但没有运气：
using (BinaryReader reader = new BinaryReader(File.Open(nFilePath, FileMode.Open), Encoding.Default))
用于尝试读取COMP字段的代码：
public class DateFromUIntExtractor : LineExtractor
{
public DateFromUIntExtractor() : base(4)
{
}
public override string ExtractText(BinaryReader nReader)
{
// e.g 32,237,44,44, included but commented out things i've tried
byte[] data = nReader.ReadBytes(Length); // Length = 4
//Array.Reverse(data); - Makes num = 552414212
//data = ConvertAsciiToEbcdic(data);
int num = BitConverter.ToUInt32(data, 0);
// in this example num = 70053152
DateTime date = new DateTime(1601,1,1);
date = date.AddDays(num); // Error : num is too big
Extract = date.ToString("yyyyMMdd");
return Extract;
}
}
数据是否格式错误？还是我错过了什么？
我要完成的任务是复制一个COBOL程序，当程序输出一个.dat文件时，该程序将数据从一个定义转换为另一个定义，但以CSV格式。
我对源代码定义的经验不足，是文本文件中的数据是PUA-ICGROUP或PUA-PUGROUP。在COBOL程序中，当PUA-ICGROUP时，它选择PUA-HEADER>PUA-KEY>PUA-RTYPE = "03"，其他都是PUA-PUGROUP。
C-WRITE-START.
IF PUA-RTYPE = 3 THEN
PERFORM C-WRITE-A
ELSE
PERFORM C-WRITE-B
END-IF.
C-WRITE-EXIT.
EXIT.
