Is it possible to search in a file using shell and then replace a value? When I install a service I would like to be able to search out a variable in a config file and then replace/insert my own settings in that value.
Sure, you can do this using sed or awk. sed example:
sed -i 's/Andrew/James/g' /home/oleksandr/names.txt
You can use sed to perform search/replace. I usually do this from a bash shell script, and move the original file containing values to be substituted to a new name, and run sed writing the output to my original file name like this:
#!/bin/bash
mv myfile.txt myfile.txt.in
sed -e 's/PatternToBeReplaced/Replacement/g' myfile.txt.in > myfile.txt.
If you don't specify an output, the replacement will go to stdout.
sed -i 's/variable/replacement/g' *.conf
You can use sed to do this:
sed -i 's/toreplace/yoursetting/' configfile
sed is probably available on every unix like system out there. If you want to replace more than one occurence you can add a g to the s-command:
sed -i 's/toreplace/yoursetting/g' configfile
Be careful since this can completely destroy your configfile if you don't specify your toreplace-value correctly. sed also supports regular expressions in searching and replacing.
Look at the UNIX power tools awk, sed, grep and in-place edit of files with Perl.
filepath="/var/start/system/dir1"
searchstring="test"
replacestring="test01"
i=0;
for file in $(grep -l -R $searchstring $filepath)
do
cp $file $file.bak
sed -e "s/$searchstring/$replacestring/ig" $file > tempfile.tmp
mv tempfile.tmp $file
let i++;
echo "Modified: " $file
done
Generally a tool like awk or sed are used for this.
$ sed -i 's/ugly/beautiful/g' /home/bruno/old-friends/sue.txt
Related
I've 95 files that looks like :
2019-10-29-18-00/dev/xx;512.00;0.4;/var/x/xx/xxx
2019-10-29-18-00/dev/xx;512.00;0.68;/xx
2019-10-29-18-00/dev/xx;512.00;1.84;/xx/xx/xx
2019-10-29-18-00/dev/xx;512.00;80.08;/opt/xx/x
2019-10-29-18-00/dev/xx;20480.00;83.44;/var/x/x
2019-10-29-18-00/dev/xx;3584.00;840.43;/var/xx/x
2019-10-30-00-00/dev/xx;2048.00;411.59;/
2019-10-30-00-00/dev/xx;7168.00;6168.09;/usr
2019-10-30-00-00/dev/xx;3072.00;1036.1;/var
2019-10-30-00-00/dev/xx;5120.00;348.72;/tmp
2019-10-30-00-00/dev/xx;20480.00;2033.19;/home
2019-10-30-12-00;/dev/xx;5120.00;348.72;/tmp
2019-10-30-12-00;/dev/hd1;20480.00;2037.62;/home
2019-10-30-12-00;/dev/xx;512.00;0.43;/xx
2019-10-30-12-00;/dev/xx;3584.00;794.39;/xx
2019-10-30-12-00;/dev/xx;512.00;0.4;/var/xx/xx/xx
2019-10-30-12-00;/dev/xx;512.00;0.68;/xx
2019-10-30-12-00;/dev/xx;512.00;1.84;/var/xx/xx
2019-10-30-12-00;/dev/xx;512.00;80.08;/opt/xx/x
2019-10-30-12-00;/dev/xx;20480.00;83.44;/var/xx/xx
2019-10-30-12-00;/dev/x;3584.00;840.43;/var/xx/xx
For some lines I've 2019-10-29-18-00/dev and for some other lines, I've 2019-10-30-12-00;/dev/
I want to add the ; before the /dev/ where it is missing, so for that I use this sed command :
sed 's/\/dev/\;\/dev/'
But How I can apply this command for each lines where the ; is missing ? I try this :
for i in $(cat /home/xxx/xxx/xxx/*.txt | grep -e "00/dev/")
do
sed 's/\/dev/\;\/dev/' $i > $i
done
But it doesn't work... Can you help me ?
