I need to search for each instance of a colon ":" and then prepend a string to the word before that colon.
Example:
some data here word:number
Desired outcome:
some data here prepend_word:number
I've tried:
sed "s/:/s/^/prepend_/g"
This adds prepend_ to the beginning of the line: prepend_some data here word:number
sed "s/:/prepend_&/g"
this adds prepend_ right before the colon: some data here wordprepend_:number
You need to use
sed 's/[^[:space:]]*:/prepend_&/g'
The [^[:space:]]*: pattern searches for 0 or more non-whitespace chars and a : after them, and the prepend_& replacement pattern will replace the match with itself (see &) and insert prepend_ before it.
See an online sed demo:
sed 's/[^[:space:]]*:/prepend_&/g' <<< "some data here word:number more:here"
Output: some data here prepend_word:number prepend_more:here.
Related
I would like to sed to find and replace every occurrence of - with _ but only before the first occurrence of = on every line.
Here is a dataset to work with:
ke-y_0-1="foo"
key_two="bar"
key_03-three="baz-jazz-mazz"
key-="rax_foo"
key-05-five="craz-"
In the end the dataset should look like this:
ke_y_0_1="foo"
key_two="bar"
key_03_three="baz-jazz-mazz"
key_="rax_foo"
key_05_five="craz-"
I found this regex will match properly.
\-(?=.*=)
However the regex uses positive lookaheads and it appears that sed (even with -E, -e or -r) dose not know how to work with positive lookaheads.
I tried the following but keep getting Invalid preceding regular expression
cat dataset.txt | sed -r "s/-(?=.*=)/_/g"
Is it possible to convert this in a usable way with sed?
Note, I do not want to use perl. However I am open to awk.
You can use
sed ':a;s/^\([^=]*\)-/\1_/;ta' file
See the online demo:
#!/bin/bash
s='ke-y_0-1="foo"
key_two="bar"
key_03-three="baz-jazz-mazz"
key-="rax_foo"
key-05-five="craz-"'
sed ':a; s/^\([^=]*\)-/\1_/;ta' <<< "$s"
Output:
ke_y_0_1="foo"
key_two="bar"
key_03_three="baz-jazz-mazz"
key_="rax_foo"
key_05_five="craz-"
Details:
:a - setting a label named a
s/^\([^=]*\)-/\1_/ - find any zero or more chars other than a = char from the start of string (while capturing into Group 1 (\1)) and then matches a - char, and replaces with Group 1 value (\1) and a _ (that replaces the found - char)
ta - jump to lable a location upon successful replacement. Else, stop.
You might also use awk setting the field separator to = and replace all - with _ for the first field.
To print only the replaced lines:
awk 'BEGIN{FS=OFS="="}gsub("-", "_", $1)' file
Output
ke_y_0_1="foo"
key_03_three="baz-jazz-mazz"
key_="rax_foo"
key_05_five="craz-"
If you want to print all lines:
awk 'BEGIN{FS=OFS="="}{gsub("-", "_", $1);print}' file
original line in file sed.txt:
outer_string_PATTERN_string(PATTERN_And_PATTERN_PATTERN_i)PATTERN_outer_string(i_PATTERN_inner)_outer_string
only need to replace PATTERN to pattern which in brackets, not lowercase, it could replace to other word.
expect result:
outer_string_PATTERN_string(pattern_And_pattern_pattern_i)PATTERN_outer_string(i_pattern_inner)_outer_string
I could use ([^)]*) pattern to find the substring which would be replace some worlds in. But I can't use this pattern to index the substring's position, and it will replace the whole line's PATTERN to pattern.
:/tmp$ sed 's/([^)]*)/---/g' sed.txt
outer_string_PATTERN_string---PATTERN_outer_string---_outer_string
:/tmp$ sed '/([^)]*)/s/PATTERN/pattern/g' sed.txt
outer_string_pattern_string(pattern_And_pattern_pattern_i)pattern_outer_string(i_pattern_inner)_outer_string
I also tried to use the regex group in sed to capture and replace the words, but I can't figure out the command.
Can sed implement that? And how to achieve that? THX.
Can sed implement that?
It can be done using GNU sed and basic regular expressions
(BRE):
sed '
s/)/)\n/g
:1
s/\(([^)]*\)PATTERN\([^)]*)\n\)/\1pattern\2/
t1
s/\n//g
' < file
where
1st s inserts a newline after each )
2nd s replaces the last (* is greedy) PATTERN inside ()s with pattern
t loops back if a substitution was made
3rd s strips all inserted newlines
EDIT
2nd substitute command edited according to OP's suggestion
since there is no need to match \n inside ().
