I want to replace '_v' with a whitespace and the last dot . into a dash "-". I tried using
sed 's/_v/ /' and tr '_v' ' '
Original Text
src-env-package_v1.0.1.18
output
src-en -package 1.0.1.18
Expected Output
src-env-package 1.0.1-18
This might work for you (GNU sed):
sed -E 's/(.*)_v(.*)\./\1 \2-/' file
Use the greed of the .* regexp to find the last occurrence of _v and likewise . and substitute a space for the former and a - for the latter.
If one of the conditions may occur but not necessarily both, use:
sed -E 's/(.*)_v/\1 /;s/(.*)\./\1-/' file
With your shown samples please try following sed code. Using sed's capability to store matched regex values into temp buffer(called capturing groups) here. Also using -E option here to enable ERE(extended regular expressions) for handling regex in better way.
Here is the Online demo for used regex.
sed -E 's/^(src-env-package)_v([0-9]+\..*)\.([0-9]+)$/\1 \2-\3/' Input_file
OR if its a variable value on which you want to run sed command then use following:
var="src-env-package_v1.0.1.18"
sed -E 's/^(src-env-package)_v([0-9]+\..*)\.([0-9]+)$/\1 \2-\3/' <<<"$var"
src-env-package 1.0.1-18
Bonus solution: Adding a perl one-liner solution here, using capturing groups concept(as explained above) in perl and getting the values as per requirement.
perl -pe 's/^(src-env-package)_v((?:[0-9]+\.){1,}[0-9]+)\.([0-9]+)$/\1 \2-\3/' Input_file
Related
I want to substring the File name in unix using sed command.
File name : Test_Test1_Test2_10082019_030013.csv.20191008-075740
I want the characters after the 3rd underscore or (all the characters after Test2 ) i need to be printed .
Can this be done using sed command?
I have tried this command
sed 's/^.*_\([^_]*\)$/\1/' <<< 'Test_Test1_Test2_10082019_030013.csv.20191008-075740'
but this is giving result as 030013.csv.20191008-075740
I need it from 10082019_030013.csv.20191008-075740
Thanks
Neha
To remove from the beginning up to including the 3rd underscore you can use
sed 's/^\([^_]*_\)\{3\}//' <<< 'Test_Test1_Test2_10082019_030013.csv.20191008-075740'
This removes the initial part that consists of 3 groups of (any number of non-underscore characters followed by an underscore). The result is
10082019_030013.csv.20191008-075740
If you use GNU sed you can switch it to extended regular expressions and omit the backslashes.
sed -r 's/^([^_]*_){3}//' <<< 'Test_Test1_Test2_10082019_030013.csv.20191008-075740'
Could you please try following.
sed 's/\([^_]*\)_\([^_]*\)_\([^_]*\)_\(.*\)/\4/' Input_file
Or as per Bodo's nice suggestion:
sed 's/[^_]*_[^_]*_[^_]_\(.*\)/\1/' Input_file
This might work for you (GNU sed):
sed 's/_/\n/3;s/.*\n//;t;s/Test2/\n/;s/.*\n//;t;d' file
Replace the third _ by a newline and then remove everything upto and including the first newline. If this succeeds, bail out and print the result. Otherwise, try the same method with Test2 and if this fails delete the entire line.
I have one file named `config_3_setConfigPW.ldif? containing the following line:
{pass}
on terminal, I used following commands
SLAPPASSWD=Pwd&0011
sed -i "s#{pass}#$SLAPPASSWD#" config_3_setConfigPW.ldif
It should replace {pass} to Pwd&0011 but it generates Pwd{pass}0011.
The reason is that the SLAPPASSWD shell variable is expanded before sed sees it. So sed sees:
sed -i "s#{pass}#Pwd&0011#" config_3_setConfigPW.ldif
When an "&" is on the right hand side of a pattern it means "copy the matched input", and in your case the matched input is "{pass}".
The real problem is that you would have to escape all the special characters that might arise in SLAPPASSWD, to prevent sed doing this. For example, if you had character "#" in the password, sed would think it was the end of the substitute command, and give a syntax error.
