Using to delete - regex

I need to Write a ‘sed’ command that would delete the first field from every line of a file (that is, everything up to and including the first spaces in the line.)
I think it should look something like this but I'm not quite sure:
sed '^[^:]*/d file

In sed /d means delete. Your code will delete lines that match the regex.
sed 's/^[^ ]* //g' file

This might work for you (GNU sed):
sed -r 's/^\S+\s+//' file
This removes the first non-space(s) followed by space(s).

d command in sed deletes the whole line.
You need to use s command like this:
sed -i.bak 's/^[^ ]* //' file

Assuming by spaces you mean, you want to remove the entirety of the first block of whitespace and everything proceeding it. In which case do something like
sed 's/^\w*[ \t]*//' file.txt
e.g.
$ printf "string1 \t string2\n\tstring3 string4\n"
string1 string2
string3 string4
$ printf "string1 \t string2\n\tstring3 string4\n" | sed 's/^\w*[ \t]*//'
string2
string3 string4

Related

Deleting everything between two string matches in a file

I got this text in file.txt:
Osmun.Prez#mail.com:c7lB2m6b#3.a.a:tt_webid_v2=6990226111024612869; tt_webid=6990226111024612869; tt_csrf_token=VD5Nb_TQFH4RKhoJeSe2nzLB; R6kq3TV7=AHkh4PB6AQAA3LIS90nWf2ss0Q7ZTCQjUat4axctvhQY68DdUEz92RwpmVSX|1|0|e9d6917c2fe555827dcf5ee916ba9778079ab2a9; ttwid=1%7CAFodeNF0iZM2fyy-ZeiZ6HTpZoG_MSx6SmXHgGVQ-V4%7C1627538859%7C59ca1e4a56f9f537b55e655a6dabff88e44eb48502b164ed6b4199f5a5263cb0; passport_csrf_token_default=6f7653c3ce946a6ce5444723fb0c509b; passport_csrf_token=6f7653c3ce946a6ce5444723fb0c509b; sid_guard=0483b7d37f4e4bd20ab3046e29724798%7C1627538893%7C5184000%7CMon%2C+27-Sep-2021+06%3A08%3A13+GMT; uid_tt=27b52febe6222486b9f6b6a90ef4ffeace5ea25c09d29a1583be5a1ecf760996; uid_tt_ss=27b52febe6222486b9f6b6a90ef4ffeace5ea25c09d29a1583be5a1ecf760996; sid_tt=0483b7d37f4e4bd20ab3046e29724798; sessionid=0483b7d37f4e4bd20ab3046e29724798; sessionid_ss=0483b7d37f4e4bd20ab3046e29724798; store-idc=maliva; store-country-code=us; odin_tt=294845c8f7711db177f7c549a9f44edb1555031b27a2a485df809cd92c4e544ac0772bf462df5b7a100f6e488c45303cd62df3b6b950f0842520cd887850137b035d990f29cc8b752765e594560c977f; cmpl_token=AgQQAPNSF-RMpbE89z5HYF0_-2PcrxjXf4fZYP5_ZA
How can I delete everything from the string inside ( first & only instance ) from :tt_ to _ZA in file.txt keeping only Osmun.Prez#mail.com:c7lB2m6b#3.a.a using bash linux?
Thank you
Something like:
sed -i "s/:tt_.*//" file.txt
if you want to edit the file in place. If not, remove the -i switch.
The sed command means: replace (s), in each line of file.txt, all the chars (.*) starting by the pattern :tt_ with an empty string (//).
Or the command:
sed -i "s/:tt_.*_ZA//" file.txt
which is more adherent to what you ask for, but returns the same output.
Use pattern substitution:
i=$(cat file.txt)
echo "${i/:tt*_ZA}"
Assuming the general requirement is to remove everything after the 2nd : ...
Sample data:
$ cat file.txt
Osmun.Prez#mail.com:c7lB2m6b#3.a.a:tt_webid_v ... to end of line
some.one#home.com:B52_m6b#9_az.more.stuff:delete from here ... to end of line
One sed idea:
$ sed -En 's/^([^:]*:[^:]*).*$/\1/p' file.txt
Osmun.Prez#mail.com:c7lB2m6b#3.a.a
some.one#home.com:B52_m6b#9_az.more.stuff
Using awk
awk 'BEGIN{FS=OFS=":"}{print $1,$2}'
Using : as the delimiter, it is easy to extract the columns before :tt
This deletes all chars from ":tt_" to the last "_ZA", inclusive, in file.txt
Mac_3.2.57$cat file.txt | sed 's/\(\)[:]tt.*_ZA\(.*\)/\1\2/'
Osmun.Prez#mail.com:c7lB2m6b#3.a.a
Mac_3.2.57$
Or if it is always the first 2 values which are separated by colon (as per you example)
cat file.txt | cut -f1,2 -d’:’

