I need something similar to : sed: Replace part of a line
I have an IPaddres that as the pattern, I need to replace string before it.
Sample :
#stuff SSIPaddress
Needs to be
stuff SIPaddress
Ideas or at least how to turn the one from the link to work for me :)
Based on your example :
sed 's/^#stuff SS\(\([0-9]\{1,3\}\.\)\{3\}[0-9]\{1,3\}\)\(.*\)/stuff S\1\3/' <<< "#stuff SS1.2.3.4"
outputs :
stuff S1.2.3.4
CodeGnome, you're right, but let me show what I tried:
sed: Replace part of a line -this replaces after a pattern.
I tried to reverse the string to work for me ( replace after pattern, right )
echo “`grep "SS" file | sed -re '/([0-9]{1,3}\.){4}/p'
| grep -v drama`” | rev | sed -i …|rev
Problem with this idea is, it prints a duplicate of the matched line back in the file && cant double pipe sed -i without a file at the end ( err sed: no input file)
The winning code, after a lot of sed madness
sed -ri 'drama/! { /SS/ s/SS/S/;s/#// } ' file
due to
#*.* SSdrama:123
#*.* SSIP
" Do not match drama, in a line with SS pattern, for all else ( only one ocurrence in my sample) sed replace SS with S and # with '';Note that I didn't have to match the IP at the end at all.
Related
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’:’
I got my research result after using sed :
zcat file* | sed -e 's/.*text=\(.*\)status=[^/]*/\1/' | cut -f 1 - | grep "pattern"
But it only shows the part that I cut. How can I print all lines after a match ?
I'm using zcat so I cannot use awk.
Thanks.
Edited :
This is my log file :
[01/09/2015 00:00:47] INFO=54646486432154646 from=steve idfrom=55516654455457 to=jone idto=5552045646464 guid=100021623456461451463 n
um=6 text=hi my number is 0 811 22 1/12 status=new survstatus=new
My aim is to find all users that spam my site with their telephone numbers (using grep "pattern") then print all the lines to get all the information about each spam. The problem is there may be matches in INFO or id, so I use sed to get the text first.
Printing all lines after a match in sed:
$ sed -ne '/pattern/,$ p'
# alternatively, if you don't want to print the match:
$ sed -e '1,/pattern/ d'
Filtering lines when pattern matches between "text=" and "status=" can be done with a simple grep, no need for sed and cut:
$ grep 'text=.*pattern.* status='
You can use awk
awk '/pattern/,EOF'
n.b. don't be fooled: EOF is just an uninitialized variable, and by default 0 (false). So that condition cannot be satisfied until the end of file.
Perhaps this could be combined with all the previous answers using awk as well.
Maybe this is what you actually want? Find lines matching "pattern" and extract the field after text= up through just before status=?
zcat file* | sed -e '/pattern/s/.*text=\(.*\)status=[^/]*/\1/'
You are not revealing what pattern actually is -- if it's a variable, you cannot use single quotes around it.
Notice that \(.*\)status=[^/]* would match up through survstatus=new in your example. That is probably not what you want? There doesn't seem to be a status= followed by a slash anywhere -- you really should explain in more detail what you are actually trying to accomplish.
Your question title says "all line after a match" so perhaps you want everything after text=? Then that's simply
sed 's/.*text=//'
i.e. replace up through text= with nothing, and keep the rest. (I trust you can figure out how to change the surrounding script into zcat file* | sed '/pattern/s/.*text=//' ... oops, maybe my trust failed.)
The seldom used branch command will do this for you. Until you match, use n for next then branch to beginning. After match, use n to skip the matching line, then a loop copying the remaining lines.
cat file | sed -n -e ':start; /pattern/b match;n; b start; :match n; :copy; p; n ; b copy'
zcat file* | sed -e 's/.*text=\(.*\)status=[^/]*/\1/' | ***cut -f 1 - | grep "pattern"***
instead change the last 2 segments of your pipeline so that:
zcat file* | sed -e 's/.*text=\(.*\)status=[^/]*/\1/' | **awk '$1 ~ "pattern" {print $0}'**
I wanted some help on extended regular expressions.
I have been trying to figure out but in vain
I have a file conflicts.txt which looks like this please note that it is only a part of this file , there are many lines like these
Server/core/wildSetting.json
Server/core
Client/arcade/src/assets
Client/arcade/src/assets/
Client/arcade/src/assets
Client/arcade/src/Game/
i am writing a shell script which goes thorugh this file line by line :
if [ -s "$CONFLICTS" ] ; then
count=0
while read LINE
do
let count++
echo -e "\n $LINE \n"
done < $CONFLICTS
fi
the above prints the file line by line what i am trying now is to redirect the lines which have a certain text into some other file for that i have modified echo line of the code to :
echo -e "\n $LINE \n" | grep -E "Server/game" > newfile.txt
My Query :
As we can see there are many lines of the form Server/Core...
