With the sed command, is it possible to do internal string commands? in this case the actual lines are:
s/9G /9F6 09999F7 09999F8 09999F9 09999G /g
s/0G /0F6 09999F7 09999F8 09999F9 09999G /g
The number can be set using [09] but I didn't know if I could retrieve it from, say, & and use it before the F6 in something like the following:
s/[09]G /(&:0:1)F6 09999F7 09999F8 09999F9 09999G /g
This actual code does not work, by the way.
You are looking for a so called sub expression in the form of \(SUB_PATTERN\):
sed 's/\([09]\)G /\1F6 09999F7 09999F8 09999F9 09999G /g' file
From man sed:
s/regexp/replacement/
Attempt to match regexp against the pattern space. If successful, replace that portion matched with replacement.
The replacement may contain the special character & to refer to that portion of the pattern space which matched, and
the special escapes \1 through \9 to refer to the corresponding matching sub-expressions in the regexp.
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 wanted to replace underscores with hyphens in all places where the character('_') is preceded and following by uppercase letters e.g. QWQW_IOIO, OP_FD_GF_JK, TRT_JKJ, etc. The replacement is needed throughout one document.
I tried to replace this in vim using:
:%s/[A-Z]_[A-Z]/[A-Z]-[A-Z]/g
But that resulted in QWQW_IOIO with QWQ[A-Z]-[A-Z]OIO :(
I tried using a sed command:
sed -i '/[A-Z]_[A-Z]/ s/_/-/g' ./file_name
This resulted in replacement over the whole line. e.g.
QWQW_IOIO variable may contain '_' or '-' line was replaced by
QWQW-IOIO variable may contain '-' or '-'
You had the right idea with your first vim approach. But you need to use a capturing group to remember what character was found in the [A-Z] section. Those are nicely explained here and under :h /\1. As a side note, I would recommend using \u instead of [A-Z], since it is both shorter and faster. That means the solution you want is:
:%s/\(\u\)_\(\u\)/\1-\2/g
Or, if you would like to use the magic setting to make it more readable:
:%s/\v(\u)_(\u)/\1-\2/g
Another option would be to limit the part of the search that gets replaced with the \zs and \ze atoms:
:%s/\u\zs_\ze\u/-/g
This is the shortest solution I'm aware of.
This should do what you want, assuming GNU sed.
sed -i -r -e 's/([A-Z]+)_([A-Z]+)/\1-\2/g' ./file_name
Explanation:
-r flag enables extended regex
[A-Z]+ is "one or more uppercase letters"
() groups a pattern together and creates a numbered memorized match
\1, \2 put those memorized matches in the replacement.
So basically this finds a chunk of uppercase letters followed by an underscore, followed by another chunk of uppercase letters, memorizes only the letter chunks as 2 groups,
([A-Z]+)_([A-Z]+)
Then it replays those groups, but with a hyphen in between instead of an underscore.
\1-\2
The g flag at the end says to do this even if the pattern shows up multiple times on one line.
Note that this falls apart a little in this case:
QWQW_IOIO_ABAB
Because it matches the first time, but not the second; the second part won't match because IOIO was consumed by the first match. So that would result in
QWQW-IOIO_ABAB
This version drops the + so it only matches one uppercase letter, and won't break in the same way:
sed -i -r -e 's/([A-Z])_([A-Z])/\1-\2/g'
It still has a small flaw, if you have a string like this:
A_B_C
Same issue as before, just one letter now instead of multiple.
How do I substitute &&&&&&&&& with &? I have tried :%s/&&&&&&&&&/&/gbut i only get more &.
Because & is special in the replacement part (it means "the whole matched string"), you have to escape it:
:%s/&&&&&&&&&/\&/g
(Note the backslash before & in the replacement part.)
You must escape & in the replacement section. Unescaped, & refers to the whole match.
:%s/&&&&&&&&&/\&/g
You need to escape & with \ in replace string
:%s/&&&&&&&&&/\&/g
In replace string, & has a special meaning and contains matching string.
Therefore in you case, you are replacing nothing but match itself hence no change.
If your intension is to replace multiple & with a single one, then try following
:%s/&\+/\&/g
You need a quantifier in the expression:
:%s/&\+/&/g
not sure if it works like that in vim, it is default regex, the plus sign tells it should capture any combination of one or multiple '&' signs, using &{9} should find exactly nine
I have a string separated by dot in Linux Shell,
$example=This.is.My.String
I want to
1.Add some string before the last dot, for example, I want to add "Good.Long" before the last dot, so I get:
This.is.My.Goood.Long.String
2.Get the part after the last dot, so I will get
String
3.Turn the dot into underscore except the last dot, so I will get
This_is_My.String
If you have time, please explain a little bit, I am still learning Regular Expression.
Thanks a lot!
I don't know what you mean by 'Linux Shell' so I will assume bash. This solution will also work in zsh, etcetera:
example=This.is.My.String
before_last_dot=${example%.*}
after_last_dot=${example##*.}
echo ${before_last_dot}.Goood.Long.${after_last_dot}
This.is.My.Goood.Long.String
echo ${before_last_dot//./_}.${after_last_dot}
This_is_My.String
The interim variables before_last_dot and after_last_dot should explain my usage of the % and ## operators. The //, I also think is self-explanatory but I'd be happy to clarify if you have any questions.