01 DLRPUARC.
03 PUA-HEADER.
05 PUA-KEY.
07 PUA-CDELIM PIC 99.
07 PUA-SUPNO PIC 9(7).
07 PUA-RTYPE PIC 99.
07 PUA-REF PIC 9(9).
07 PUA-SEQ PIC 999.
05 PUA-ALTKEY.
07 PUA-ACDELIM PIC 99.
07 PUA-ASUPNO PIC 9(7).
07 PUA-ATRNDATE PIC 9(6).
07 PUA-ARTYPE PIC 99.
07 PUA-AREF PIC 9(9).
07 PUA-ASEQ PIC 999.
05 FILLER PIC X(82).
03 PUA-ICGROUP REDEFINES PUA-HEADER.
05 FILLER PIC X(52).
05 PUA-ICEXTREF PIC X(10).
05 PUA-ICORDNO PIC 9(11).
05 PUA-ICVALDATE PIC 9(6) COMP.
05 PUA-ICPAYDATE PIC 9(6) COMP.
05 PUA-ICSYSDATE PIC 9(6) COMP.
05 PUA-ICTRNVAL PIC S9(9).
05 PUA-ICCLRREF PIC 9(6).
05 PUA-ICDELDATE PIC 9(6) COMP.
05 PUA-ICOTHQRY PIC X.
05 PUA-ICPRCQRY PIC X.
05 PUA-ICMRSQRY PIC X.
05 PUA-ICDSCTYPE PIC 9.
05 PUA-ICDSCVAL PIC S9(9) COMP.
05 PUA-ICVATCODE PIC 9.
05 PUA-ICVATAMT PIC S9(8) COMP.
05 PUA-ICTAXAMT PIC S9(8) COMP.
05 PUA-ICMRSREF PIC 9(6).
05 PUA-ICSUBDIV PIC 9.
05 PUA-ICCOSTCTR PIC X(5).
05 PUA-ICSEQIND PIC X.
05 FILLER PIC X(4).
03 PUA-PUGROUP REDEFINES PUA-HEADER.
05 FILLER PIC X(52).
05 PUA-PUEXTREF PIC X(10).
05 PUA-PUORDNO PIC 9(11).
05 PUA-PUVALDATE PIC 9(6) COMP.
05 FILLER PIC XXX.
05 PUA-PUSYSDATE PIC 9(6) COMP.
05 PUA-PUTRNVAL PIC S9(9).
05 PUA-PUCLRREF PIC 9(6).
05 PUA-PUDELDATE PIC 9(6) COMP.
05 PUA-PUOTHQRY PIC X.
05 PUA-PUSUBDIV PIC 9.
05 FILLER PIC X(32).
01 OUT-A-REC.
03 OUT-A-PUA-CDELIM PIC 99.
03 OUT-A-PUA-SUPNO PIC 9(7).
03 OUT-A-PUA-RTYPE PIC 99.
03 OUT-A-PUA-REF PIC 9(9).
03 OUT-A-PUA-SEQ PIC 999.
03 OUT-A-PUA-ATRNDATE PIC 9(8).
03 OUT-A-PUA-ICEXTREF PIC X(10).
03 OUT-A-PUA-ICORDNO PIC 9(11).
03 OUT-A-PUA-ICVALDATE PIC 9(8).
03 OUT-A-PUA-ICPAYDATE PIC 9(8).
03 OUT-A-PUA-ICSYSDATE PIC 9(8).
03 OUT-A-PUA-ICTRNVAL PIC S9(9) SIGN LEADING SEPARATE.
03 OUT-A-PUA-ICCLRREF PIC 9(6).
03 OUT-A-PUA-ICDELDATE PIC 9(8).
03 OUT-A-PUA-ICOTHQRY PIC X.
03 OUT-A-PUA-ICPRCQRY PIC X.
03 OUT-A-PUA-ICMRSQRY PIC X.
03 OUT-A-PUA-ICDSCTYPE PIC 9.
03 OUT-A-PUA-ICDSCVAL PIC S9(9) SIGN LEADING SEPARATE.
03 OUT-A-PUA-ICVATCODE PIC 9.
03 OUT-A-PUA-ICVATAMT PIC S9(8) SIGN LEADING SEPARATE.
03 OUT-A-PUA-ICTAXAMT PIC S9(8) SIGN LEADING SEPARATE.
03 OUT-A-PUA-ICMRSREF PIC 9(6).
03 OUT-A-PUA-ICSUBDIV PIC 9.
03 OUT-A-PUA-ICCOSTCTR PIC X(5).
03 OUT-A-PUA-ICSEQIND PIC X.
03 OUT-A-CTRL-M PIC X.
03 OUT-A-NL PIC X.
FD F-OUTPUTB
LABEL RECORDS OMITTED.