I want to write a regular expression and use it in grep, which matches two kind of lines
1) line s containing the ONLY the string "Server/core" preceeded and suceeded by any number of spaces
2) all the lines containing the string "assets"
I have written a regular expression for the same but it doesn't work
here my regEx:
grep -E '[^' '*Server/core$] | [assets]'
can you please tell me what is the right way of doing it ?
Please note that there can be any number of spaces before and after "Server/core" as this file is a result of parsing a previous file.
Thanks !
Based on what's asked in the comments:
1) the lines containing the string "assets"
$ grep "assets" file
Client/arcade/src/assets
Client/arcade/src/assets/
Client/arcade/src/assets
2) lines that contain only the sting "Server/core" preceeded and succeed by any amount of space
$ grep "^[ ]*Server/core[ ]*$" file
Server/core
sed (Stream EDitor) can solve your problem perfectly.
Try this command sed -n '/^ *Server\/core\|assets/p' conflicts.txt.
There is something wrong with your grep -E '[^' '*Server/core$] | [assets]'.
The ^ in a squared brackets omits all the strings containing any of the subsequent characters in the brackets.
If you want to perform in-place modification, add the -i option to the sed command like
sed -in '/^ *Server\/core\|assets/p' conflicts.txt
Your regex just needs to be this:
assets|^\s*Server/Core\s*$
I think sed or awk would be a better tool than grep - you would need to escape the forward slash if you used one of these.
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
I'm searching for a way to replace the first occurrence of a certain text in a text file with a value ${A} and the second occurrence of the same text, on a different line, with ${B}. Can this be achieved with sed or awk or any other UNIX tool?
The toolset is fairly limited: bash, common UNIX tools like sed, grep, awk etc. Perl, Python, Ruby etc. cannot be used...
Thanks in advance for any advice
Robert
Example:
...
Text
Text
Text
Text
TEXT_TO_BE_REPLACED
Text
Text
Text
TEXT_TO_BE_REPLACED
Text
Text
Text
...
should be replaced with
...
Text
Text
Text
Text
REPLACEMENT_TEXT_A
Text
Text
Text
REPLACEMENT_TEXT_B
Text
Text
Text
...
Sed with one run:
sed -e 's/\(TEXT_TO_BE_REPLACED\)/REPLACEMENT_TEXT_A/1' \
-e 's/\(TEXT_TO_BE_REPLACED\)/REPLACEMENT_B/1' < input_file > output_file
Just run your script twice - once to replace the first occurrence with ${A}, once to replace the (now first) occurence with ${B}.
To replace just one occurence:
sed '0,/RE/s//to_that/' file
(shamelessly stolen from How to use sed to replace only the first occurrence in a file?)
Here is a possible solution using awk:
#!/usr/bin/awk -f
/TEXT_TO_BE_REPLACED/ {
if ( n == 0 ) {
sub( /TEXT_TO_BE_REPLACED/, "REPLACEMENT_TEXT_A", $0 );
n++;
}
else if ( n == 1 ) {
sub( /TEXT_TO_BE_REPLACED/, "REPLACEMENT_TEXT_B", $0 );
n++;
}
}
{
print
}
awk 'BEGIN { a[0]="REPLACEMENT_A"; a[1]="REPLACEMENT_B"; } \
/TEXT_TO_BE_REPLACED/ { gsub( "TEXT_TO_BE_REPLACED", a[i++]); i%=2 }; 1'
So, you can use sed to do this like so:
First, I made a file named test.txt that contained:
well here is an example text example
and here is another example text
I choose to use the word "example" to be the value to change.
Here is the command: cat test.txt | sed -e 's/(example)/test2/2' -e 's/(example)/test1/1'
which provides the following output:
well here is an test1 text test2
and here is another test1 text
Now the sed command broken down:
s - begins search + replace
/ - start search ended with another /
The parentheses group our text ie example
/test2/ what we are putting in place of example
The number after the slashes is the occurrence we want to replace.
the -e allows you to run both commands on one command line.
You may also use the text editor ed:
# cf. http://wiki.bash-hackers.org/howto/edit-ed
cat <<-'EOF' | sed -e 's/^ *//' -e 's/ *$//' | ed -s file
H
/TEXT_TO_BE_REPLACED/s//REPLACEMENT_TEXT_A/
/TEXT_TO_BE_REPLACED/s//REPLACEMENT_TEXT_B/
wq
EOF