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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