Could you please try following with GNU awkif you are ok with it.
awk -i inplace '/00\/dev\//{gsub(/00\/dev\//,"/00;/dev/")} 1' *.txt
sed solution: Tested with GNU sed for few files and it worked fine.
sed -i.bak '/00\/dev/s/00\/dev/00\;\/dev/g' *.txt
This might work for you (GNU sed & parallel):
parallel -q sed -i 's#;*/dev#;/dev#' ::: *.txt
or if you prefer:
sed -i 's#;*/dev#;/dev#' *.txt
Ignore lines with ;/dev.
sed '/;\/dev/{p;d}; s^/dev^;/dev^'
The /;\/dev/ check if the line has ;/dev. If it has ;/dev do: p - print the current line and d - start from the beginning.
You can use any character with s command in sed. Also, there is no need in escaping \;, just ;.
How I can apply this command for each lines where the ; is missing ? I try this
Don't edit the same file redirecting to the same file $i > $i. Think about it. How can you re-write and read from the same file at the same time? You can't, the resulting file will be in most cases empty, as the > $i will "execute" first making the file empty, then sed $i will start running and it will read an empty file. Use a temporary file sed ... "$i" > temp.txt; mv temp.txt "$i" or use gnu extension -i sed option to edit in place.
What you want to do really is:
grep -l '00/dev/' /home/xxx/xxx/xxx/*.txt |
xargs -n1 sed -i '/;\/dev/{p;d}; s^/dev^;/dev^'
grep -l prints list of files that match the pattern, then xargs for each single one -n1 of the files executes sed which -i edits files in place.
grep for filtering can be eliminated in your case, we can accomplish the task with a single sed command:
for f in $(cat /home/xxx/xxx/xxx/*.txt)
do
[[ -f "$f" ]] && sed -Ei '/00\/dev/ s/([^;])(\/dev)/\1;\2/' "$f"
done
The easiest way would be to adjust your regex so that it's looking a bit wider than '/dev/', e.g.
sed -i -E 's|([0-9])/dev|\1;/dev|'
(note that I'm taking advantage of sed's flexible approach to delimiters on substitute. Also, -E changes the group syntax)
Alternatively, sed lets you filter which lines it handles:
sed -i '/[0-9]\/dev/ s/\/dev/;/dev/'
This uses the same substitution you already have but only applied on lines that match the filter regex
I am attempting to use sed to delete a line, read from user input, from a file whose name is stored in a variable. Right now all sed does is print the line and nothing else.
This is a code snippet of the command I am using:
FILE="/home/devosion/scripts/files/todo.db"
read DELETELINE
sed -e "$DELETELINE"'d' "$FILE"
Is there something I am missing here?
Edit: Switching out the -e option with -i fixed my woes!
You need to delimit the search.
#!/bin/bash
read -r Line
sed "/$Line/d" file
Will delete any line containing the typed input.
Bear in mind that sed matches on regex though and any special characters will be seen as such.
For example searching for 1* will actually delete lines containing any number of 1's not an actual 1 and a star.
Also bear in mind that when the variable expands, it cannot contain the delimiters or the command will break or have unexpexted results.
For example if "$Line" contained "/hello" then the sed command will fail with
sed: -e expression #1, char 4: extra characters after command.
You can either escape the / in this case or use different delimiters.
Personally i would use awk for this
awk -vLine="$Line" '!index($0,Line)' file
Which searches for an exact string and has none of the drawbacks of the sed command.
You might have success with grep instead of sed
read -p "Enter a regex to remove lines: " filter
grep -v "$filter" "$file"
Storing in-place is a little more work:
tmp=$(mktemp)
grep -v "$filter" "$file" > "$tmp" && mv "$tmp" "$file"
or, with sponge (apt install moreutils)
grep -v "$filter" "$file" | sponge "$file"
Note: try to get out of the habit of using ALLCAPSVARS: one day you'll accidentally use PATH=... and then wonder why your script is broken.
