How to read a binary file written in LabVIEW 6.1 using VB ?

How to read a file saved in binay forrmat in LabVIEW 6.1 platform through VB.
Is there any code ?
Thanks

Hello,
LabVIEW files stored in binary can take on just about any format that
you choose, so there will not be a single piece of code that will give
you what you are looking for. You can use VB's standard file
operation functions to read in a binary file created by LabVIEW, just
as you would read in any other binary file. See this page
on Microsoft's web site for more information on the binary file read
operation. How you stored the data in the binary file will
determine how you need to extract it from the byte array. So, if
the first element stored in the file was a 32-bit integer, you could
extract an int from the byte stream to recover that piece of your
data. You would need to know in advance how the file was
formatted in order to successfully retrieve the information. If
you need a reference for the VB functions, the microsoft link above
will get you to some good information.
Cheers,
Matt Pollock
National Instruments

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Read a binary file written by a C program

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The question is in the link below.( Just in case of reference)
http://forum.java.sun.com/thread.jsp?forum=31&thread=509414
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So, what is the possible solution for this. I am badly
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How to read excel binary file in ABAP application server

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How can I read a binary file stream with many data type, as with AcqKnowledge physio binary data file?

I would like to read in and write physiological data files which were saved by BioPac�s AcqKnowledge 3.8.1 software, in conjunction with their MP150 acquisition system. To start with, I�d like to write a converter from different physiodata file format into the AcqKnowledge binary file format for version 3.5 � 3.7 (including 3.7.3). It will allow us to read different file format into an analysis package which can only read in file written by AcqKnowledge version 3.5 � 3.7 (including 3.7.3).
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Two samples *.acq binary files are attach to this post to the list. Demo.acq is for version 3.7-3.7.2, while SCR_EKGtest1b.acq was recorded and saved with AcqKnowledge 3.8.1, which version number is 41.
I would be grateful if you someone could explain how to handle such binary file stream with LabView and send an example to i
llustrate it.
Many thanks in advance for your help.
Donat-Pierre
Attachments:
Demo.acq ‏248 KB
SCR_EKG_test1b.acq ‏97 KB

The reading of double is also straight forward : just use a dble float wired to the type cast node, after inverting the string (indian conversion).
See the attached example.
The measure of skin thickness is based on OCT (optical coherent tomography = interferometry) : an optical fiber system send and received light emitted to/back from the skin at a few centimeter distance. A profile of skin structure is then computed from the optical signal.
CC
Chilly Charly (aka CC)
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Attachments:
Read_AK_time_info.vi.zip ‏9 KB

How can I read a binary file stream with many data type, as with AcqKnowledge physio binary data file?

I would like to read in and write physiological data which was saved by Biopac�s AcqKnowledge 3.8.1 software, in conjunction with their MP150 acquisition system. To start with, I�d like to write a converter from different physiodata file format into the AcqKnowledge binary file format for version 3.5 � 3.7 (including 3.7.3). It will allow us to read different file format into an analysis package which can only read in file written by AcqKnowledge version 3.5 � 3.7 (including 3.7.3).
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tm ). Note the link for the Mac File format is very instructive too - it is presented in a different style and might make sense to some people with C library like look (http://biopac.com/AppNotes/app155macffmt/macff.htm) .
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I would be grateful if you someone could explain how to handle such binary file stream with LabView and send an example to illustrate it.
Many thanks in advance for your help.
Donat-Pierre

One more step...
short are U16 integer
double are double precision float
bool seem to be 2 bytes (= U16)
char are string (variable length)
rgb are U16 integer, with high order byte = 0
rect should be 4 x U16 (top, left, bottom, right)
Chilly Charly (aka CC)
E-List Master - Kudos glutton - Press the yellow button on the left...

How can I read a binary file?