Because of this, I wouldn't use sed for this. You could try gawk or perl?
eg, this will print out the modified file in awk (though it still assumes that SLAPPASSWD contains no " character
awk -F \{pass\} ' { print $1"'${SLAPPASSWD}'"$2 } ' config_3_setConfigPW.ldif
That's because$SLAPPASSWD contains the character sequences & which is a metacharacter used by sed and evaluates to the matched text in the s command. Meaning:
sed 's/{pass}/match: &/' <<< '{pass}'
would give you:
match: {pass}
A time ago I've asked this question: "Is it possible to escape regex metacharacters reliably with sed". Answers there show how to reliably escape the password before using it as the replacement part:
pwd="Pwd&0011"
pwdEscaped="$(sed 's/[&/\]/\\&/g' <<< "$pwd")"
# Now you can safely pass $pwd to sed
sed -i "s/{pass}/$pwdEscaped/" config_3_setConfigPW.ldif
Bear in mind that sed NEVER operates on strings. The thing sed searches for is a regexp and the thing it replaces it with is string-like but has some metacharacters you need to be aware of, e.g. & or \<number>, and all of it needs to avoid using the sed delimiters, / typically.
If you want to operate on strings you need to use awk:
awk -v old="{pass}" -v new="$SLAPPASSWD" 's=index($0,old){ $0 = substr($0,1,s-1) new substr($0,s+length(old))} 1' file
Even the above would need tweaked if old or new contained escape characters.
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 file 1.htm. I want to replace a letter ṣ (s with dot below). I tried with both sed and perl and it does not replace.
sed -i 's/ṣ/s/g' "1.htm"
perl -i -pe 's/ṣ/s/g' "1.htm"
can anyone suggest what to do
1.html (not replacing ṣ)
Also i have found another strange thing. Sed (same command as above) replaces in one file but not the other I am putting the links
replacable.html
unreplacable.html same as 1.html
Why is it happening so. sed is able to replace ṣ in one file but not the other.
You have combined characters in the html file. That is, the "ṣ" is really a "s" followed by a " ̣" (a COMBINING DOT BELOW). One possibility to fix the oneliner is:
perl -C -i -pe 's/s\x{0323}/s/g' "1.htm"
That is, turn utf8 mode for stdout/stdin on (-C) and explicitely write the two characters in the left side of the s///.
Another possibility is to normalize all the combining characters using Unicode::Normalize, e.g.:
perl -C -MUnicode::Normalize -Mutf8 -i -pe '$_=NFKC($_); s/ṣ/s/g' "1.htm"
But this would also normalize all the other characters in the input file, which may or may not be OK for you.
This might work for you (GNU sed):
sed 's/\o341\o271\o243/s/g' file
To find seds octal interpretation of a character use:
echo 'ṣ'| sed l
This returns (for me):
\341\271\243$
ṣ
Then use \onnn (or combinations of) to find the correct pattern in the lefthandside (LFH) of the substitute command.
N.B. \onnn may also be used in the RHS of the substitute command.
I have a data file that needs to have several uniq identifiers stripped of hyphens.
So I have:
(Special_Section "data-values")
and I want to have it replaced with:
(Special_Section "datavalues")
I wanted to use a simple sed find/replace, but the data and values are different each time. Preferably, I'd run this in-place since the file has a lot of other information I want to keep in tact.
Does sed or awk have a way to remove the hyphen from the matched portion only?
Currently I can match with: sed -i 's/Special_Section "[a-zA-Z0-9]*-[a-zA-Z0-9]*"/&/g *myfiles*
But I would like to then run s/-// on & if it's possible.
You seems to be using GNU sed, so something like this might work:
sed -ri '
s/(Special_Section [^-]*)-([^)]*)/\1\2/g
' <your_filename_glob>
does this work?
sed -i '/(Special_Section ".*-.*")/{s/-//}' yourFile
Close - scan for the lines and then substitute on those that match:
sed -i '/Special_Section "[a-zA-Z0-9]*-[a-zA-Z0-9]*"/s/\( "[a-zA-Z0-9]*\)-\([a-zA-Z0-9]*\)"/\1\2/' *myfiles*
You can split that over several lines to avoid the scroll bar in SO:
sed -i '/Special_Section "[a-zA-Z0-9]*-[a-zA-Z0-9]*"/{
s/\( "[a-zA-Z0-9]*\)-\([a-zA-Z0-9]*\)"/\1\2/
}' *myfiles*
And on further thoughts, you can also do:
sed -i 's/\(Special_Section "[a-zA-Z0-9]*\)-\([a-zA-Z0-9]*"\)/\1\2/' *myfiles*
This is more compact. You can add the g qualifier if you need it. Both solutions use the special \(...\) notation to capture parts of the regular expression.