I found this, it allows for a range deletion with variables:
#!/bin/bash
lastline=$(whatever you need to do to find the last line)` //or any variation
lines="1,$lastline"
sed -i "$lines"'d' yourfile
keeps it all one util.
Please try this :
sed -i "${DELETELINE}d" $FILE
I copied a directory structure from a windows box to a Linux box and I would like to use sed to replace c:\IBM\WebSphere with /opt/IBM/WebSphere in all files under this directory.
Any thoughts?
I think sed is a little inconvenient for that purpose, if you want to change the actual files you can use perl one-liner
perl -pi -e 's/c:\\IBM\\/\/opt\/IBM\//g' *
Add or adjust the paths according to what you need (add WebSphere if you want the replacement to change only these dirs)
Since sed can take any character as a delimiter, use
sed -e 's_\\_/_g'
to replace the \ to /.
sed -e 's_[Cc]:_/opt_g'
to replace the c: with /opt
You can string those together:
echo "C:\\IBM\\WebSphere" | sed -e 's_\\_/_g' -e 's_[Cc]:_/opt_g'
Output:
/opt/IBM/WebSphere
I don't see an awk solution, just add one:
awk -F'\' -v OFS='/' '$1=/^c:/?"/opt":$1'
test:
kent$ awk -F'\' -v OFS='/' '$1=/^c:/?"/opt":$1' <<<'c:\IBM\WebSphere'
/opt/IBM/WebSphere
echo "C:\Users\San.Tan\My Folder\project1" | sed -e 's/C:\\/mnt\/c\//;s/\\/\//g'
replaces
C:\Users\San.Tan\My Folder\project1
to
mnt/c/Users/San.Tan/My Folder/project1
in case someone needs to replace windows paths to Windows Subsystem for Linux(WSL) paths
I have file, with lines, contains ip with netmask
a.b.c.d/24
w.x.y.z/32
etc
How to delete delete specific row?
i'm using
sed -ie "s#a.b.c.d/24##g" %filname%
but after the removal is an empty string in file.
It should run inside a script, with ip as parameter and also work in freebsd under sh.
Sed solution
sed -i '/<pattern-to-match-with-proper-escape>/d' data.txt
-i option will change the original file.
Awk solution
awk '!/<pattern-to-match-with-proper-escape>/' data.txt
Using sed:
sed -i '\|a.b.c.d/24|d' file
Command line arg:
For the input being command line argument, say 1st argument($1):
sed -i "\|$1|d" file
Replace $1 with appropriate argument number as is your case.
You should use d (delete) not g. Also do not use s (replacement).
sed -ie '/a.b.c.d\/24/d' %filename%
In a script you should using it in this way
IP=$1
IPA=${IP////\\/}
sed -i /"${IPA}"/d %filename%
And the script parameter should be called in this way:
./script.sh a.b.c.d/24
perl -i -lne 'print unless(/a.b.c.d\/24/)' your_file
or in awk if you donot want to do inplace editing:
awk '$0!~/a.b.c.d\/24/' your_file
I have a text file with a bunch of file paths such as -
web/index.erb
web/contact.erb
...
etc. I need to append before the
</head>
a line of code, to every single file, I'm trying to figure out how to do this without opening each file of course. I've heard sed, but I've never used it before..was hoping there would be a grep command maybe?
Thanks
xargs can be used to apply sed (or any other command) to each filename or argument in a list. So combining that with Rom1's answer gives:
xargs sed -i 's/<\/html>/myline\n<\/html>/g' < fileslist.txt
while read f ; do
sed -i '/<\/head>/i*iamthelineofcode*' "$f"
done <iamthefileoffiles.list
or
sed -i '/<\/head>/i*iamthelineofcode*' $(cat iamthefileoffiles.list)