In my AIR application, I'm trying to read a binary file and then post it via an HTTPService to a remote ColdFusion component that accepts a "binary" type argument. The result of the post is always:
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fs.open(fileObj, FileMode.READ);
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One more step...
short are U16 integer
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Read an avi file using "Read from binary file" vi

My question is how to read an avi file using "Read from binary file" vi .
My objective is to create a series of small avi files by using IMAQ AVI write frame with mpeg-4 codec of 2 second long (so 40 frames in each file with 20 fps ) and then send them one by one so as to create a stream of video. The image are grabbed from USB camera. If I read those frames using IMAQ AVI read frame then compression advantage would be lost so I want to read the whole file itself.
I read the avi file using "Read from binary file" with unsigned 8 bit data format and then sent to remote end and save it and then display it, however it didnt work. I later found that if I read an image file using "Read from binary file" with unsigned 8 bit data format and save it in local computer itself , the format would be changed and it would be unrecognizable. Am I doing wrong by reading the file in unsined 8 bit integer format or should I have used any other data types.
I am using Labview 8.5 and Labview vision development module and vision acquisition module 8.5
Your help would be highly appreciated.
Thank you.
Solved!
Go to Solution.
Attachments:
read avi file in other data format.JPG ‏26 KB

Hello,
Check out the (full) help message for "write to binary file"
The "prepend array or string size" input defaults to true, so in your example the data written to the file will have array size information added at the beginning and your output file will be (four bytes) longer than your input file. Wire a False constant to "prepend array or string size" to prevent this happening.
Rod.
Message Edited by Rod on 10-14-2008 02:43 PM

What's the best way for reading this binary file?

I've written a program that acquires data from a DAQmx card and writes it on a binary file (attached file and picture). The data that I'm acquiring comes from 8 channels, at 2.5MS/s for, at least, 5 seconds. What's the best way of reading this binary file, knowing that:
-I'll need it also on graphics (only after acquiring)
-I also need to see these values and use them later in Matlab.
I've tried the "Array to Spreadsheet String", but LabView goes out of memory (even if I don't use all of the 8 channels, but only 1).
LabView 8.6
Solved!
Go to Solution.
Attachments:
AcquireWrite02.vi ‏15 KB
myvi.jpg ‏55 KB

But my real problem, at least now, is how can I divide the information to get not only one graphic but eight?
I can read the file, but I get this (with only two channels):
So what I tried was, using a for loop, saving 250 elements in different arrays and then writing it to the .txt file. But it doesn't come right... I used 250 because that's what I got from the graphic: at each 250 points it plots the other channel.
Am I missing something here? How should I treat the information coming from the binary file, if not the way I'm doing?
(attached are the .vi files I'm using to save in the .txt format)
(EDITED. I just saw that I was dividing my graph's data in 4 just before plotting it... so It isn't 250 but 1000 elements for each channel... Still, the problem has not been solved)
Message Edited by Danigno on 11-17-2008 08:47 AM
Attachments:
mygraph.jpg ‏280 KB
Read Binary File and Save as txt - 2 channels - with SetFilePosition.vi ‏14 KB
Read Binary File and Save as txt - with SetFilePosition_b_save2files_with_array.vi ‏14 KB

Reading a binary file (byte swapping?): HELP!

Hi,
Below are 2 simple programs, one in C and the other in Java that read a binary file. The first 3 integer numbers read by the C program are:
910, 1024437292, 100002
The first 3 integer numbers read by the Java program are:
-1912406016, 749473597, -1568276224
The file was written with another C program. The values read by the C program listed here are correct.
I supect that I have a byte swapping problem. Do you know how I can solve this problem? I really can't change the format of the input file.
Thanks for your help,
Miguel
#include <stdio.h>
void main(int argc, char *argv[]) {
long buff[600];
FILE *fp = fopen("c:\\inetpub\\wwwroot\\super\\wvdot3.awg","r");
fread(buff,sizeof(int),512,fp);
int sz = sizeof(int);
import java.io.*;
public class test {
public static void main(String[] args) {
try {
DataInputStream din = new DataInputStream(new BufferedInputStream
(new FileInputStream("c:\\inetpub\\wwwroot\\super\\wvdot3.awg")));
int nrecs = din.readInt();
int version = din.readInt();
int routetype = din.readInt();
System.out.println("read: " + nrecs + " " + version + " " + routetype);
} catch (IOException exc) {
System.out.println("IOException: " + exc);