01 OUT-B-REC.
03 OUT-B-PUA-CDELIM PIC 99.
03 OUT-B-PUA-SUPNO PIC 9(7).
03 OUT-B-PUA-RTYPE PIC 99.
03 OUT-B-PUA-REF PIC 9(9).
03 OUT-B-PUA-SEQ PIC 999.
03 OUT-B-PUA-ATRNDATE PIC 9(8).
03 OUT-B-PUA-PUEXTREF PIC X(10).
03 OUT-B-PUA-PUORDNO PIC 9(11).
03 OUT-B-PUA-PUVALDATE PIC 9(8).
03 OUT-B-PUA-PUSYSDATE PIC 9(8).
03 OUT-B-PUA-PUTRNVAL PIC S9(9) SIGN LEADING SEPARATE.
03 OUT-B-PUA-PUCLRREF PIC 9(6).
03 OUT-B-PUA-PUDELDATE PIC 9(8).
03 OUT-B-PUA-PUOTHQRY PIC X.
03 OUT-B-PUA-PUSUBDIV PIC 9.
03 OUT-B-CTRL-M PIC X.
03 OUT-B-NL PIC X.
以下是cobol程序对日期所做的一小部分摘录，无论其天气来源是否为COMP。 （我没有写这段代码）。似乎正在尝试解决2kY问题。
IF PUA-ATRNDATE IS ZERO THEN
MOVE ZERO TO OUT-A-PUA-ATRNDATE
ELSE
MOVE PUA-ATRNDATE TO W-DATE-6DIGIT
MOVE W-DATE-SEG1 TO W-DATE-YY
MOVE W-DATE-SEG2 TO W-DATE-MM
MOVE W-DATE-SEG3 TO W-DATE-DD
IF W-DATE-YY > 50 THEN
MOVE "19" TO W-DATE-CC
ELSE
MOVE "20" TO W-DATE-CC
END-IF
MOVE W-DATE-CCYYMMDD TO OUT-A-PUA-ATRNDATE
END-IF.
MOVE PUA-ICEXTREF TO OUT-A-PUA-ICEXTREF.
MOVE PUA-ICORDNO TO OUT-A-PUA-ICORDNO.
IF PUA-ICVALDATE IS ZERO THEN
MOVE ZERO TO OUT-A-PUA-ICVALDATE
ELSE
MOVE PUA-ICVALDATE TO W-DATE-6DIGIT
MOVE W-DATE-SEG1 TO W-DATE-YY
MOVE W-DATE-SEG2 TO W-DATE-MM
MOVE W-DATE-SEG3 TO W-DATE-DD
IF W-DATE-YY > 50 THEN
MOVE "19" TO W-DATE-CC
ELSE
MOVE "20" TO W-DATE-CC
END-IF
MOVE W-DATE-CCYYMMDD TO OUT-A-PUA-ICVALDATE
END-IF.
01 W-DATE-6DIGIT PIC 9(6).
01 W-DATE-6DIGIT-REDEF REDEFINES W-DATE-6DIGIT.
03 W-DATE-SEG1 PIC 99.
03 W-DATE-SEG2 PIC 99.
03 W-DATE-SEG3 PIC 99.
01 W-DATE-CCYYMMDD PIC 9(8).
01 W-DATE-CCYYMMDD-REDEF REDEFINES W-DATE-CCYYMMDD.
03 W-DATE-CC PIC 99.
03 W-DATE-YY PIC 99.
03 W-DATE-MM PIC 99.
03 W-DATE-DD PIC 99.
从Notepad ++复制，每行从'220 ...'开始，结束行为135，然后转到下一行，这意味着长度为134（？）
2200010010300005463400022000100106062003000054634000062703 09720200000 í,! íƒ00056319D001144ÕšNNN0 1 G¨ 000000197202G
2200010010300005463500022000100106062903000054635000062858 09720200000 í, í" íƒ00082838{050906±RNNN0 1 áð 000000197202G
2200010010300005465500022000100106073003000054655000063378 09720200000 í í† í00179637A050906±RNNN0 1 000000197202G
注意到上面缺少一些符号：
2200010010300005463400022000100106062003000054634000062703 09720200000 í,[EOT]![SO] íƒ00056319D001144[SOH]ÕšNNN0 1 [SOH]G¨ 000000197202G
2200010010300005463500022000100106062903000054635000062858 09720200000 í, í" íƒ00082838{050906[SOH]±RNNN0 1 [SOH]áð 000000197202G
2200010010300005465500022000100106073003000054655000063378 09720200000 í í† í00179637A050906[SOH]±RNNN0 1 [EOT][NAK][EM] 000000197202G
我接受了里克·史密斯（Rick Smith）的以下回答，因为当我将其数据放入正确的日期时间值时。因此，或者我的数据被弄乱了，或者它的其他东西被弄乱了，因为我的数据将引发错误或将来几千年的日期时间值。
我已经能够获得这些日期时间实际应该是的输出CSV：
[使用：int n =（（b [0] << 16）+（b [1] << 8）+ b [2]）;]
HEX: 0x20 0xED 0x2C
BIN: 32 237 44
INT: 2157868 (longer than 6 digit)
Actual DATE: 2006-07-16
HEX: 0x04 0x21 0x0e
BIN: 4 33 14
INT: 270606 (correct but segments are in reverse)
Actual DATE: 2006-06-27
HEX: 0x20 0xED 0x83
BIN: 32 237 131
INT: 2157955 (longer than 6 digits)
Actual DATE: 2006-08-03
原来是坏数据...
