Please show/attach the SELECT statement and the record description.  (you can delete the data names for privacy, but we need to see the structure.)

Thanks Tom.

<<Please show/attach the SELECT statement and the record description.>>

 SELECT AUDIT
              ASSIGN TO DISC
              ORGANIZATION IS SEQUENTIAL
              ACCESS MODE IS SEQUENTIAL
              FILE STATUS IS FG4AUD-STATUS-CODE

FD AUDIT
      RECORD VARYING 0 TO 4096 CHARACTERS DEPENDING ON REC-LEN
      LABEL RECORD IS STANDARD

01 AUD--REC. |
    03 AUD--RECNAME   PIC X(12).
    03 AUD--TERMNO     PIC X(2).
    03 AUD--DATE           PIC 9(6) COMP-6.
    03 AUD--TIME            PIC 9(6) COMP-6.
    03 AUD--FUNC          PIC X.
    03 AUD--USER          PIC X(12).
    03 AUD--RELKEY      PIC 9(6) COMP-6.
    03 AUD--DATA            PIC X(4058).

OK, thank, Swapnil.

It is a binary sequential, variable record length file.

The structure of these files is:
       4 octet binary record length
       binary data for record
       4 octet binary record length

The record length is both before and after the data record; this allows both READ and READ PREVIOUS.

My guess is that, for this particular record , something has corrupted the record length counts so they do not match.  If the counts are not equal then the structure of the file is not correct for variable-length binary sequential ==> 98 errror.  Since the structure is unreliable, there is no way to read beyond the record using normal COBOL I/O.

Depending on how valuable this data might be, you can write a COBOL program to recover the data one byte at a time, using a select/file description of a fixed-length, binary file with record length = 1.  With a bit of COBOL logic, you can read through the file to check the binary record lengths (read 4 records, which determines the number of reads to pass the record, then read 4 more records, to compare the record lengths.  At some point, you will come to a record where the 4-octet record lengths do not match.  Depending on what you see, you can devise some means to pass the record, or rewrite (one byte at a time) the corrupted record length.

Good luck!

OK, thank, Swapnil.

It is a binary sequential, variable record length file.

The structure of these files is:
       4 octet binary record length
       binary data for record
       4 octet binary record length

The record length is both before and after the data record; this allows both READ and READ PREVIOUS.

My guess is that, for this particular record , something has corrupted the record length counts so they do not match.  If the counts are not equal then the structure of the file is not correct for variable-length binary sequential ==> 98 errror.  Since the structure is unreliable, there is no way to read beyond the record using normal COBOL I/O.

Depending on how valuable this data might be, you can write a COBOL program to recover the data one byte at a time, using a select/file description of a fixed-length, binary file with record length = 1.  With a bit of COBOL logic, you can read through the file to check the binary record lengths (read 4 records, which determines the number of reads to pass the record, then read 4 more records, to compare the record lengths.  At some point, you will come to a record where the 4-octet record lengths do not match.  Depending on what you see, you can devise some means to pass the record, or rewrite (one byte at a time) the corrupted record length.

Good luck!

In fact, you can determine the exact place in the file where the structure error occurs .  Using the REC-LEN value returned after each READ, you can calculate the zero-relative disk address in the physical file.  Just accumulate all the REC-LEN values, and adding 8 for each record read.  When you get the 98 error, your accumulated value is the address.  If you have some sort of binary file dump capability, you may be able to inspect the file contents directly.