Yeah, the problem is byte ordering. I had to deal with the same thing, since Java is always big endian, and intel and DEC Alpha are both little endian. I assume you're original program runs on Intel.
I ended up creating a new LittleEndianDataInputStream. Unfortunately, you can't just extend DataInputStream since all it's methods are final. You can however copy the source, rename it, and then change all the multi-byte numeric methods to reverse the byte ordering.
For example, here is the code for the original and little endian version of readInt()
<code>
// DataInputStream
public final int readInt() throws IOException {
InputStream in = this.in;
int ch1 = in.read();
int ch2 = in.read();
int ch3 = in.read();
int ch4 = in.read();
if ((ch1 | ch2 | ch3 | ch4) < 0)
throw new EOFException();
return ((ch1 << 24) + (ch2 << 16) + (ch3 << 8) + (ch4 << 0));
// LittleEndianDataInputStream
public int readInt() throws IOException
int ch1 = in.read();
int ch2 = in.read();
int ch3 = in.read();
int ch4 = in.read();
if ((ch1 | ch2 | ch3 | ch4) < 0)
throw new EOFException();
// here's the change
return ( (ch4 << 24) + (ch3 << 16) + (ch2 << 8) + (ch1 << 0) );
</code>

How to convert proprietary binary file

Hi All,
I've got a problem that i can only tell you a little about due to the fact that i've signed a non-disclosure agreement which prevents me from telling you all the information that i have. I know my way around LV reasonably well but have no experience of programming C or C++. I have a data capture device that saves files in a properitary binary format.
I've got partially there - i have a Read From Binary File.vi working to read in a string. I then typecast this to U8 and hey presto i get out most of the data that i am looking for. The problem is that there is a mix of ints, chars, and others and i don't have access to that data (which is what i want).
The manufacturer has given me a bit of code that explains the file format, but the problem is that i cannot quite understand it (nor can i post it). However, it goes something like this:
#define blah blah
struct blah blah
      union
unsigned char blah;
struct
                       unsigned char blah     :1;
                       unsigned char blah     :3;
                       unsigned char blah     :2;
}    blah;
   }blah;
unsigned char blah;
double blah;
int blah;
int blah;
Blah stand for various variables.
Could someone have a stab at explaining what is going on. I am guessing that if i use some sort of cluster (=struct??) then i might be able to get the info that i need.
Thanks.
Phil

Easy part out of the way first: You do not need to read the file into a string and then typecast to a U8. You can simply read as a byte array by just doing this:
OK. Next part. What's probably stumbling you is the use of struct and union. Yes, a struct is basically a cluster. A structure is an ordered set of items. Each member is located in memory one after the other (I'm ignoring memory alignment here for the sake of illustration). A union, on the other hand describes an object in which all of the members start at the same memory location. However, only one member is actually present in terms of value. The amount of storage is the size of the largest member. In the example you provided what's being described is the following (again, ignoring memory alignment considerations here for the sake of illustration):
"something" that is 3 bytes long (since the struct member is longer than the single unsigned char)
a 1 byte unsigned char
a double (8 bytes - most likely)
an int (4 bytes - most likely)
an int (4 bytes - most likely)
The "something" is the union, and it can be represent either a single 1 byte value (with the other 2 bytes being meaningless), or a structure of 3 separate 1 byte values. Which one it is depends on the variable values that were written to file, so you need to know something about what was written.
Now I said that I was ignoring memory alignment issues for illustration. Well, you can't. This just means that the amount of memory used will be larger than a simple addition of the size of each datatype. Let's take a simple example:
struct y {
   char a;
   short b;
   char c;
How many bytes is y? The direct addition would tell you 4 (1 + 2 + 1). However, the more likely answer is 6 since memory alignment consideration dictate that you need to start on an "even" memory address. A C compiler will tell you the correct answer because this is platform dependent.
So what does all this mean to you? Well, my suggestion is that you write a wrapper DLL to convert the file since you will be able to cast the data you read from file directly into a variable of the type of the overall struct. Trying to deal with this in LabVIEW is probably not worth the effort.
Caveat: I'm no C expert, so I hope that I didn't say anything that was outrageously wrong. If I did, someone please correct me!
Message Edited by smercurio_fc on 11-07-2008 10:00 AM
Attachments:
Example_VI_BD.png ‏1 KB

