Fastest way of submitting shelved changelist

Question

I have shelved a changelist(say X) and I have ensured its correctness. Now i want to submit the changelist. I can

1. p4 unshelve -s X -c Y where Y is a new changelist i create before executing this command. Then p4 shelve -c X -d and lastly p4 submit -c Y
2. Otherwise I can unshelve X, delete the shelve X and move the local copies to changelist X(p4 reopen -c X ) and submit X.
3. The last option is to unshelve X, delete shelve X, p4 change(and edit description) and finally p4 submit

What are your suggestions with regards to :-

1. Use 1/3 but copy paste the description that X already had to new changelists
2. Use 2 but copy each opened file manually in command p4 reopen.

The only advantage of 2 over 1 and 3 is that I don't have to delete X(because it turned out to be my submitted changelist, while in other cases it still is "pending").

Answer 1

The shortest method is:

$ p4 shelve -c #changelist -d $ p4 submit -c #changelist 

Answer 2

In the new 2013.1 release of the Perforce server, you can now do 'p4 submit -e X', which directly submits the shelf without needing to unshelve it first.

Try it, it's a great feature!