This doesn't use sed (or even regular expressions), but bash's inbuilt parameter substitution. I prefer to stick to just one language per script, with as few forks as possible :-)
Other users have given good answers for #1 and #2. There are some disadvantages to some of the answers for #3. In one case, you have to run the substitution twice. In another, if your string has other underscores they might get clobbered. This command works in one go and only affects dots:
sed 's/\(.*\)\./\1\n./;h;s/[^\n]*\n//;x;s/\n.*//;s/\./_/g;G;s/\n//'
It splits the line before the last dot by inserting a newline and copies the result into hold space:
s/\(.*\)\./\1\n./;h
removes everything up to and including the newline from the copy in pattern space and swaps hold space and pattern space:
s/[^\n]*\n//;x
removes everything after and including the newline from the copy that's now in pattern space
s/\n.*//
changes all dots into underscores in the copy in pattern space and appends hold space onto the end of pattern space
s/\./_/g;G
removes the newline that the append operation adds
s/\n//
Then the sed script is finished and the pattern space is output.
At the end of each numbered step (some consist of two actual steps):
Step Pattern Space Hold Space
This.is.My\n.String This.is.My\n.String
This.is.My\n.String .String
This.is.My .String
This_is_My\n.String .String
This_is_My.String .String
Solution
Two versions of this, too:
Complex: sed 's/\(.*\)\([.][^.]*$\)/\1.Goood.Long\2/'
Simple: sed 's/.*\./&Goood.Long./' - thanks Dennis Williamson
What do you want?
Complex: sed 's/.*[.]\([^.]*\)$/\1/'
Simpler: sed 's/.*\.//' - thanks, glenn jackman.
sed 's/\([^.]*\)[.]\([^.]*[.]\)/\1_\2/g'
With 3, you probably need to run the substitute (in its entirety) at least twice, in general.
Explanation
Remember, in sed, the notation \(...\) is a 'capture' that can be referenced as '\1' or similar in the replacement text.
Capture everything up to a string starting with a dot followed by a sequence of non-dots (which you also capture); replace by what came before the last dot, the new material, and the last dot and what came after it.
Ignore everything up to the last dot followed by a capture of a sequence of non-dots; replace with the capture only.
Find and capture a sequence of non-dots, a dot (not captured), followed by a sequence of non-dots and a dot; replace the first dot with an underscore. This is done globally, but the second and subsequent matches won't touch anything already matched. Therefore, I think you need ceil(log2N) passes, where N is the number of dots to be replaced. One pass deals with 1 dot to replace; two passes deals with 2 or 3; three passes deals with 4-7, and so on.
Here's a version that uses Bash's regex matching (Bash 3.2 or greater).
[[ $example =~ ^(.*)\.(.*)$ ]]
echo ${BASH_REMATCH[1]//./_}.${BASH_REMATCH[2]}
Here's a Bash version that uses IFS (Internal Field Separator).
saveIFS=$IFS
IFS=.
array=($e) # * split the string at each dot
lastword=${array[#]: -1}
unset "array[${#array}-1]" # *
IFS=_
echo "${array[*]}.$lastword" # The asterisk as a subscript when inside quotes causes IFS (an underscore in this case) to be inserted between each element of the array
IFS=$saveIFS
* use declare -p array after these steps to see what the array looks like.
1.
$ echo 'This.is.my.string' | sed 's}[^\.][^\.]*$}Good Long.&}'
This.is.my.Good Long.string
before: a dot, then no dot until the end. after: obvious, & is what matched the first part
2.
$ echo 'This.is.my.string' | sed 's}.*\.}}'
string
sed greedy matches, so it will extend the first closure (.*) as far as possible i.e. to the last dot.
3.
$ echo 'This.is.my.string' | tr . _ | sed 's/_\([^_]*\)$/\.\1/'
This_is_my.string
convert all dots to _, then turn the last _ to a dot.
(caveat: this will turn 'This.is.my.string_foo' to 'This_is_my_string.foo', not 'This_is_my.string_foo')
You don't need regular expressions at all (those complex things hurt my eyes!) if you use Awk and are a little creative.
1. echo $example| awk -v ins="Good.long" -F . '{OFS="."; $NF = ins"."$NF;print}'
What this does:
-v ins="Good.long" tells awk to create a variable called 'ins' with "Good.long" as content,
-F . tells awk to use the dot as a separator for your fields for input,
-OFS tells awk to use the dot as a separator for your fields as output,
NF is the number of fields, so $NF represents the last field,
the $NF=... part replaces the last field, it appends the current last string to what you want to insert (the variable called "ins" declared earlier).
2. echo $example| awk -F . '{print $NF}'
$NF is the last field, so that's all!
3. echo $example| awk -F . '{OFS="_"; $(NF-1) = $(NF-1)"."$NF; NF=NF-1; print}'
Here we have to be creative, as Awk AFAIK doesn't allow deleting fields. Of course, we set the output field separateor to underscore.
$(NF-1) = $(NF-1)"."$NF: First, we replace the second last field with the last glued to the second last, with a dot between.
Then, we fool awk to make it think the Number of fields is equal to the number of fields minus one, hence deleting the last field!
Note you can't say $NF="", because then it would display two underscores.