问题
Objective
Change these filenames:
- F00001-0708-RG-biasliuyda
- F00001-0708-CS-akgdlaul
- F00001-0708-VF-hioulgigl
to these filenames:
- F0001-0708-RG-biasliuyda
- F0001-0708-CS-akgdlaul
- F0001-0708-VF-hioulgigl
Shell Code
To test:
ls F00001-0708-*|sed \'s/\\(.\\).\\(.*\\)/mv & \\1\\2/\'
To perform:
ls F00001-0708-*|sed \'s/\\(.\\).\\(.*\\)/mv & \\1\\2/\' | sh
My Question
I don\'t understand the sed code. I understand what the substitution command
$ sed \'s/something/mv\'
means. And I understand regular expressions somewhat. But I don\'t understand what\'s happening here:
\\(.\\).\\(.*\\)
or here:
& \\1\\2/
The former, to me, just looks like it means: \"a single character, followed by a single character, followed by any length sequence of a single character\"--but surely there\'s more to it than that. As far as the latter part:
& \\1\\2/
I have no idea. I really want to understand this code. Please help me out here, guys.
回答1:
First, I should say that the easiest way to do this is to use the prename or rename commands.
On Ubuntu, OSX (Homebrew package rename, MacPorts package p5-file-rename), or other systems with perl rename (prename):
rename s/0000/000/ F0000*
or on systems with rename from util-linux-ng, such as RHEL:
rename 0000 000 F0000*
That's a lot more understandable than the equivalent sed command.
But as for understanding the sed command, the sed manpage is helpful. If you run man sed and search for & (using the / command to search), you'll find it's a special character in s/foo/bar/ replacements.
s/regexp/replacement/
Attempt to match regexp against the pattern space. If success‐
ful, replace that portion matched with replacement. The
replacement may contain the special character & to refer to that
portion of the pattern space which matched, and the special
escapes \1 through \9 to refer to the corresponding matching
sub-expressions in the regexp.
Therefore, \(.\) matches the first character, which can be referenced by \1.
Then . matches the next character, which is always 0.
Then \(.*\) matches the rest of the filename, which can be referenced by \2.
The replacement string puts it all together using & (the original
filename) and \1\2 which is every part of the filename except the 2nd
character, which was a 0.
This is a pretty cryptic way to do this, IMHO. If for some reason the rename command was not available and you wanted to use sed to do the rename (or perhaps you were doing something too complex for rename?), being more explicit in your regex would make it much more readable. Perhaps something like:
ls F00001-0708-*|sed 's/F0000\(.*\)/mv & F000\1/' | sh
Being able to see what's actually changing in the s/search/replacement/ makes it much more readable. Also it won't keep sucking characters out of your filename if you accidentally run it twice or something.
回答2:
you've had your sed explanation, now you can use just the shell, no need external commands
for file in F0000*
do
echo mv "$file" "${file/#F0000/F000}"
# ${file/#F0000/F000} means replace the pattern that starts at beginning of string
done
回答3:
I wrote a small post with examples on batch renaming using sed couple of years ago:
http://www.guyrutenberg.com/2009/01/12/batch-renaming-using-sed/
For example:
for i in *; do
mv "$i" "`echo $i | sed "s/regex/replace_text/"`";
done
If the regex contains groups (e.g. \(subregex\) then you can use them in the replacement text as \1\,\2 etc.
回答4:
The easiest way would be:
for i in F00001*; do mv "$i" "${i/F00001/F0001}"; done
or, portably,
for i in F00001*; do mv "$i" "F0001${i#F00001}"; done
This replaces the F00001 prefix in the filenames with F0001.
credits to mahesh here: http://www.debian-administration.org/articles/150
回答5:
The sed command
s/\(.\).\(.*\)/mv & \1\2/
means to replace:
\(.\).\(.*\)
with:
mv & \1\2
just like a regular sed command. However, the parentheses, & and \n markers change it a little.
The search string matches (and remembers as pattern 1) the single character at the start, followed by a single character, follwed by the rest of the string (remembered as pattern 2).
In the replacement string, you can refer to these matched patterns to use them as part of the replacement. You can also refer to the whole matched portion as &.
So what that sed command is doing is creating a mv command based on the original file (for the source) and character 1 and 3 onwards, effectively removing character 2 (for the destination). It will give you a series of lines along the following format:
mv F00001-0708-RG-biasliuyda F0001-0708-RG-biasliuyda
mv abcdef acdef
and so on.
回答6:
The backslash-paren stuff means, "while matching the pattern, hold on to the stuff that matches in here." Later, on the replacement text side, you can get those remembered fragments back with "\1" (first parenthesized block), "\2" (second block), and so on.
回答7:
The parentheses capture particular strings for use by the backslashed numbers.
回答8:
If all you're really doing is removing the second character, regardless of what it is, you can do this:
s/.//2
but your command is building a mv command and piping it to the shell for execution.
This is no more readable than your version:
find -type f | sed -n 'h;s/.//4;x;s/^/mv /;G;s/\n/ /g;p' | sh
The fourth character is removed because find is prepending each filename with "./".
回答9:
ls F00001-0708-*|sed 's|^F0000\(.*\)|mv & F000\1|' | bash
回答10:
Here's what I would do:
for file in *.[Jj][Pp][Gg] ;do
echo mv -vi \"$file\" `jhead $file|
grep Date|
cut -b 16-|
sed -e 's/:/-/g' -e 's/ /_/g' -e 's/$/.jpg/g'` ;
done
Then if that looks ok, add | sh to the end. So:
for file in *.[Jj][Pp][Gg] ;do
echo mv -vi \"$file\" `jhead $file|
grep Date|
cut -b 16-|
sed -e 's/:/-/g' -e 's/ /_/g' -e 's/$/.jpg/g'` ;
done | sh
来源：https://stackoverflow.com/questions/2372719/using-sed-to-mass-rename-files