Linux Sed command replace after special character - linux

How can I use sed command in Linux to replace key value pair. I want to replace characters that occur after “:”
For example
App.log.level: “xyz”

It sounds like you just want something like sed 's/:.*$/: YOURTEXTHERE/' where the general format is sed 's/REPLACE_THIS/WITH_THIS/g'
The /:.*$/ bit means I want to replace all text from a colon to the end of the line. The : YOURTEXTHERE is what you're replacing with. (I'm putting the colon back in and putting the extra text.) Since I'm only doing one replacement per line, I don't need the g at the end (although it wouldn't hurt anything.)
A real example:
>> echo App.log.level: \"xyz\" | sed 's/:.*$/: YOURTEXTHERE/'
App.log.level: YOURTEXTHERE

Related

Sed - delete all chars before dash

I have below in file with contents
devtools-cloudformation
devtools-common-rpm
I want to remove devtools- and keep the rest of characters same, I tried below command but its removing all dashes
sed 's/.*-//' projects.txt
This worked for me
sed 's/-/\n/;s/.*\n//' projects.txt
If I understood well, you want to delete everything up till the first dash.
Try this:
sed 's/[^\-]*-//'
This deletes this first dash as well.
In case you want to maintain that first dash:
sed 's/[^\-]*-/-/'
The reason your solution doesn't work is the fact that the regular expression .*- means 'anything followed by a dash'.
The string devtools-common- matches this criterion and is therefore removed.
The expression I suggest says 'anything but a dash, followed by a dash'.

I am trying to replace a text for example

Example:
"word" -nothing
To
word" - nothing
in gvim.
I tried
:%s/^.*\"/
But what I get is: -nothing
Well I am new to scripting so I would like to know if it can be done in any other way like using gvim or awk or sed.
In vim... Check for \(word + quote + space + hyphen\) as first reference, followed directly by another \(word\) as second reference... replace by first reference + space + second reference... Make sure the find/replace can happen multiple times on a line with g suffix.
:%s/\(\w" -\)\(\w\)/\1 \2/g
Note that I left out the leading quote... I suppose it is possible you might have spaces in the quoted text - and I think this form might be better for you. Now in sed, that is the really cool thing about the relationship between *nix tools - they all use similar (or the same) regular expressions pattern language. So, the same exact pattern as above can be done in sed (using : as delimiter for clarity).
sed 's:\(\w" -\)\(\w\):\1 \2:g'
Awk doesn't do back references; so, not to say it can't be done, but it is not so convenient.
Could you please try following and let me know if this helps you.
awk '{sub(/^"/,"");sub(/-/,"- ")} 1' Input_file
Solution 2nd: With sed.
sed 's/^"//;s/-/- /' Input_file
Since you also tagged grep: GNU grep has the -P switch for PCRE (Perl compatible reg ex) which has \K: Keep the stuff left of the \K, don't include it in $&, so:
$ echo \"word\" | grep -oP "\"\Kword\""
word"
If I understand your question correctly you want to replace first " in each line with empty string. So in sed it is just:
sed 's/"//'
Without g flag it will replace only first occurrence in each line.
EDIT:
The same way it will work in Vim (unless you have 'gdefault' option set), so in Vim you can:
:%s/"//
try this :
:%s/\"(.*)\"/\1\"/gc

