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.
Related
I am currently using sed to delete lines and subsequent line with various patterns from a file using the following the following code:
sed -i -e"/String1/,+1d" -e"/String2/,+1d," filename.txt
Works very well however I have a lot of patterns which vary from time to time.
Is it possible to put all patterns in another text file and make sed to delete all entries for patterns found in such file ?
Thanks
Here is an awk version
awk 'NR==FNR {a[$0]++;next} {for (i in a) if ($0~i) f=2} --f<0' list yourfile
NR==FNR {a[$0]++;next} store the list of lines to remove for file list in array a
for (i in a) for every line, loop through all lines in list
if ($0~i) f=2 if trigger line is found, set flag f to 2
--f<0 decrease flag f by one and test if it less than 0, if yes, print the line.
example
cat yourfile
one
two
three
four
five
six
seven
eight
nine
ten
eleven
cat list
three
eight
awk 'NR==FNR {a[$0]++;next} {for (i in a) if ($0~i) f=2} --f<0' list yourfile
one
two
five
six
seven
ten
eleven
Trying to stick with sed - at all cost, and being creative :-)
Consider using sed itself to generate the sed script that will perform the substitutions, based on the patterns file.
Important to note that this is solution will process each input file with one-pass, making it possible to use on large files/many patterns.
Proposed Solution:
sed -i -e "$(sed -e '/\//d;s/^/\//;s/$/\/,+1d/' < patterns.txt)" filename.txt
The embedded sed program (sed -e '/\//d;s/^/\//;s/$/\/,+1d/ ...) will convert the patterns.txt to a small sed script:
pattern.txt:
three
eight
foo/bar
Output: (noticed foo/bar ignored - contains '/')
/three/,+1d
/eight/,+1d
Notes, Limitations, etc:
One limit (of above implementation) is the delimiter, code remove any pattern with '/' to simplify generation of sed script, and to avoid potential injection. Possible to work around this limitation and allow for alternate delimiter (by escaping special characters in the pattern, or leveraging the '\%' addresses). May need additional testing.
Code assumes that the patterns are valid RE.
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 have a file containing many blocks of lines. In each block, I have one numeric character of multiple digits (15353580 for instance). I need to extract all these numbers and put them as a column in a new file.
I came across this thread. The sed command does the job but does not separate the numbers from each other. Using the second example ("123 he23llo") of the most voted response, I would like to have 123-23 instead of 12323, where my '-' stands for a line break. How can I do so ?
You can use this sed,
sed -e 's/[^0-9]\+/-/g' -e 's/-$//'
Example:
$ echo "123 hel43lo 23fds" | sed -e 's/[^0-9]\+/-/g' -e 's/-$//'
123-43-23
I have a file with a lot of text, but I want to print only words that contain "#" at the beginning. Ex:
My name is #Laura and I live in #London. Name=#Laura. City=#London
How can I print all words that start with #?.I did this the following and it worked, but I want to do it using sed. I tried several patters, but I cannot make it print anything.
grep -o -E "#\w+" file.txt
Thanks
Use this sed command:
sed 's/[^#]*\(#[^ .]*\)/\1\n/g' file.txt
Explanation: we invoke the substitution command of sed. This has following structure: sed 's/regex/replace/options'. We will search for a regex and replace it using the g option. g makes sure the match is made multiple times per line.
We look for a series of non at chars followed by an # and a number of non-spaces #[^ ]*. We put this last part in a group \(\) and sub it during the replacement \1.
Note that we add a newline at the end of each match, you can also get the output on a single line by omitting the \n.
I'm trying to extract a list of CentOS domain names only from http://mirrorlist.centos.org/?release=6.4&arch=x86_64&repo=os
Truncating prefix "http://" and "ftp://" to the first "/" character only resulting a list of
yum.phx.singlehop.com
mirror.nyi.net
bay.uchicago.edu
centos.mirror.constant.com
mirror.teklinks.com
centos.mirror.netriplex.com
centos.someimage.com
mirror.sanctuaryhost.com
mirrors.cat.pdx.edu
mirrors.tummy.com
I searched stackoverflow for the sed method but I'm still having trouble.
I tried doing this with sed
curl "http://mirrorlist.centos.org/?release=6.4&arch=x86_64&repo=os" | sed '/:\/\//,/\//p'
but doesn't look like it is doing anything. Can you give me some advice?
Here you go:
curl "http://mirrorlist.centos.org/?release=6.4&arch=x86_64&repo=os" | sed -e 's?.*://??' -e 's?/.*??'
Your sed was completely wrong:
/x/,/y/ is a range. It selects multiple lines, from a line matching /x/ until a line matching /y/
The p command prints the selected range
Since all lines match both the start and end pattern you used, you effectively selected all lines. And, since sed echoes the input by default, the p command results in duplicated lines (all lines printed twice).
In my fix:
I used s??? instead of s/// because this way I didn't need to escape all the / in the patterns, so it's a bit more readable this way
I used two expressions with the -e flag:
s?.*://?? matches everything up until :// and replaces it with nothing
s?/.*?? matches everything from / until the end replaces it with nothing
The two expressions are executed in the given order
In modern versions of sed you can omit -e and separate the two expressions with ;. I stick to using -e because it's more portable.