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更新一堆檔案中的某字串

發表於 : 2015-03-17 15:53:29
yehlu

Re: 更新一堆檔案中的某字串

發表於 : 2015-04-24 19:51:41
yehlu
http://vasir.net/blog/ubuntu/replace_st ... iple_files

代碼: 選擇全部

grep -rl matchstring somedir/ | xargs sed -i 's/string1/string2/g'

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grep -rl 'windows' ./ | xargs sed -i 's/windows/linux/g'

Re: 更新一堆檔案中的某字串

發表於 : 2015-04-24 20:03:52
yehlu
http://unix.stackexchange.com/questions ... in-a-files

1. Replacing all occurrences of one string with another in all files in the current directory:

These are for cases where you know that the directory contains only regular files and that you want to process all non-hidden files. If that is not the case, use the approaches in 2.

All sed solutions in this answer assume GNU sed. If using FreeBSD or OS/X, replace -i with -i ''.

Non recursive, files in this directory only:

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sed -i -- 's/foo/bar/g' *
perl -Ti -pe 's/foo/bar/g' ./* 
(the perl one will fail for file names ending in | or space)).

Recursive, regular files (including hidden ones) in this and all subdirectories

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find . -type f -exec sed -i 's/foo/bar/g' {} +
If you are using zsh:

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sed -i -- 's/foo/bar/g' **/*(D.)
(may fail if the list is too big, see zargs to work around).

If you are using bash, bash having no support for glob qualifiers, you can't check for regular files:

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shopt -s globstar
shopt -s dotglob
Then:

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sed -i -- 's/foo/bar/g' **/*
2. Replace only if the file name matches another string / has a specific extension / is of a certain type etc:

Non-recursive, files in this directory only:

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sed -i -- 's/foo/bar/g' *baz*    ## all files whose name contains baz
sed -i -- 's/foo/bar/g' *.baz    ## files ending in .baz
Recursive, regular files in this and all subdirectories

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find . -type f -name "*baz*" -exec sed -i 's/foo/bar/g' {} +
If you are using bash:

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shopt -s globstar
shopt -s dotglob
Then:

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sed -i -- 's/foo/bar/g' **/*baz*
sed -i -- 's/foo/bar/g' **/*.baz
If you are using zsh:

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sed -i -- 's/foo/bar/g' **/*baz*(D.)
sed -i -- 's/foo/bar/g' **/*.baz(D.)
The -- serves to tell sed that no more flags will be given in the command line. This is useful to protect against file names starting with -.

If a file is of a certain type, for example, executable (see man find for more options):

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find . -type f -executable -exec sed -i 's/foo/bar/g' {} +
zsh:

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sed -i -- 's/foo/bar/g' **/*(D*)
3. Replace only if the string is found in a certain context

Replace foo with bar only there is a baz later on the same line:

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sed -i ':1;s/foo\(.*baz\)/bar\1/;t1' file
In sed, using \( \) saves whatever is in the parentheses and you can then access it with \1. There are many variations of this theme, to learn more about such regular expressions, see here. We need to repeat the operation for all foo occurrences, which is done with the t conditional branching.

Replace foo with bar only if foo is found on the 3d column (field) of the input file (assuming whitespace-separated fields):

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gawk -i inplace 'gsub(/foo/,"baz",$3)' file
(need gawk 4.1.0 or newer).

For a different field just use $N where N is the number of the field of interest. For a different field separator (: in this example) use:

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gawk -i inplace -F':' 'gsub(/foo/,"baz",$3)' file
Another solution using perl:

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perl -i -ane '$F[2]=~s/foo/baz/g; $" = " "; print "@F\n"' foo 
NOTE: both the awk and perl solutions will print space separated fields even if the input file had tabs. For a different field use $F[N-1] where N is the field umber you want and for a different field separator use (the $"=":" sets the output field separator to :):

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perl -i -F':' -ane '$F[2]=~s/foo/baz/g; $"=":";print "@F"' foo 
Replace foo with bar only on the 4th line:

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sed -i '4s/foo/bar/g' file
gawk -i inplace 'NR==4{gsub(/foo/,"baz")};1' file
perl -i -pe 's/foo/bar/g if $.==4' file
4. Multiple replace operations: replace with different strings

You can combine sed commands:

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sed -i 's/foo/bar/g; s/baz/zab/g; s/Alice/Joan/g' file
or Perl commands

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perl -i -pe 's/foo/bar/g; s/baz/zab/g; s/Alice/Joan/g' file
If you have a large number of patterns, it is easier to save your patterns and their replacements in a sed script file:

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#! /usr/bin/sed -i
s/foo/bar/g
s/baz/zab/g
Or, if you have too many pattern pairs for the above to be feasible, you can read pattern pairs from a file (two space separated patterns, $pattern and $replacement, per line):

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while read -r pattern replacement; do   
   sed -i "s/$pattern/$replacement/" file
done < patterns.txt
That will be quite slow for long lists of patterns and large data files so you might want to read the patterns and create a sed script from them instead:

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sed -f <(awk '{printf "s/%s/%s/g\n", $1, $2}' patterns.txt) -i -- file.txt
Then, run the sed script on your input file(s):

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sed -f sedscript.txt inputfile.txt
5. Multiple replace operations: replace multiple patterns with the same string

Replace any of foo, bar or baz with foobar

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sed -Ei 's/foo|bar|baz/foobar/g' file
or

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perl -i -pe 's/foo|bar|baz/foobar/g' file

Re: 更新一堆檔案中的某字串

發表於 : 2015-04-28 09:02:13
yehlu

代碼: 選擇全部

sed -i '/amount/s/ccsInteger/ccsFloat/g' stock_return_maint.php
multi file

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grep -l amount *.php | xargs sed -i '/amount/s/ccsInteger/ccsFloat/g'
grep -l price *.php | xargs sed -i '/price/s/ccsInteger/ccsFloat/g'