I know a line number in a file, wherein I want to keep the first word and delete the rest till the end of the line. How do I do this using sed ?
So lets say, I want to go to line no 10 in a file, which looks like this -
goodword "blah blah"\
and what i want is
goodword
I have tried this - sed 's/([a-z])./\1/'
But this does it on all the lines in a file. I want it only on one specified line.
If by "first word" you mean "everything up to the first space", and if by "retain this change in the file itself" you mean that you don't mind creating a new file with the same name as the previous file, and if you have a sed that supports -i, you can probably just do:
sed -i '10s/ .*//' input-file
If you want to be more restrictive in the definition of a word, you can use '10s/\([a-z]*\).*/\1/'
Can you use grep or awk to grab just one line, and then pipe it into sed (if grep or awk couldn't do the entire job for you) to work on just one line? I think the key here is isolating that one line first, and then worrying about extracting something from it.
Using awk
awk 'NR==10 {print $1}' file
goodword
Related
I'm trying to substitute the first empty line in my input file with a multiline block, i. e. out of
one
two
three
four
five
six
I want to create
one
two
foo
three
four
five
six
For this I tried this sed script:
sed '/^$/i\
\
foo'
But it inserts at /each/ empty line.
How can I tweak this call to sed so that it inserts just at the first occurrence of an empty line? Is there a way to tell sed that now the rest of the input should just be copied from to the output?
I do not want to switch to awk or other shell tools like read in a loop or similar. I'm just interested in the use of sed for this task.
You can loop and print lines until the end of the file:
sed '/^$/{i\
\
foo
:a;n;ba}' file
I found a way by replacing the i with a s command:
sed '0,/^$/s//\
foo\
/'
But I would prefer a solution using the i command because not everything I could want to do after the search might be easily replaceable with an s.
I want to insert a range of lines from a file, say something like 210,221r before the first occurrence of a pattern in a bunch of other files.
As I am clearly not a GNU sed expert, I cannot figure how to do this.
I tried
sed '0,/pattern/{210,221r file
}' bunch_of_files
But apparently file is read from line 210 to EOF.
Try this:
sed -r 's/(FIND_ME)/PUT_BEFORE\1/' test.text
-r enables extendend regular expressions
the string you are looking for ("FIND_ME") is inside parentheses, which creates a capture group
\1 puts the captured text into the replacement.
About your second question: You can read the replacement from a file like this*:
sed -r 's/(FIND_ME)/`cat REPLACEMENT.TXT`\1/' test.text
If replace special characters inside REPLACEMENT.TXT beforehand with sed you are golden.
*= this depends on your terminal emulator. It works in bash.
In https://stackoverflow.com/a/11246712/4328188 CodeGnome gave some "sed black magic" :
In order to insert text before a pattern, you need to swap the pattern space into the hold space before reading in the file. For example:
sed '/pattern/ {
h
r file
g
N
}' in
However, to read specific lines from file, one may have to use a two-calls solution similar to dummy's answer. I'd enjoy knowing of a one-call solution if it is possible though.
I'm very much a junior when it comes to the sed command, and my Bruce Barnett guide sits right next to me, but one thing has been troubling me. With a file, can you filter it using sed to select only specific items? For example, in the following file:
alpha|november
bravo|october
charlie|papa
alpha|quebec
bravo|romeo
charlie|sahara
Would it be possible to set a command to return only the bravos, like:
bravo|october
bravo|romeo
With sed:
sed '/^bravo|/!d' filename
Alternatively, with grep (because it's sort of made for this stuff):
grep '^bravo|' filename
or with awk, which works nicely for tabular data,
awk -F '|' '$1 == "bravo"' filename
The first two use a regular expression, selecting those lines that match it. In ^bravo|, ^ matches the beginning of the line and bravo| the literal string bravo|, so this selects all lines that begin with bravo|.
The awk way splits the line across the field separator | and selects those lines whose first field is bravo.
You could also use a regex with awk:
awk '/^bravo|/' filename
...but I don't think this plays to awk's strengths in this case.
Another solution with sed:
sed -n '/^bravo|/p' filename
-n option => no printing by default.
If line begins with bravo|, print it (p)
2 way (at least) with sed
removing unwanted line
sed '/^bravo\|/ !d' YourFile
Printing only wanted lines
sed -n '/^bravo\|/ p' YourFile
if no other constraint or action occur, both are the same and a grep is better.
If there will be some action after, it could change the performance where a d cycle directly to the next line and a p will print then continue the following action.
Note the escape of pipe is needed for GNU sed, not on posix version
I have a list of usernames and i would like add possible combinations to it.
Example. Lets say this is the list I have
johna
maryb
charlesc
Is there is a way to use sed to edit it the way it looks like
ajohn
bmary
ccharles
And also
john_a
mary_b
charles_c
etc...
