Unzip Microsoft Infopath .xsn file with Java

Question

I am facing one issue which is related to unzipping .xsn file from java code. I am stuck up with and looking for some resolution.

Guys can you please help me out from this problem?

I have tried with java traditional code to ZipFile class.

score 2 · Answer 1 · answered Sep 19 '14 at 13:56

Below is the answer for my requirement which might be useful to you.

        String command = "expand \"C:\\Users\\amishra\\Desktop\\backup\\BOM.xsn\" \"C:\\Users\\amishra\\Desktop\\backup\" -F:*";            
        Process process = Runtime.getRuntime().exec(command);

        BufferedReader stdInput = new BufferedReader(new InputStreamReader(process.getInputStream()));
        BufferedReader stdError = new BufferedReader(new InputStreamReader(process.getErrorStream()));

        String s;

        while ((s = stdInput.readLine()) != null) {
            System.out.println(s);
        }

        // Read command errors
        System.out.println("Standard error: ");
        while ((s = stdError.readLine()) != null) {
            System.out.println(s);
        }

score 1 · Accepted Answer · answered Sep 19 '14 at 12:28

1

The XSN file is really a CAB file. Try checking out the Microsoft CAB SDK here

http://support.microsoft.com/?scid=kb;EN-US;310618

answered Sep 19 '14 at 12:28

Mohit

1,185
2
11
25

Thanks Mohit for your quick response. But i need to extract the same with the use of java technologies. – Ashish Mishra Sep 19 '14 at 12:34
XSN file is the archive containing many files (you can extract them , for example, by changing .xsn extension to .cab). For CAB files i already posted the url. – Mohit Sep 19 '14 at 12:45
Thanks a lot Mohit for your quick response again. I have tried with microsoft native command 'expand'[Win7 onwards] and 'extract'[Lower than Win7]. Its perfectly works for my requirement. Thanks a lot. Below is my code snippet for achieving same. – Ashish Mishra Sep 19 '14 at 13:51

Unzip Microsoft Infopath .xsn file with Java

2 Answers2