Sed is not matching a backslash literal or am I doing something wrong?

I have a file which has 3 lines:
%fddfdffd
\%dffdfd
hello %12345678
I need to remove anything after "%" character (inlcuding the "%" character) but not if the "%" begins with a "\".
I tried this but it didn't work:
sed -i "s/[^\\]%.*//g"
The task is actually working on a latex file to remove the comments using sed
The file after using sed should be:
\%dffdfd
hello
I suggest with your three cases:
sed '/^%/d; /\\%/b; s/%.*//' file
Output:
\%dffdfd
hello
See: man sed
This might work for you (GNU sed):
sed -E 's/(^|[^\])%.*/\1/' file
If the line starts with a % or % follows any character other than \, delete the rest of the line.
If as a consequence the line is empty and is also to be deleted, use:
sed -E '/^%/d;s/([^\])%.*/\1/' file

sed delete trailing pattern of digits

I have a .txt file where the last column includes a number pattern after the text like 'Baker 2-13' or 'Charlie 03-144.' I would like to remove all the digits at the end of the line, and just be left with Baker and Charlie. I have tried piping the sed command at the end of my awk statement, with no success.
sed -E 's/[0-9]{1,2}"-"[0-9]{1,3}$//'
I've tried adding the space and carriage returns to my sed command, but still no luck.
sed -E 's/[0-9]{1,2}"-"[0-9]{1,3}\s\r$//'
I've also tried this, but it only works when I echo a text sample, it doesn't work on each line of my .txt file
echo "CHARLIE 02-157" | sed -E 's/[0-9]*([0-9])+\-[0-9]*([0-9])+$//'
Any ideas?
This should work:
sed -i.bak -E 's/[0-9]{1,2}-[0-9]{1,3}$//' file
cat file
Baker
Charlie
You don't need to quote hyphen in the pattern.
Simple sed solution
sed 's/[- 0-9]*$//'
This will delete trailing dashes, blanks and numbers!

Using sed to find and replace within matched substrings

I'd like to use sed to process a property file such as:
java.home=/usr/bin/java
groovy-home=/usr/lib/groovy
workspace.home=/build/me/my-workspace
I'd like to replace the .'s and -'s with _'s but only up to the ='s token. The output would be
java_home=/usr/bin/java
groovy_home=/usr/lib/groovy
workspace_home=/build/me/my-workspace
I've tried various approaches including using addresses but I keep failing. Does anybody know how to do this?
What about...
$ echo foo.bar=/bla/bla-bla | sed -e 's/\([^-.]*\)[-.]\([^-.]*=.*\)/\1_\2/'
foo_bar=/bla/bla-bla
This won't work for the case where you have more than 1 dot or dash one the left, though. I'll have to think about it further.
awk makes life easier in this case:
awk -F= -vOFS="=" '{gsub(/[.-]/,"_",$1)}1' file
here you go:
kent$ echo "java.home=/usr/bin/java
groovy-home=/usr/lib/groovy
workspace.home=/build/me/my-workspace"|awk -F= -vOFS="=" '{gsub(/[.-]/,"_",$1)}1'
java_home=/usr/bin/java
groovy_home=/usr/lib/groovy
workspace_home=/build/me/my-workspace
if you really want to do with sed (gnu sed)
sed -r 's/([^=]*)(.*)/echo -n \1 \|sed -r "s:[-.]:_:g"; echo -n \2/ge' file
same example:
kent$ echo "java.home=/usr/bin/java
groovy-home=/usr/lib/groovy
workspace.home=/build/me/my-workspace"|sed -r 's/([^=]*)(.*)/echo -n \1 \|sed -r "s:[-.]:_:g"; echo -n \2/ge'
java_home=/usr/bin/java
groovy_home=/usr/lib/groovy
workspace_home=/build/me/my-workspace
In this case I would use AWK instead of sed:
awk -F"=" '{gsub("\\.|-","_",$1); print $1"="$2;}' <file.properties>
Output:
java_home/usr/bin/java
groovy_home/usr/lib/groovy
workspace_home/build/me/my-workspace
This might work for you (GNU sed):
sed -r 's/=/\n&/;h;y/-./__/;G;s/\n.*\n//' file
"You wait ages for a bus..."
This works with any number of dots and hyphens in the line and does not require GNU sed:
sed 'h; s/.*=//; x; s/=.*//; s/[.-]/_/g; G; s/\n/=/' < data
Here's how:
h: save a copy of the line in the hold space
s: throw away everything before the equal sign in the pattern space
x: swap the pattern and hold
s: blow away everything after the = in the pattern
s: replaces dots and hyphens with underscores
G: join the pattern and hold with a newline
s: replace that newline with an equal to glue it all back together
Other way using sed
sed -re 's/(.*)([.-])(.*)=(.*)/\1_\3=\4/g' temp.txt
Output
java_home=/usr/bin/java
groovy_home=/usr/lib/groovy
workspace_home=/build/me/my-workspace
In case there are more than .- on left hand side then this
sed -re ':a; s/^([^.-]+)([\.-])(.*)=/\1_\3=/1;t a' temp.txt

