I am trying to convert a .png file into binary code in Visual COBOL. Here is what I have
IDENTIFICATION DIVISION.
PROGRAM-ID. ImageToBinary.
ENVIRONMENT DIVISION.
INPUT-OUTPUT SECTION.
FILE-CONTROL.
SELECT IMAGE-FILE ASSIGN TO 'image.jpg'
ORGANIZATION IS LINE SEQUENTIAL.
DATA DIVISION.
FILE SECTION.
FD IMAGE-FILE.
01 IMAGE-RECORD PIC X(1024).
WORKING-STORAGE SECTION.
01 WS-END-OF-FILE PIC X(1) VALUE 'N'.
01 WS-BINARY-DATA PIC X(1024).
01 WS-IMAGE-BUFFER PIC X(1024) OCCURS 100 TIMES.
01 WS-BUFFER-INDEX PIC 9(4) VALUE 1.
01 WS-IMAGE-SIZE PIC 9(9) VALUE 0.
PROCEDURE DIVISION.
BEGIN.
OPEN INPUT IMAGE-FILE.
PERFORM UNTIL WS-END-OF-FILE = 'Y'
READ IMAGE-FILE INTO IMAGE-RECORD
AT END
MOVE 'Y' TO WS-END-OF-FILE
NOT AT END
MOVE IMAGE-RECORD TO WS-BINARY-DATA
MOVE WS-BINARY-DATA TO WS-IMAGE-BUFFER(WS-BUFFER-INDEX)
ADD 1 TO WS-BUFFER-INDEX
ADD 1024 TO WS-IMAGE-SIZE
END-READ
END-PERFORM.
CLOSE IMAGE-FILE
DISPLAY "Image Size: " WS-IMAGE-SIZE " bytes"
DISPLAY "Binary data stored in WS-IMAGE-BUFFER."
STOP RUN.
This code, however, does not work. When I read the image file into IMAGE-RECORD, it is blank. I am using Visual COBOL v9 on a Windows 10 pro. Thanks in advance.
I am trying to convert a .png file into binary code in Visual COBOL. Here is what I have
IDENTIFICATION DIVISION.
PROGRAM-ID. ImageToBinary.
ENVIRONMENT DIVISION.
INPUT-OUTPUT SECTION.
FILE-CONTROL.
SELECT IMAGE-FILE ASSIGN TO 'image.jpg'
ORGANIZATION IS LINE SEQUENTIAL.
DATA DIVISION.
FILE SECTION.
FD IMAGE-FILE.
01 IMAGE-RECORD PIC X(1024).
WORKING-STORAGE SECTION.
01 WS-END-OF-FILE PIC X(1) VALUE 'N'.
01 WS-BINARY-DATA PIC X(1024).
01 WS-IMAGE-BUFFER PIC X(1024) OCCURS 100 TIMES.
01 WS-BUFFER-INDEX PIC 9(4) VALUE 1.
01 WS-IMAGE-SIZE PIC 9(9) VALUE 0.
PROCEDURE DIVISION.
BEGIN.
OPEN INPUT IMAGE-FILE.
PERFORM UNTIL WS-END-OF-FILE = 'Y'
READ IMAGE-FILE INTO IMAGE-RECORD
AT END
MOVE 'Y' TO WS-END-OF-FILE
NOT AT END
MOVE IMAGE-RECORD TO WS-BINARY-DATA
MOVE WS-BINARY-DATA TO WS-IMAGE-BUFFER(WS-BUFFER-INDEX)
ADD 1 TO WS-BUFFER-INDEX
ADD 1024 TO WS-IMAGE-SIZE
END-READ
END-PERFORM.
CLOSE IMAGE-FILE
DISPLAY "Image Size: " WS-IMAGE-SIZE " bytes"
DISPLAY "Binary data stored in WS-IMAGE-BUFFER."
STOP RUN.
This code, however, does not work. When I read the image file into IMAGE-RECORD, it is blank. I am using Visual COBOL v9 on a Windows 10 pro. Thanks in advance.