Can sed implement that?
Yes. But you do not want to do it in sed. Use other programming language, like Python, Perl, or awk.
how to achieve that?
Implementing non-greedy regex is not simple in sed. Basically, generally, it consists of:
taking chunk of the input
process the chunk
put it in hold space
shuffle hold with pattern space - extract what been already processed, what's not
repeat
shuffle with hold space
output
Anyway, the following script:
#!/bin/bash
sed <<<'outer_string_PATTERN_string(PATTERN_i_PATTERN_PATTERN_i)PATTERN_outer_string(i_PATTERN_inner)_outer_string' '
:loop;
/\([^(]*\)\(([^)]*)\)\(.*\)/{
# Lowercase the second part.
s//\1\L\2\E\n\3/;
# Mix with hold space.
G;
s/\(.*\)\n\(.*\)\n\(.*\)/\3\1\n\2/;
# Put processed stuff into hold spcae
h; s/\n.*//; x;
# Process the other stuff again.
s/.*\n//;
bloop;
};
# Is hold space empty?
x; /^$/!{
# Pattern space has trailing stuff - add it.
G; s/\n//;
# We will print it.
h;
# Clear hold space
s/.*//
};x;
'
outputs:
PATTERN_outer_string(i_pattern_inner)outer_string_PATTERN_string(pattern_i_pattern_pattern_i)_outer_string
As an alternative, it is easier to do this in gnu awk with RS that matches (...) substring:
awk -v RS='\\([^)]+)' '{gsub(/PATTERN/, "pattern", RT); ORS=RT} 1' file
outer_string_PATTERN_string(pattern_i_pattern_pattern_i)PATTERN_outer_string(i_pattern_inner)_outer_string
Steps:
RS='\\([^)]+)' captures a (...) string as record separator
gsub function then replaces PATTERN with pattern in matched text i.e. RT
ORS=RT sets ORS as the new modified RT
1 prints each record to stdout
Another alternative solution using lookahead assertion in a perl regex:
perl -pe 's/PATTERN(?=[^()]*\))/pattern/g' file
Solved by this:
:/tmp$ sed 's/(/\n(/g' sed.txt | sed 's/)/)\n/g' | sed '/([^)]*)/s/PATTERN/pattern/g' | sed ':a;N;$!ba;s/\n//g'
outer_string_PATTERN_string(pattern_And_pattern_pattern_i)PATTERN_outer_string(i_pattern_inner)_outer_string
make pattern () in a new line
find the () lines and replace the PATTERN to pattern
merge multiple lines in one line
thanks for How can I replace a newline (\n) using sed?
Because the question can be misleading, here is a little example. I have this kind of file:
some text
some text ##some-text-KEY-some-other-text##
text again ##some-text-KEY-some-other-text## ##some-text-KEY-some-other-text##
again ##some-text-KEY-some-other-text-KEY-text##
some text with KEY ##KEY-some-text##
blabla ##KEY##
In this example, I want to replace each occurrence of KEY- inside a pair of ## by VALUE-. I started with this sed command:
sed -i 's/\(##[^#]*\)KEY-\([^#]*##\)/\1VALUE-\2/g'
Here is how it works:
\(##[^#]*\): create a first group composed of two # and any characters except # ...
KEY-: ... until the last occurrence of KEY- on that line
\([^#]*##\): and create a second group with all the characters except # until the next pair of #.
The problem is my command can't handle correctly the following line because there are multiple KEY- inside my pair of ##:
again ##some-text-KEY-some-other-text-KEY-text##
Indeed, I get this result:
again ##some-text-KEY-some-other-text-VALUE-text##
If I want to replace all the occurrences of KEY- in that line, I have to run my command multiple times and I prefer to avoid that. I also tried with lazy operators but the problem is the same.
How can I create a regex and a sed command who can handle correctly all my file?
The problem is rather complex: you need to replace all occurrences of some multicharacter text inside blocks of text between identical multicharacter delimiters.
The easiest and safest way to solve the task is using Perl:
perl -i -pe 's/(##)(.*?)(##)/$end_delim=$3; "$1" . $2=~s|KEY-|VALUE-|gr . "$end_delim"/ge' file
See the online demo.
The (##)(.*?)(##) pattern will match strings between two adjacent ## substrings capturing the start delimiter into Group 1, end delimiter in Group 3, and all text in between into Group 2. Since the regex substitution re-sets all placeholders, the temporary variable is used to keep the value of the end delimiter ($end_delim=$3), then, "$1" . $2=~s|KEY-|VALUE-|gr . "$end_delim" replaces the match with the value in the Group 1 of the first match (the first ##), then the Group 2 value with all KEY- replaced with VALUE-, and then the end delimiter.