Change CSV Delimiter with sed

I've got a CSV file that looks like:
1,3,"3,5",4,"5,5"
Now I want to change all the "," not within quotes to ";" with sed, so it looks like this:
1;3;"3,5";5;"5,5"
But I can't find a pattern that works.
If you are expecting only numbers then the following expression will work
sed -e 's/,/;/g' -e 's/\("[0-9][0-9]*\);\([0-9][0-9]*"\)/\1,\2/g'
e.g.
$ echo '1,3,"3,5",4,"5,5"' | sed -e 's/,/;/g' -e 's/\("[0-9][0-9]*\);\([0-9][0-9]*"\)/\1,\2/g'
1;3;"3,5";4;"5,5"
You can't just replace the [0-9][0-9]* with .* to retain any , in that is delimted by quotes, .* is too greedy and matches too much. So you have to use [a-z0-9]*
$ echo '1,3,"3,5",4,"5,5",",6","4,",7,"a,b",c' | sed -e 's/,/;/g' -e 's/\("[a-z0-9]*\);\([a-z0-9]*"\)/\1,\2/g'
1;3;"3,5";4;"5,5";",6";"4,";7;"a,b";c
It also has the advantage over the first solution of being simple to understand. We just replace every , by ; and then correct every ; in quotes back to a ,
You could try something like this:
echo '1,3,"3,5",4,"5,5"' | sed -r 's|("[^"]*),([^"]*")|\1\x1\2|g;s|,|;|g;s|\x1|,|g'
which replaces all commas within quotes with \x1 char, then replaces all commas left with semicolons, and then replaces \x1 chars back to commas. This might work, given the file is correctly formed, there're initially no \x1 chars in it and there're no situations where there is a double quote inside double quotes, like "a\"b".
Using gawk
gawk '{$1=$1}1' FPAT="([^,]+)|(\"[^\"]+\")" OFS=';' filename
Test:
[jaypal:~/Temp] cat filename
1,3,"3,5",4,"5,5"
[jaypal:~/Temp] gawk '{$1=$1}1' FPAT='([^,]+)|(\"[^\"]+\")' OFS=';' filename
1;3;"3,5";4;"5,5"
This might work for you:
echo '1,3,"3,5",4,"5,5"' |
sed 's/\("[^",]*\),\([^"]*"\)/\1\n\2/g;y/,/;/;s/\n/,/g'
1;3;"3,5";4;"5,5"
Here's alternative solution which is longer but more flexible:
echo '1,3,"3,5",4,"5,5"' |
sed 's/^/\n/;:a;s/\n\([^,"]\|"[^"]*"\)/\1\n/;ta;s/\n,/;\n/;ta;s/\n//'
1;3;"3,5";4;"5,5"