A line sequential file is a set of records delimited by CRLF characters. This cannot be used to read in a binary file.
I would recommend using the byte stream library routines, CBL_OPEN_FILE and CBL_READ_FILE to accomplish this.
I am trying to convert a .png file into binary code in Visual COBOL. Here is what I have
IDENTIFICATION DIVISION.
PROGRAM-ID. ImageToBinary.
ENVIRONMENT DIVISION.
INPUT-OUTPUT SECTION.
FILE-CONTROL.
SELECT IMAGE-FILE ASSIGN TO 'image.jpg'
ORGANIZATION IS LINE SEQUENTIAL.
DATA DIVISION.
FILE SECTION.
FD IMAGE-FILE.
01 IMAGE-RECORD PIC X(1024).
WORKING-STORAGE SECTION.
01 WS-END-OF-FILE PIC X(1) VALUE 'N'.
01 WS-BINARY-DATA PIC X(1024).
01 WS-IMAGE-BUFFER PIC X(1024) OCCURS 100 TIMES.
01 WS-BUFFER-INDEX PIC 9(4) VALUE 1.
01 WS-IMAGE-SIZE PIC 9(9) VALUE 0.
PROCEDURE DIVISION.
BEGIN.
OPEN INPUT IMAGE-FILE.
PERFORM UNTIL WS-END-OF-FILE = 'Y'
READ IMAGE-FILE INTO IMAGE-RECORD
AT END
MOVE 'Y' TO WS-END-OF-FILE
NOT AT END
MOVE IMAGE-RECORD TO WS-BINARY-DATA
MOVE WS-BINARY-DATA TO WS-IMAGE-BUFFER(WS-BUFFER-INDEX)
ADD 1 TO WS-BUFFER-INDEX
ADD 1024 TO WS-IMAGE-SIZE
END-READ
END-PERFORM.
CLOSE IMAGE-FILE
DISPLAY "Image Size: " WS-IMAGE-SIZE " bytes"
DISPLAY "Binary data stored in WS-IMAGE-BUFFER."
STOP RUN.
This code, however, does not work. When I read the image file into IMAGE-RECORD, it is blank. I am using Visual COBOL v9 on a Windows 10 pro. Thanks in advance.
What do you want to do with this file?
Chris give a correct answer!
there are other programs to works/integrate with grafic file, a example is irfanview
A line sequential file is a set of records delimited by CRLF characters. This cannot be used to read in a binary file.
I would recommend using the byte stream library routines, CBL_OPEN_FILE and CBL_READ_FILE to accomplish this.
Chris,
I have the following code:
WORKING-STORAGE
01 IMAGE-FILE PIC X(16) VALUE 'C:\\image.PNG'.
PROCEDURE DIVISION.
CALL "CBL_OPEN_FILE" USING IMAGE-FILE
RETURNING STATUS-CODE
END-CALL.
However, it returns a strange status code that looks like a 'u'. The hex display for the status code is H"B5". The CALL statement is unable to read the image file.
What do you want to do with this file?
Chris give a correct answer!
there are other programs to works/integrate with grafic file, a example is irfanview
Greiner,
I am trying to print an image file using PCL language but in order to do that, I need to get the binary code for the image which I am trying to do with COBOL.
Chris,
I have the following code:
WORKING-STORAGE
01 IMAGE-FILE PIC X(16) VALUE 'C:\\image.PNG'.
PROCEDURE DIVISION.
CALL "CBL_OPEN_FILE" USING IMAGE-FILE
RETURNING STATUS-CODE
END-CALL.
However, it returns a strange status code that looks like a 'u'. The hex display for the status code is H"B5". The CALL statement is unable to read the image file.
It is not so simple, do you have read the help and See the example for this
you must define more and read the Bytes you had define until eof!
Chris,
I have the following code:
WORKING-STORAGE
01 IMAGE-FILE PIC X(16) VALUE 'C:\\image.PNG'.
PROCEDURE DIVISION.