If there are no KEY-s in between matches on the same line you may use a branch with sed by enclosing your command with :A and tA:
sed -i ':A; s/\(##[^#]*\)KEY-\([^#]*##\)/\1VALUE-\2/g; tA' file
Note you missed the first placeholder in \VALUE-\2, it should be \1VALUE-\2.
See the online demo:
s="some KEY- text
some text ##some-text-KEY-some-other-text##
text again ##some-text-KEY-some-other-text## ##some-text-KEY-some-other-text##
again ##some-text-KEY-some-other-text-KEY-text##
some text with KEY ##KEY-some-text##
blabla ##KEY##"
sed ':A; s/\(##[^#]*\)KEY-\([^#]*##\)/\1VALUE-\2/g; tA' <<< "$s"
Output:
some KEY- text
some text ##some-text-VALUE-some-other-text##
text again ##some-text-VALUE-some-other-text## ##some-text-VALUE-some-other-text##
again ##some-text-VALUE-some-other-text-VALUE-text##
some text with KEY ##VALUE-some-text##
blabla ##KEY##
More details:
sed allows the usage of loops and branches. The :A in the code above is a label, a special location marker that can be "jumped at" using the appropriate operator. t is used to create a branch, this "command jumps to the label only if the previous substitute command was successful". So, once the pattern matched and the replacement occurred, sed goes back to where it was and re-tries a match. If it is not successful, sed goes on to search for the matches further in the string. So, tA means go back to the location marked with A if there was a successful search-and-replace operation.
This might work for you (GNU sed):
sed -E 's/##/\n/g;:a;s/^([^\n]*(\n[^\n]*\n[^\n]*)*\n[^\n]*)KEY-/\1VALUE-/;ta;s/\n/##/g' file
Convert ##'s to newlines. Using a loop, replace VAL- between matched newlines to VALUE-. When all done replace newlines by ##'s.
I have a file which I want to modify into a new file using cat.
So the file contains lines like:
name "myName"
place "xyz"
and so on....
I want these lines to be changed to
name "Jon"
place "paris"
I tried to do it like this but its not working:
cat originalFile | sed 's/^name\*/name "Jon"/' > tempFile
I tried using all sorts of special characters and it did not work. I am unable to recognize the space characters after name and then "myName".
You may match the rest of the line using .*, and you may match a space with a space, or [[:blank:]] or [[:space:]]:
sed 's/^\(name[[:space:]]\).*/\1"Jon"/;s/^\(place[[:space:]]\).*/\1"paris"/' originalFile > tempFile
Note there are two replace commands here joined with s semicolon. The first parts are wrapped with a capturing group that is necessary because the space POSIX character class is not literal and in order to keep it after replacing the \1 backreference should be used (to insert the text captured with Group 1).
See the online demo:
s='name "myName"
place "xyz"'
sed 's/^\(name[[:space:]]\).*/\1"Jon"/;s/^\(place[[:space:]]\).*/\1"paris"/' <<< "$s"
Output:
name "Jon"
place "paris"
An awk alternative:
awk '$1=="name"{$0="name \"Jon\""} $1=="place"{$0="place \"paris\""} 1' originalFile
It will work when there're space(s) before name or place.
It's not regex match here but just string compare.
awk separates fields by space characters which including \n or .
Append > tempFile to it when the results seems correct to you.
I've been trying to clean up the data in a csv file which contain data similar to this:
8979880, Number One : Exclusive Mix, 387387, http://www.smashhits.com
4844404, Top 40 : 1988, 3893938, http://www.best80s.com
48094940, Highlander:The Return, 489494, http://www.instantaccess.com
My goal is to replace the colon in field 2 with a space. Initially I used sed to replace the : with a spacelike so:
sed i "s/:/ /g" file.csv
This works in removing the colon but unfortunately this also removes the colon in the url which is not what I want. How can I specify that I only want the command to affect the data in field 2?
Using awk you can do
awk '/:/{sub(/:/, " ")} 1' file.csv
With /:/ you match the first occurrence of :
With {sub(/:/, " ")} you replace : with a space
1 simply prints the line.
You can use gnu sed like this:
sed -r 's/^([^,]*,[^,]*):/\1 /g' file.csv
Explanation
^ anchors the expression at the start of each line
now [^,]*, matches the first field including the separator
and then [^,]*: matches from the second field to the :
the parenthises ^(...): take care that everything up to but not including the : in the second field is captured into \1
finally the replacement with \1 (there is a space after the \1 does the replacement of the : with space on line where the regex matched