"Read From Binary File" function Help ambiguity

I must be getting tired, but for some reason a doubt crept in my mind as I was designing a new piece of code this morning:
"is the "Read From Binary File" using the last file position or is it starting from the beginning of the file?"
"That's a stupid question", I told myself.
"I used this function a million times and have always assumed it is reusing the last file position. Moreover, there is no file offset input to that function, so WTH am I afraid of?"
So, for kicks, I fired up the Help window and read the following description (*):
Reads binary data from a file and returns it in data. How the data is read depends on the format of the specified file. This function does not work for files inside an LLB.
(*) BTW, has anybody ever complained that you can't select and copy anything from the floating Help Window?
Not much there. I particularly admire the phrasing of the second sentence... What about: "This function can do a lot of things, but it would much to complex to describe this is extensive details, so if you are asking, you probably can't afford using it"?
Anyhow, I clicked on the "Detailed Help" and got this (among other things):
Use the Set File Position function if you need to perform random access.
WHAT? I am pretty darn sure I DO NOT USE the Set File Position when I read a file in successive and contiguous chunks. I just pass the file refnum into a shift register and back to the function and that's it.
Now, the description of the "Refnum Out" ouput says: If file is a refnum or if you wire refnum out to another function, LabVIEW assumes that the file is still in use until you close it. Translated in plain English, is that supposed to mean that if the file is not closed it is open, or is that implying that it contains more info that just "the file is open and can be found here"?
I started searching around and finally ended up with the entry for "refnums, file I/O". Down the bottom of the (long) article, I found this under the heading "References to Objects or Applications" (but nothing specific to files, BTW):
...LabVIEW creates a refnum associated with that file, device, or network connection...
[...] LabVIEW remembers information associated with each refnum, such as the current location for reading from or writing to the object and the degree of user access, so you can perform concurrent but independent operations on a single object. If a VI opens an object multiple times, each open operation returns a different refnum. LabVIEW automatically closes refnums for you when a VI finishes running, but it is a good programming practice to close refnums as soon as you are finished with them to most efficiently use memory and other resources.
So it seems that my recollection was correct. I do not know what the "degree of user access" for a file is, but that's not the topic of today's post.
So, my point is: the Help File for this function is incomplete or ambiguous at best. Please correct it. And provide a link to the "refnum, file I/O" Help entry in its detailed Help. It would H E L P...
Thanks for reading.

Reading in succesive chunks is *NOT* random access. An open file always has
a current position, which is updated with each read or write operation.
You only need to set the file position if you want to start elsewhere.
LabVIEW Champion . Do more with less code and in less time .

"Read from Binary File" and efficiency

For the first time I have tried using Read from Binary File on sizable data files, and I'm seeing some real performance problems. To prevent possible data loss, I write data as I receive them from DAQ, 10 times per second. What I write is a 2-D array of double, 1000 data points x 2-4 channels. When reading in the file, I wish I could read it as a 3-D array, all at once. That doesn't seem supported, so I repeatedly do reads of 2-D array, and use a shift register with Build Array to assemble the 3-D array that I really need. But it's incredibly slow! It seems that I can only read a few hundred of these 2-D items per second.
It also occurred to me that the Build Array being used in a shift register to keep adding on to the array, could be a quadratic-time operation depending on how it is implemented. Continually and repeatedly allocating bigger and bigger chunks of memory, and copying the growing array at every step.
So I'm looking for suggestions on how to efficiently store, efficiently read, and efficiently reassemble my data into the 3-D array that I need. Perhaps I could simplify life if I had "write raw data" and "read raw data" operations that write and read only the numbers and no metadata.then I could write the file and read it back in any size chunks I please -- and read it with other programs besides. But I don't see them in the menus.
Suggestions?
Ken
Solved!
Go to Solution.

