views:

44

answers:

3

I work with another developer in the same working copy (I know that is a bad idea), we usually do updated of individual files, and now we have files in some revision and others in another. How can I see a list of files with their respectives revision numbers? (The working copy is in a linux box, and we're using svn command line.

Thanks in advance for any help

+2  A: 

Try this in your working copy

svn info *

or

svn info -R *

to see all files and directories recursively

You may type svn help info to see other options

Dmitry Yudakov
Yes, but I looking for the versions of all subdirectories and files recursively.That only show me the versions of the files in the parent directory.
Castro
ok, see my edit
Dmitry Yudakov
A: 

The svnversion command may be what you need as it will show the range of revisions in the working copy.

the_mandrill
A: 

Finally I used combined solution using the command posted by Dmitry Yudakov and a litle script in js-rhino. Now I can find all the files with a different revision number doing something like:

svn info -R > tmp_info rhino read-svn.js | grep -v 295

/* The script */ 
lines = readFile("tmp_info").split("\n");  
lines.pop();
String.prototype.trim = function() {
  return this.replace(/^\s+|\s+$/g,"");
}
var idx = 0;
var files = [];
files[0] = {};
var line;
for (i in lines) {
  line = lines[i].toString();
  if(line.length) { 
    key = line.split(':')[0];
    if(key == 'Name' || key == 'Revision' || key == 'Path')
      files[idx][key] = line.split(':')[1];
  } else {
    idx++;
    files[idx] = {};
  }
}

print( 'files : ' + files.length + "\n");
for (i = 0; i< files.length ; i++) {
  var file = files[i];
  if(typeof(file.Name) !== "undefined")
    print(" REVISION: " + file.Revision.trim() + ' -  ' + file.Path.trim() +'/' + file.Name.trim() );
}
Castro