CALL "CBL_OPEN_FILE" USING IMAGE-FILE
RETURNING STATUS-CODE
END-CALL.
However, it returns a strange status code that looks like a 'u'. The hex display for the status code is H"B5". The CALL statement is unable to read the image file.
Here is ab example:
identification division.
program-id. Program1.
environment division.
configuration section.
data division.
working-storage section.
01 data-buffer pic x(1000) value spaces.
01 filename pic x(30) value "myfile.jpg".
01 access-mode pic x comp-x value 67. *> read-write and > 4gb.
01 deny-mode pic x comp-x value 3.
01 device pic x comp-x value 0.
01 file-handle pic x(4) comp-5 value zeroes.
01 file-offset pic x(8) comp-x.
01 byte-count pic x(8) comp-x.
01 flags pic x comp-x.
procedure division.
call "CBL_OPEN_FILE"
using filename
access-mode
deny-mode
device
file-handle
end-call
if return-code not = 0
display "error on open = " return-code
stop run
end-if
move 128 to flags *> File-offset will contain the filesize
call "CBL_READ_FILE"
using file-handle
file-offset
byte-count
flags
data-buffer
end-call
if return-code not = 0
display "error on get filesize = " return-code
stop run
end-if
move 0 to flags
move file-offset to byte-count *> read the entire file
*> buffer must be large enough
move 1 to file-offset
call "CBL_READ_FILE" using file-handle
file-offset
byte-count
flags
data-buffer
end-call
if return-code not = 0
display "error on get filesize = " return-code
stop run
end-if
goback.
end program Program1.
Here is ab example:
identification division.
program-id. Program1.
environment division.
configuration section.
data division.
working-storage section.
01 data-buffer pic x(1000) value spaces.
01 filename pic x(30) value "myfile.jpg".
01 access-mode pic x comp-x value 67. *> read-write and > 4gb.
01 deny-mode pic x comp-x value 3.
01 device pic x comp-x value 0.
01 file-handle pic x(4) comp-5 value zeroes.
01 file-offset pic x(8) comp-x.
01 byte-count pic x(8) comp-x.
01 flags pic x comp-x.
procedure division.
call "CBL_OPEN_FILE"
using filename
access-mode
deny-mode
device
file-handle
end-call
if return-code not = 0
display "error on open = " return-code
stop run
end-if
move 128 to flags *> File-offset will contain the filesize
call "CBL_READ_FILE"
using file-handle
file-offset
byte-count
flags
data-buffer
end-call
if return-code not = 0
display "error on get filesize = " return-code
stop run
end-if
move 0 to flags
move file-offset to byte-count *> read the entire file
*> buffer must be large enough
move 1 to file-offset
call "CBL_READ_FILE" using file-handle
file-offset
byte-count
flags
data-buffer
end-call
if return-code not = 0
display "error on get filesize = " return-code
stop run
end-if
goback.
end program Program1.
Chris,
The code you provided works as it does not return an error. But I am still trying to understand how to obtain the binary code of my image. Where does it store the binary code when the call "CBL_READ_FILE" executes?
Chris,
The code you provided works as it does not return an error. But I am still trying to understand how to obtain the binary code of my image. Where does it store the binary code when the call "CBL_READ_FILE" executes?
It reads the number of bytes specified in byte-count into the data-buffer data-item. The first call to cbl_read_file with a flag value of 128 simply returns the size of the file into the file-offset data-item. So I move the size of the file from file-offset to byte-count and then the second call to cbl_read_file will read the contents of the file into data-buffer. This example is very simple and assumes that the file is not greater than 1000 bytes which is the size of data-buffer.
Are you sure that the PCL command requires the actual image to be embedded in the PCL stream and not just a filename of an image to load?
It reads the number of bytes specified in byte-count into the data-buffer data-item. The first call to cbl_read_file with a flag value of 128 simply returns the size of the file into the file-offset data-item. So I move the size of the file from file-offset to byte-count and then the second call to cbl_read_file will read the contents of the file into data-buffer. This example is very simple and assumes that the file is not greater than 1000 bytes which is the size of data-buffer.