I quote the detailed help from Read from Binary File:
data type sets the type of data the function uses to read from
the binary file. The function interprets the data starting at the current file
position to be count instances of data type.
If the type is an array, string, or cluster containing an array or string, the
function assumes that each instance of that data type contains size information.
If an instance does not include size information, the function misinterprets the
data. If LabVIEW determines that the data does not match the type, it sets data
to the default for the specified type and returns an error.
So I see how I could write data without any array metadata by turning off "prepend array or string size information", but I don't see any way to read it back in such bare form. If I did, I'd have to tell it how long an array to read, and I don't see where to do that. If I could overcome this, I could indeed read in much larger chunks.
I'll try the auto-indexing tunnel anyway. I didn't tell you the whole truth, the 3-D array is actually further sliced up based on metadata that I carry, and ends up as a 4-D array of "runs". But I can do that after the fact instead of trying to do it with shift registers as I'm reading.
Thanks,
Ken

How to read from text file?

I would like to read data (frequencies) already written in a text file. I will need read these frequencies one at a time to set the function generator (as part of my data acquisition application), acquire data that is in turn written to a file and then go back and read the next frequency from the file to repeat the process again. I also have another idea of doing the same, which is read all the frequencies from the text file and populate a table and a frequency value is picked from the table each time to go through the process mentioned above.
Can anyone suggest the following: (1) How to read from a text file, (2) What could be the most efficient way of solving my above problem.
I am a new LabVIEW user and any help will be appreciated.

Hi Research,
Depending on the format of the data file, there are a few options for reading it. If it is tab delimited, you may want to use the Read from SpreadSheet File VI which will read the file into an Array. You can then use the Index Array VI to pull out individual entries. If the files is ASCII but not tab delimited, you could use the regular Open File and Read File VIs. You can either read the file out piece by piece, or read the entire file into a string and then use the Match Pattern VI to parse out the different elements (there are actually many ways to do this - check out the Strings subpalette).
Since you're new to LabVIEW, you may want to check out these resources:
Three Hour Introduction to LabVIEW
Six Hour Introduction to LabVIEW
Getting Started with LabVIEW
I hope this helps! Let us know if you have more questions,
Megan B.
National Instruments

Reading in binary files

Hello all,
I'm attempting to read in a file written out by a c program into my java console app. The c program wrote out bytes in this fashion:
0x1234 written as ox12 ox34 and java reads this in via the DataInputStream class as ox3412, so I wrote my own readShort function to do this correctly as follows:
public short readShort(DataInputStream dis) throws IOException{
try {
byte[] b = new byte[2];
dis.read(b);
short s = (short)((short)b[0] + (short)(b[1]<<8));
return s;
}catch(IOException ioe) {
ioe.printStackTrace();
throw ioe;
This works great except when reading in bytes starting with letters. i.e. bytes 0xca 0x00 come out of this function as 0xffca. I don't understand why when I cast the first byte to a short, this is how it is interpreted, any help is appreciated! thanks!

Masking the high bits means:
short s = (short)((b[0] & 0xff) + ((b[1]<<8) & 0xff00));
java upcasts to int when you perform operations like add, and it only uses signed primitives.
If you have a whole file that needs this kind of conversion, I wonder if it might be more efficient to convert bigger chunks. You could read in, say, 1024 bytes into an array. Then copy these bytes into another array while sorting them into the right order for java. Finally, read them back into shorts using a DataInputStream wrapped around a ByteArrayInputStream.

How to read a binary file written in LabVIEW 6.1 using VB ?

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