答案 0 :(得分：2)
我使用示例数据的第一条记录中包含的值创建了一个具有三个日期字段的COBOL文件。第一个和第三个日期是YYMMDD，第二个日期是DDMMYY。
日期在代码中给出，并且与您尝试读取的日期具有相同的格式，即3字节大尾数二进制。
<?php
$alert = "";
include('class/class.upload.php');
$handle = new upload($_FILES['image_input']);
if ($handle->uploaded) {
$handle->image_resize = true;
$handle->image_x = 1000;
$handle->image_y = 1000;
$handle->image_ratio_fill = true;
$handle->image_convert = 'jpg';
$handle->image_background_color = '#FFFFFF';
$handle->file_new_name_body = '10001';
$handle->process('img/');
if ($handle->processed) {
$alert = '<div class="alert">OK</div>';
$handle->clean();
} else {
$alert = '<div class="alert">ERROR: '.$handle->error.'</div>';
}
}
echo $alert;
?>
<form enctype="multipart/form-data" method="post">
<input type="file" name="image_input">
<input type="submit" name="Submit" value="upload">
</form>
然后，此C＃程序使用 environment division.
input-output section.
file-control.
select out-file assign "dates.dat"
organization sequential
.
data division.
file section.
fd out-file.
01 date-rec.
02 date-1 comp pic 9(6).
02 date-2 comp pic 9(6).
02 date-3 comp pic 9(6).
procedure division.
begin.
open output out-file
move 060716 to date-1
move 270606 to date-2
move 060803 to date-3
write date-rec
close out-file
stop run
.
读取这些日期，并显示字节，十进制值和转换后的日期。请注意，由于没有必要，我评论了BinaryReader。
/*, Encoding.Default*/这是输出：
using System;
using System.Globalization;
using System.IO;
namespace ConsoleApp1
{
class Program
{
static void Main(string[] args)
{
byte[] b = { 0, 0, 0 };
string s;
DateTime d = new DateTime();
using (BinaryReader reader = new BinaryReader(File.Open(@"y:\dates.dat", FileMode.Open)/*, Encoding.Default*/))
{
for (int i = 0; i < 3; i++) // Three dates in file
{
b = reader.ReadBytes(b.Length);
Console.WriteLine("Bytes: {0}, {1}, {2}", b[0].ToString("X2"), b[1].ToString("X2"), b[2].ToString("X2"));
int n = ((b[0] << 16) + (b[1] << 8) + b[2]);
s = n.ToString("D6");
switch (i)
{
case 0:
case 2:
Console.WriteLine("Date(YYMMDD): {0}", s);
d = DateTime.ParseExact(s, "yyMMdd", CultureInfo.InvariantCulture);
Console.WriteLine("Date(yyyyMMdd): {0}", d.ToString("yyyyMMdd"));
break;
case 1:
Console.WriteLine("Date(DDMMYY): {0}", s);
d = DateTime.ParseExact(s, "ddMMyy", CultureInfo.InvariantCulture);
Console.WriteLine("Date(yyyyMMdd): {0}", d.ToString("yyyyMMdd"));
break;
default:
break;
}
Console.WriteLine("");
}
}
}
}
}
Bytes: 00, ED, 2C
Date(YYMMDD): 060716
Date(yyyyMMdd): 20060716
Bytes: 04, 21, 0E
Date(DDMMYY): 270606
Date(yyyyMMdd): 20060627
Bytes: 00, ED, 83
Date(YYMMDD): 060803
Date(yyyyMMdd): 20060803
来自这个问题的答案，String to DateTime conversion as per specified format。