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
Related
I want to replace A with T and T with A
sed -e 's/T/A/g;s/A/T/g
as an example above line changes A:T to T:T
I am hoping to get T:A.
How do I do this?
If you want to change single characters, it is simply:
sed 'y/TA/AT/'
If you want to change longer (non-overlapping) strings, you need a temporary value that you know is never used. Conveniently, newline can never appear. So:
sed '
s/T/\n/g
s/A/T/g
s/\n/A/g
'
I'm not a SED expert - so not sure if that can be done as a single command. Just wondering if you've thought about doing that swap like you would in a programming language that would need a temporary variable to do the switch?
Maybe like change the A to a value you know you don't have in the string like Y for example. Then change the T to A and then change Y to T. Would something like that work?
Edit: I did a quick search just out of curiosity. Found this: https://unix.stackexchange.com/questions/528994/swapping-words-with-sed
In case that helps, but with regex stuff, the result is highly dependent on how structured and unique your inputs are. Not sure how to just swap two arbitrary sub-strings or characters throughout an entire string if there's no particular structure that tells you when you're about to get that sub-string or character like the answer above looking for the parenthesis.
Use this Perl one-liner for case-sensitive replacement:
echo 'TATAtata' | perl -pe 'tr{AT}{TA}'
ATATtata
Or this one-liner for case-insensitive replacement:
echo 'TATAtata' | perl -pe 'tr{ATat}{TAta}'
ATATatat
The Perl one-liner uses these command line flags:
-e : Tells Perl to look for code in-line, instead of in a file.
-p : Loop over the input one line at a time, assigning it to $_ by default. Add print $_ after each loop iteration.
SEE ALSO:
perldoc perlrun: how to execute the Perl interpreter: command line switches
perldoc tr
perldoc tr/SEARCHLIST/REPLACEMENTLIST/cdsr
How can I use sed command in Linux to replace key value pair. I want to replace characters that occur after â:â
For example
App.log.level: âxyzâ
It sounds like you just want something like sed 's/:.*$/: 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
I need to match all instances of strings in one file, with a master list in another. However, if my string is abc I want only that, not abcdef, abc1234 and so on.
So, a word boundary for the regex? Right now, I'm using a simple awk one liner:
cat results_file| sort -k 1| awk -F" " '{ print $1" /home/owner/file_2_search"}'|
xargs -L 1 /bin/grep -i
However, to force a word boundary, I'd need to grep string\b and the quotes (single or double) seem to be required.
In awk, \b is a special character, you need \\b ... And the quoted quotes ... (arg) ... Or am I missing something and overdoing this?
This is a Linux box, so presumably gawk. I have gone over quoting rules for awk, and realize this has got to be simple (and not complex ... but), but am not seeing it.
Had meant to post as an answer, not a comment. Will try to pose a more readable question, but confess to having second thoughts about doing this as a one-liner in the first place -- may be best to follow an alternate method. Appreciate the willingness to help.
--Joe
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]\+/$((&+1))/g;s/.*/echo "&"/e' file
N.B. This method is fraught with problems and may cause unexpected results.
I'm playing around with a script that updates a file replacing a line of text with one that is stored in a variable. After trying a whole bunch of things, I'm guessing the issue is with the text inserted, it's probably using characters that sed doesn't like. An example of how I'm using the command, with what the replacement text looks like is below:
sed 's/^define.*$/'define('AUTH_KEY', 'r*v8]Wic;#Y4{|0EQ9Z?~W,-P}k:d{k)ylAFHm-d(tY6v?U,5{hn].e9eH%/Xmdy');'/' change.html
I've read somewhere else on here that you need to escape the characters but was unable to get it working with the answer I found here: Escape a string for a sed replace pattern
Any help, or a pointer in the right direction is greatly appreciated, thank you! :)
Be aware that sed can take any character as the separator, and that / is only conventional. If you can guarantee that your key is free of spaces, you could try:
sed 's ^define.*$ &(yourkey) '
Try something like this -
Assign your replacement text to a variable -
[jaypal:~/Temp] abc="'define('AUTH_KEY', 'r*v8]Wic;#Y4{|0EQ9Z?~W,-P}k:d{k)ylAFHm-d(tY6v?U,5{hn].e9eH%/Xmdy');'"
Use that variable in awk's sub function.
[jaypal:~/Temp] awk -v rep="$abc" '{sub(/^define.*$/,rep,$0); print}' change.html