Can anyone assist me into getting the commands to do so, any explanation will be awesome as well. I would like to understand how it works if possible. I usually get confused when I see things like 's/\.(.*.... without knowing what some of those mean... anyway thanks in advance.
EDIT ... I change the username
sed s/\(user\)\(.\)/\2\1/
Breakdown:
sed s/string/replacement/ will replace all instances of string with replacement.
Then, string in that sed expression is \(user\)\(.\). This can be broken down into two
parts: \(user\) and \(.\). Each of these is a capture group - bracketed by \( \). That means that once we've matched something with them, we can reuse it in the replacement string.
\(user\) matches, surprisingly enough, the user part of the string. \(.\) matches any single character - that's what the . means. Then, you have two captured groups - user and a (or b or c).
The replacement part just uses these to recreate the pattern a little differently. \2\1 says "print the second capture group, then the first capture group". Which in this case, will print out auser - since we matched user and a with each group.
ex:
$ echo "usera
> userb
> userc" | sed "s/\(user\)\(.\)/\2\1/"
auser
buser
cuser
You can change the \2\1 to use any string you want - ie. \2_\1 will give a_user, b_user, c_user.
Also, in order to match any preceding string (not just "user"), just replace the \(user\) with \(.*\). Ex:
$ echo "marya
> johnb
> alfredc" | sed "s/\(.*\)\(.\)/\2\1/"
amary
bjohn
calfred
here's a partial answer to what is probably the easy part. To use sed to change usera to user_a you could use:
sed 's/user/user_/' temp
where temp is the name of the file that contains your initial list of usernames. How this works: It is finding the first instance of "user" on each line and replacing it with "user_"
Similarly for your dot example:
sed 's/user/user./' temp
will replace the first instance of "user" on each line with "user."
Sed does not offer non-greedy regex, so I suggest perl:
perl -pe 's/(.*?)(.)$/$2$1/g' file
ajohn
bmary
ccharles
perl -pe 's/(.*?)(.)$/$1_$2/g' file
john_a
mary_b
charles_c
That way you don't need to know the username before hand.
Simple solution using awk
awk '{a=$NF;$NF="";$0=a$0}1' FS="" OFS="" file
ajohn
bmary
ccharles
and
awk '{a=$NF;$NF="";$0=$0"_" a}1' FS="" OFS="" file
john_a
mary_b
charles_c
By setting FS to nothing, every letter is a field in awk. You can then easy manipulate it.
And no need to using capturing groups etc, just plain field swapping.
This might work for you (GNU sed):
sed -r 's/^([^_]*)_?(.)$/\2\1/' file
This matches any charactes other than underscores (in the first back reference (\1)), a possible underscore and the last character (in the second back reference (\2)) and swaps them around.
I saw from
How to use sed to replace only the first occurrence in a file?
How to do most of what I want:
sed -n '0,/.*\(something\).*/s//\1/p'
This finds the first match of something and extracts it (of course my real example is more complicated). I know sed has a 'quit' command which will make it stop early, but I don't know how to combine the 'q' with the above line to get my desired behavior.
I've tried replacing the 'p' with {p;q;} as I've seen in some examples, but that is clearly not working.
The "print and quit" trick works, if you also put the substitution command into the block (instead of only putting the print and quit there).
Try:
sed -n '/.*\(something\).*/{s//\1/p;q}'
Read this like: Match for the pattern given between the slashes. If this pattern is found, execute the actions specified in the block: Replace, print and exit. Typically, match- and substitution-patterns are equal, but this is not required, so the 's' command could also contain a different RE.
Actually, this is quite similar to what ghostdog74 answered for gawk.
My specific use case was to find out via 'git log' on a git-p4 imported tree which perforce branch was used for the last commit. git log (when called without -n will log every commit that every happened (hundreds of thousands for me)).
We can't know a-priori what value to give git for '-n.' After posting this, I found my solution which was:
git log | sed -n '0,/.*\[git-p4:.*\/\/depot\/blah\/\([^\/]*\)\/.*/s//\1/p; /\[git-p4/ q'
I'd still like to know how to do this with a non '/' separator, and without having to specify the 'git-p4' part twice, once for the extraction and once for the quit. There has got to be a way to combine both on the same line...
Sed usually has an option to specify more than one pattern to execute (IIRC, it's the -e option). That way, you can specify a second pattern that quits after the first line.
Another approach is to use sed to extract the first line (sed '1q'), then pipe that to a second sed command (what you show above).
use gawk
gawk '/MATCH/{
print "do something with "$0
exit
}' file
To "return something from first line and quit" you could also use grep:
grep --max-count 1 'something'
grep is supposed to be faster than sed and awk [1]. Even if not, this syntax is easier to remember (and to type: grep -m1 'something').
If you don't want to see the whole line but just the matching part, the --only-matching (-o) option might suffice.