wow a bunch of new (to me) tricks up0 and seq20 plus the replace
feature.
Great instead of an app I got a bash tutorial, this group is great.
Thanks
Aaron
On Wed, 2004-10-27 at 00:37, Meir Kriheli wrote:
> Aaron wrote:
> >>You can use bash replacement feature:
> >>
> >># A="Hello This is a test"
> >>
> >># echo $A
> >>Hello This is a test
> >>
> >># echo ${A// /}
> >>HelloThisisatest
> >>
> >>A simple for loop and mv should handle it, example:
> >>
> >>for i in *; do mv "$i" "${i// /}"; done
> >>
> >>If you have lots files in a dir, it could pose a problem, using find and
> >> -exec can help with that.
> >>or ls *.txt or find I guess.
> >
> >
> > I will check out the bash replacement feature before I try this so I
> > understand what I am doing. History has proven the danger of blindly
> > copying a script without knowing what I am doing. (my linux history that
> > is )
>
> The syntax is:
>
> ${var_name/search/replace}
> ${var_name//search/replace}
>
> Note that the 2nd one 2 backslashes after var_name. The former replaces
> only the 1st occurrence, while the latter replaces all.
>
> You can use this method to handle some quick search and replace. Here's
> another example:
>
> Let's create some files to simulate a test scenario:
>
> $ for i in `seq 20`; do touch backup$i.log; done
> $ ls
>
> backup10.log backup14.log backup18.log backup2.log backup6.log
> backup11.log backup15.log backup19.log backup3.log backup7.log
> backup12.log backup16.log backup1.log backup4.log backup8.log
> backup13.log backup17.log backup20.log backup5.log backup9.log
>
>
> Note the sort order. Assuming you want all of them to be in the format
> of backupXX.log
>
> $ for i in backup?.log; do mv $i ${i/up/up0}; done
> $ ls
> backup01.log backup05.log backup09.log backup13.log backup17.log
> backup02.log backup06.log backup10.log backup14.log backup18.log
> backup03.log backup07.log backup11.log backup15.log backup19.log
> backup04.log backup08.log backup12.log backup16.log backup20.log
>
> HTH
> --
> Meir Kriheli
> http://mksoft.co.il
>
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