sed regex with variables to replace numbers in a file

Im trying to replace numbers in my textfile by adding one to them. i.e.
sed 's/3/4/g' path.txt
sed 's/2/3/g' path.txt
sed 's/1/2/g' path.txt
Instead of this, Can i automate it, i.e. find a /d and add one to it in the replace.
Something like
sed 's/\([0-8]\)/\1+1/g' path.txt
Also wanted to capture more than one digit i.e. ([0-9])\t([0-9]) and change each one keeping the tab inbetween
Thanks
edited #2
Using the perl example,
I also would like it to work with more digits i.e.
perl -pi~ -e 's/(\d+)\.(\d+)\.(\d+)\.(\d+)/ ($1+1)\.($2+1)\.($3+1)\.($4+1) /ge' output.txt
Any tips on making the above work?
There is no support for arithmetic in sed, but you can easily do this in Perl.
perl -pe 's/(\d+)/ $1+1 /ge'
With the /e option, the replacement expression needs to be valid Perl code. So to handle your final updated example, you need
perl -pi~ -e 's/(\d+)\.(\d+)\.(\d+)\.(\d+)/ $1+1 . "." $2+1 . "." . $3+1 . "." . $4+1 /ge'
where strings are properly quoted and adjacent strings are concatenated together with the . Perl string concatenation operator. (The arithmetic numbers are coerced into strings as well when they are concatenated with a string.)
... Though of course, the first script already does that more elegantly, since with the /g flag it already increments every sequence of digits with one, anywhere in the string.
Triplee's perl solution is the more generic answer, but Michal's sed solution works well for this particular case. However, Michal's sed solution is more easily written:
sed y/12345678/23456789/ path.txt
and is better implemented as
tr 12345678 23456789 < path.txt
This utterly fails to handle 2 digit numbers (as in the edited question).
You can do it with sed but it's not easy, see this thread.
And it's hard with awk too, see this.
I'd rather use perl for this (something like this can be seen in action # ideone):
perl -pe 's/([0-8])/$1+1/e'
(The ideone.com example must have some looping as ideone does not sets -pe by default.)
You can't do addition directly in sed - you could do it in awk by matching numbers using a regex in each line and increasing the value, but it's quite complicated. If do not need to handle arbitrary numbers but a limited set, like only single-digit numbers from 0 to 8, you can just put several replacement commands on a single sed command line by separating them with semicolons:
sed 's/8/9/g ; s/7/8/g; s/6/7/g; s/5/6/g; s/4/5/g; s/3/4/g; s/2/3/g; s/1/2/g; s/0/1/g' path.txt
This might work for you (GNU sed & Bash):
sed 's/[0-9]/$((&+1))/g;s/.*/echo "&"/e' file
This will add one to every individual digit, to increment numbers:
sed 's/[0-9]\+/$((&+1))/g;s/.*/echo "&"/e' file
N.B. This method is fraught with problems and may cause unexpected results.

Replacing comma on specific lines only

I have a dataset that is comma separated. But I have a little problem with its format. I want everything to be in the form x,x,x
Below is a sample of my dataset:
995970,16779453
995971,16828069
995972,
995973,16828069
995974,16827226
As you can see, most of my dataset is in the proper format but I have those commas on single id#'s also (my data is in form id#, connection#). How would I go about removing the commas on those single id#'s? I can't seem to figure it out just using a text editor. Any suggestions?
Edit: can I use some sort of regex expression to only remove it from those ids that have a specified length?
Edit2: Ok I figured it out using some regex, thanks for all the help!
In vi one would do something like
:%s/,$//
This means
: (enter a line mode command)
% (try the command on every line)
s (substitute)
,$ (match a comma at the end of a line)
(empty replacement text)
Sometimes you need something like /, *$/ do match a comma followed by 0 or more trailing spaces. You can get vi on windows in various different ways; one way is to install Cygwin.
You can select regular expression mode in Notepad++ and do find and replace using the following regex ,$. Leave the replace field blank.
With the sed command:
sed 's/, *//' < FILE
or inplace (requires GNU sed):
sed -ie 's/, *//' FILE

sed: Find pattern over two lines, not replace after that pattern

Wow, this one has really got me. Gonna need some tricky sed skill here I think. Here is the output value of command text I'm trying to replace:
...
fast
n : abstaining from food
The value I'd like to replace it with, is:
...
Noun
: abstaining from food
This turns out to be tricker that I thought. Because 'fast' is listed a number of times and because it is listed in other places at the beginning of the line. So I came up with this to define the range:
sed '/fast/,/^ n : / s/fast/Noun/'
Which I thought would do, but... Unfortunately, this doesn't end the replacement and the rest of the output following this match are replaced with Noun. How to get sed to stop replacement after the match? Even better, can I find a two line pattern match and replace it?
Try this:
sed "h; :b; \$b ; N; /^${1}\n n/ {h;x;s//Noun\n/; bb}; \$b ; P; D"
Unfortunately, Paul's answer reads the whole file in which makes any additional processing you might want to do difficult. This version reads the lines in pairs.
By enclosing the sed script in double quotes instead of single quotes, you can include shell variables such as positional parameters. I would recommend surrounding them with curly braces so they are set apart from the adjacent characters. When using double quotes, you'll have to be careful of the shell wanting to do its various expansions. In this example, I've escaped the dollar signs that signify the last line of the input file for the branch commands. Otherwise the shell will try to substitute the value of a variable $b which is likely to be null thus making sed unhappy.
Another technique would be to use single quotes and close and open them each time you have a shell variable:
sed 'h; :b; $b ; N; /^'${1}'\n n/ {h;x;s//Noun\n/; bb}; $b ; P; D'
# ↑open close↑ ↑open close↑
I'm assuming that the "[/code]" in your expected result is a typo. Let me know if it's not.
This seems to do what you want:
sed -e ':a;N;$!ba;s/fast\n n/Noun\n/'
I essentially stole the answer from here.
This might work for you:
sed '$!N;s/^fast\n\s*n :/Noun\n :/;P;D' file
...
Noun
: abstaining from food

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