Solved

Linux command to view a file

Posted on 2014-03-04
11
468 Views
Last Modified: 2014-05-14
I have a mywebapp.war  file  in ubuntu production server.
it has a web.xml file inside.

I want to  view this web.xml file.

What is the command to run ?   I dont want to unzip ....just want to view the content of web.xml file
0
Comment
Question by:cofactor
  • 4
  • 2
  • 2
  • +3
11 Comments
 
LVL 19

Expert Comment

by:Patricksr1972
ID: 39905602
Hi

It is an archive which needs some kind of extraction like unzip before you can read the content.
0
 
LVL 84

Expert Comment

by:ozo
ID: 39905617
unzip -p mywebapp.war  web.xml
0
 

Author Comment

by:cofactor
ID: 39905623
It is an archive which needs some kind of extraction like unzip before you can read the content.

I am trying to avoid it  because of big file size.
0
 

Author Comment

by:cofactor
ID: 39905687
unzip -p mywebapp.war  web.xml


two issues ...

1.   Dont  i require any editor to view files ?

2.  also  folder path is  mywebapp.war > WEB-INF > web.xml   .....what would be the correct command ?
0
 
LVL 5

Expert Comment

by:Dave Gould
ID: 39905736
You can extract a single file from the .war as a .war is just a zipped archive.

unzip file.war web.xml

Yopu might need to add the full path to the xml file though
0
VMware Disaster Recovery and Data Protection

In this expert guide, you’ll learn about the components of a Modern Data Center. You will use cases for the value-added capabilities of Veeam®, including combining backup and replication for VMware disaster recovery and using replication for data center migration.

 
LVL 5

Accepted Solution

by:
Dave Gould earned 400 total points
ID: 39905739
Just refrreshed and saw that I missed your last post

unzip -p mywebapp.war WEB-INF/web.xml

And no you dont need an editor unless you are really stuck with vi or vim.
If you only want to look at the file, you can just use cat or more
0
 
LVL 21

Expert Comment

by:Mazdajai
ID: 39906727
Use zipinfo to list the files in the war file then use unzip -p to pipe the contents on standard output.
0
 
LVL 37

Expert Comment

by:Gerwin Jansen
ID: 39908077
Your mywebapp.war is most likely unpacked on the webserver that runs your application, like Tomcat or JBoss for example. Do you know the location of your webserver? If you do, go there using cd and find your mywebapp.war - it should be unpacked and you could just view the web.xml using view or less.
0
 

Author Comment

by:cofactor
ID: 39943521
I'll give an update to this soon.
0
 

Author Comment

by:cofactor
ID: 40012241
what does  -p  does in unzip ?
0
 
LVL 37

Expert Comment

by:Gerwin Jansen
ID: 40012262
From the manpage:

-p

extract files to pipe (stdout). Nothing but the file data is sent to stdout, and the files are always extracted in binary format, just as they are stored (no conversions).

So if  you use "unzip -p mywebapp.war web.xml" as ozo suggests, unzip would try and get web.xml from the .war file and display it on your console

But I just suggest you change to this:

unzip -p mywebapp.war WEB-INF/web.xml

Open in new window

as web.xml will be in the WEB-INF subfolder and the command suggested by ozo will give an error.
0

Featured Post

Ransomware-A Revenue Bonanza for Service Providers

Ransomware – malware that gets on your customers’ computers, encrypts their data, and extorts a hefty ransom for the decryption keys – is a surging new threat.  The purpose of this eBook is to educate the reader about ransomware attacks.

Question has a verified solution.

If you are experiencing a similar issue, please ask a related question

Suggested Solutions

Introduction Regular patching is part of a system administrator's tasks. However, many patches require that the system be in single-user mode before they can be installed. A cluster patch in particular can take quite a while to apply if the machine…
Java performance on Solaris - Managing CPUs There are various resource controls in operating system which directly/indirectly influence the performance of application. one of the most important resource controls is "CPU".   In a multithreaded…
Learn several ways to interact with files and get file information from the bash shell. ls lists the contents of a directory: Using the -a flag displays hidden files: Using the -l flag formats the output in a long list: The file command gives us mor…
This video shows how to set up a shell script to accept a positional parameter when called, pass that to a SQL script, accept the output from the statement back and then manipulate it in the Shell.

863 members asked questions and received personalized solutions in the past 7 days.

Join the community of 500,000 technology professionals and ask your questions.

Join & Ask a Question

Need Help in Real-Time?

Connect with top rated Experts

26 Experts available now in Live!

Get 1:1 Help Now