I use a simple regular expression in file path when calling this sed command:
sed -i "s/showIntro.*/showIntro\t= false/" "c:\me\test_*\config.ini"
where * matches the suffix of a folder test_ in the path. (e.g. test_1.0, test_2.0, etc)
But for some reason the sed gives me an error:
sed: c:\me\test_*\config.ini: Invalid argument
I tried using with and without the quotes but it makes no difference.
Any idea what should I change here?
Thanks!
Try this instead:
FORFILES /M c:\me\test_*\config.ini /C "sed -i ^0x22s/showIntro.*/showIntro\t= false/^0x22 ^0x22#path^0x22"
Related
Just attempting to write a script to do a simple regex replace in php.ini, what I want to do is replace the line ;cgi.fix_pathinfo=1 with cgi.fix_pathinfo=0.
Ideally want to avoid installing any additional packages so sed seems a logical choice since it is bundled with FreeBSD. I have tried the following but doesn't seem to work:
sed 's/;cgi\.fix_pathinfo=1/cgi\.fix_pathinfo=0/' /usr/local/etc/php.ini
To change the content of a file in place with sed BSD, you can do that:
sed -i.bak -e 's/;cgi\.fix_pathinfo=1/cgi.fix_pathinfo=0/;' /usr/local/etc/php.ini
That creates a copy of the old file with a .bak extension.
Or without creating a copy:
sed -i '' -e 's/;cgi\.fix_pathinfo=1/cgi.fix_pathinfo=0/;' /usr/local/etc/php.ini
Note that in this case, a space and an empty string enclosed between quotes are mandatory. You can't simply write sed -i -e '... like with GNU sed.
I am trying to do a recursive find and replace on java files in a directory using a shell script. It works, but it is hiding all the files, and creating duplicates with a -e extension
#!/bin/bash
for file in $(find . -type f -name "*.java")
do sed -i -e 's/foo/bar/g' $file
done
From my understanding, the -e is optional - but if I do not provide it I get the following error on every file it finds
sed: 1: "./DirectoryAdapter.java": invalid command code .
Any clue as it what is happening here? For reference I am on Mac OS X running El Capitan
Here is a before and after screenshot of the directory after running the script. The replaced files still exist, they are hidden?
On OSX sed (BSD) sed requires an extension after -i option. Since it is finding -e afterwards it is adding -e to each input filename. btw you don't even need -e option here.
You can pass an empty extension like this:
sed -i '' 's/foo/bar/g' $file
Or use .bak for an extension to save original file:
sed -i.bak 's/foo/bar/g' $file
The accepted answer works for OSX but causes issues if your code is run on both GNU and OSX systems since they expect -i[SUFFIX] and -i [SUFFIX] respectively.
There are probably two reasonable solutions in this case.
Don't use -i (inplace). Instead pipe to a temporary file and overwrite the original after.
use perl.
The easiest fix for this I found was to simply use perl. The syntax is almost identical:
sed -i -e 's/foo/bar/g' $file
->
perl -pi -e 's/foo/bar/g' $file
I'm trying to grep over files which have names matching regexp. But following:
#!/bin/bash
grep -o -e "[a-zA-Z]\{1,\}" $1 -h --include=$2 -R
is working only in some cases. When I call this script like that:
./script.sh dir1/ [A-La-l]+
it doesn't work. But following:
./script.sh dir1/ \*.txt
works fine. I have also tried passing arguments within double-quotes and quotes but neither worked for me.
Do you have any ideas how to solve this problem?
grep's --include option does not accept a regex but a glob (such as *.txt), which is considerably less powerful. You will have to decide whether you want to match regexes or globs -- *.txt is not a valid regex (the equivalent regex is .*\.txt) while [A-La-l]+ is not a valid glob.
If you want to do it with regexes, you will not be able to do it with grep alone. One thing you could do is to leave the file selection to a different tool such as find:
find "$1" -type f -regex "$2" -exec grep -o -e '[a-zA-Z]\{1,\}' -h '{}' +
This will construct and run a command grep -o -e '[a-zA-Z]\{1,\}' -h list of files in $1 matching the regex $2. If you replace the + with \;, it will run the command for each file individually, which should yield the same results (very) slightly more slowly.
If you want to do it with globs, you can carry on as before; your code already does that. You should put double quotes around $1 and $2, though.
Imagine the following data stored in file data.txt
1, StringString, AnotherString 545
I want to replace "StringString" with "Strung" with the following code
sed -ir 's/String+/Strung/g' data.txt
But it won't work. This works though:
sed -ri 's/String+/Strung/g' data.txt
I don't see any reason why the order of option flags would matter. Is it a bug or is there an explanation?
Please note that I'm not looking for a workaround but rather why the order of -ir and -ri matters.
Sidenotes: The switch -i "edits the file in place" while -r allows "extended regular expression" (allowing the + operator). I'm running sed 4.2.1 Dec. 2010 on Ubuntu 12.10.
When doing -ir you are specifying that "r" should be the suffix for the backup file.
You should be able to do -i -r if you need them in that order
Did you check sed --help or man sed?
-i[SUFFIX], --in-place[=SUFFIX]
edit files in place (makes backup if extension supplied).
The default operation mode is to break symbolic and hard links.
This can be changed with --follow-symlinks and --copy.
So I want to create a script that takes 3 arguments - path to file, exact word to replace and with what to replace it. How to create such thing?
Generally I want6 it to have api like sudo script.sh "C:/myTextDoc.xml" "_WORD_TO_REPLACE_" "WordTo Use"
You don't need a script, a simple sed would do (if you're running under cygwin or a POSIX-compliant OS):
sed -i '' 's/_WORD_TO_REPLACE_/WordTo Use/' "C:/myTextDoc.xml"
Something like this?
#!/bin/bash
sed -e "s/$2/$3/g" <$1 >$1.$$ && cp $1.$$ $1 && rm $1.$$
Alternatively, you can use the single command
sed -i -e "s/$2/$3/g" $1
as Yan suggested. I generally use the first form myself. I have seen systems where -i is not supported (SunOS).
This will replace all instances of the second argument with the third, in the file passed as the first. For example, ./replace file oldword newword
Ruby(1.9+)
$ ruby -i.bak -ne 'print $_.gsub(/WORD_TO_REPLACE/,"New Word")' /path/to/file