Are you sure that the PCL command requires the actual image to be embedded in the PCL stream and not just a filename of an image to load?
Chris,
Thank you for the explanation. The image file is 14kb and I therefore increased the data-buffer to 14000. However, the data-buffer is blank when I execute both CBL_READ_FILE commands. No, I am not sure if the PCL command requires the actual image to be embedded in the PCL stream. The PCL printer language protocol was developed by Hewlett-Packard and they have documentation on PCL commands. I am still parsing through their manual, but I am not having any luck with regards to printing image files with PCL commands. The resources I have encountered suggest that printing an image in PCL requires the binary data of the image to be printed.
Chris,
Thank you for the explanation. The image file is 14kb and I therefore increased the data-buffer to 14000. However, the data-buffer is blank when I execute both CBL_READ_FILE commands. No, I am not sure if the PCL command requires the actual image to be embedded in the PCL stream. The PCL printer language protocol was developed by Hewlett-Packard and they have documentation on PCL commands. I am still parsing through their manual, but I am not having any luck with regards to printing image files with PCL commands. The resources I have encountered suggest that printing an image in PCL requires the binary data of the image to be printed.
Sorry, I had one of the data items defined incorrectly.
Please change byte-count from pic x(8) comp-x to pic x(4) comp-x.
01 byte-count pic x(4) comp-x.
Sorry, I had one of the data items defined incorrectly.
Please change byte-count from pic x(8) comp-x to pic x(4) comp-x.
01 byte-count pic x(4) comp-x.
Chris,
After making that change, the data-buffer contains the file extensions and a symbol. I am attaching a screenshot. I am assuming this is not the binary code
Chris,
After making that change, the data-buffer contains the file extensions and a symbol. I am attaching a screenshot. I am assuming this is not the binary code
That is the contents of the file. If you view the file in an editor like Notepad you will see the same thing.PNG files contain header information. I do need you to make one more change to the program though. Instead of setting the offset to a value of 1 set it to 0 as the first byte in the file is actually at offset 0.
That is the contents of the file. If you view the file in an editor like Notepad you will see the same thing.PNG files contain header information. I do need you to make one more change to the program though. Instead of setting the offset to a value of 1 set it to 0 as the first byte in the file is actually at offset 0.
Chris,
The data-buffer now looks as follows:
Greiner,
I am trying to print an image file using PCL language but in order to do that, I need to get the binary code for the image which I am trying to do with COBOL.
My understanding is the PCL language does not support PNG, so you will need to convert into a alternative binary format.
Looking at PCL spec, it says the follow formats are understood:
Compression Method
Determines how the printer interprets (decodes) the binary data in the
Transfer Raster Data command.
? * b # M
# = 0 - Unencoded (default)
1 - Run-length encoding
2 - Tagged Image File Format (TIFF) revision 4.0
3 - Delta Row
5 - Adaptive Compression
ref:
PCL 5 Printer Language Technical Quick Reference Guide (hp.com)
My understanding is the PCL language does not support PNG, so you will need to convert into a alternative binary format.
Looking at PCL spec, it says the follow formats are understood:
Compression Method
Determines how the printer interprets (decodes) the binary data in the
Transfer Raster Data command.
? * b # M
# = 0 - Unencoded (default)
1 - Run-length encoding
2 - Tagged Image File Format (TIFF) revision 4.0
3 - Delta Row
5 - Adaptive Compression
ref:
PCL 5 Printer Language Technical Quick Reference Guide (hp.com)Stephen,
Do you know of a way to convert a .png or any other image file into binary format?
Stephen,
Do you know of a way to convert a .png or any other image file into binary format?
I'm not aware of anything, so I would try to convert the .png to the .tiff format and insert it into the pcl stream.
cli tools are available for this.. if this works.. then I would look at finding a api that does the converstion.
