I am struggling with another issue, this time related to redirecting output to a file (>). I have a couple of files named like foo-ab.in foo-abc.in foo-abcd.in bar-ab.in bar-abc.in bar-abcd.in and I want to go through each file, change some small part of it with sed and then rewrite the results to something like foo-ab2.in, foo-abc2.in and foo-abcd2.in for each file. What I have so far is:
for file in foo bar; do sed -n 's/text to change/replacement/ p' $file*.in> &1; done
however I get the error:
-bash: syntax error near unexpected token `&'
I understand that I am probably doing this all wrong, but I want to accomplish this:
for file in foo bar; do sed -n 's/text to change/replacement/ p' $file*.in > $file*2.in; done
where $file*2.in has exactly the same name as $file.in with a 2 appended in the middle, but since there are multiple foo with either ab, abc, or abcd endings, I don't know how to accomplish this. I already have my sed command worked out, it is just the correct file name to output to that is tripping me up.
Apparently an answer similar to this already exists here: wildcard input files
But for clarity and the sake of closing the question, the answer is to change the command as follows:
for file in foo*.in bar*.in; do sed -n 's/text to change/replacement p' $file > "${file%.*}"2.in; done
now it inputs file foo-ab.in through file foo-abcd.in, does the sed change, and outputs file foo-ab2.in through foo-abcd2.in (and the same for bar*.in).
Related
Can't find this anywhere in my teacher's lectures and no one seems to have answered it online.
How do I use sed to take lines from one file and place them in between the lines of another file?
I know it's probably a simple question but I swear I can't find how to do it.
I attempted
sed '2r myfile.txt' mydata.txt > mydata.txt
but it deletes the mydata.txt lines entirely
If your mydata.tx is a typo and you are using mydata.txt, you cannot redirect to the input file, it will truncate it to empty before you even get started. Use sed -i '2r myfile.txt' mydata.txt or else write it to a file with another filename and rename it if necessary. (The -i will cause sed to update the input file with the output, basically automatically doing what I suggest.)
I am trying to use 'sed' to replace a list of paths in a file with another path.
An example string to process is:
/path/to/file/block
I want to replace /path/to/file with something else.
I have Tried
sed -r '/\s(\S+)\/block/s/\1/new_path/'
I know it's finding the matching string but I'm getting an invalid back reference error.
How can I do this?
This may do:
echo "/path/to/file/block" | sed -r 's|/\S*/(block)|/newpath/\1|'
/newpath/block
Test
echo "test=/path/file test2=/path/to/file/block test3=/home/root/file" | sed -r 's|/\S*/(block)|/newpath/\1|'
test=/path/file test2=/newpath/block test3=/home/root/file
Back-references always refer to the pattern of the s command, not to any address (before the command).
However, in this case, there's no need for addressing: we can apply the substitution to all lines (and it will change only lines where it matches), so we can write:
s,\s(\S+)/block/, \1/new_path,
(I added a space to the RHS, as I'm guessing you didn't mean to overwrite that; also used a different separator to reduce the need for backslashes.)
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 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
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.