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.
Related
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
I have a line like this:
"abc/x-y-z": "^1.4"
I need to replace ^1.4with * in the same file such that the output is "abc/x-y-z": "*"
The num inside the double quotes could be any variable number.
I tried this but it is highly specific to ^1.4 number:
sed -i '21s/^1.4/*/' abc.json
With your shown samples, please try following. You need to escape ^ here to make it literal character and you need to escape . dot as well to make it treat as literal character.
sed 's/\^1\.4/*/' Input_file
OR as per OP's comment to make it dynamic try:
sed 's/\^[0-9]+\.[0-9]+/*/' Input_file
Also if you are performing it on 21st line of your file then use 21s like you tried in your attempt. This code will substitute only very 1st occurrence of ^1.4 here in case you want to substitute all occurrences then use g(globally substitution) option for above code.
I have not used -i option(to do inplace update into Input_file itself) once you are happy with results then use sed -i option in above code.
When making changes to files, I prefer the file editor ed to the stream editor sed (ed is standard, sed -i isn't, and different versions have different quirks that bite people here on a regular basis).
ed -s input.txt <<EOF
21s/"^[[:digit:]]\{1,\}\(\.[[:digit:]]\{1,\}\)\{0,1\}"/"*"/
w
EOF
On line 21, matches a quote followed by a carat followed by 1 or more digits, optionally followed by a period and another sequence of 1 or more digits and finally the trailing quote character. All that is replaced by "*", and finally the changed file is written back to disk.
Posix BREs are a pain, no? GNU ed 1.17 and newer, and NetBSD ed can take EREs instead:
ed -Es input.txt <<EOF
21s/"\^[[:digit:]]+(\.[[:digit:]]+)?"/"*"/
w
EOF
which is a lot easier to read.
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 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 having issues matching strings even if they start with any number of white spaces. It's been very little time since I started using regular expressions, so I need some help
Here is an example. I have a file (file.txt) that contains two lines
#String1='Test One'
String1='Test Two'
Im trying to change the value for the second line, without affecting line 1 so I used this
sed -i "s|String1=.*$|String1='Test Three'|g"
This changes the values for both lines. How can I make sed change only the value of the second string?
Thank you
With gnu sed, you match spaces using \s, while other sed implementations usually work with the [[:space:]] character class. So, pick one of these:
sed 's/^\s*AWord/AnotherWord/'
sed 's/^[[:space:]]*AWord/AnotherWord/'
Since you're using -i, I assume GNU sed. Either way, you probably shouldn't retype your word, as that introduces the chance of a typo. I'd go with:
sed -i "s/^\(\s*String1=\).*/\1'New Value'/" file
Move the \s* outside of the parens if you don't want to preserve the leading whitespace.
There are a couple of solutions you could use to go about your problem
If you want to ignore lines that begin with a comment character such as '#' you could use something like this:
sed -i "/^\s*#/! s|String1=.*$|String1='Test Three'|g" file.txt
which will only operate on lines that do not match the regular expression /.../! that begins ^ with optional whiltespace\s* followed by an octothorp #
The other option is to include the characters before 'String' as part of the substitution. Doing it this way means you'll need to capture \(...\) the group to include it in the output with \1
sed -i "s|^\(\s*\)String1=.*$|\1String1='Test Four'|g" file.txt
With GNU sed, try:
sed -i "s|^\s*String1=.*$|String1='Test Three'|" file
or
sed -i "/^\s*String1=/s/=.*/='Test Three'/" file
Using awk you could do:
awk '/String1/ && f++ {$2="Test Three"}1' FS=\' OFS=\' file
#String1='Test One'
String1='Test Three'
It will ignore first hits of string1 since f is not true.