Regular expression to remove all punctuation except commas in regex awk
I have a variable
local= "[ 'service center','New' ]"
I have used the following awk code to remove the square braces and single quotes,
local=gensub(/[[]']+/, "", "g", local);
Expected O/P is
local =" service center,New "
The code I wrote is not working
shell awk regular-expression
asked Feb 1 at 14:01
mittumittu
475
add a comment |
I have a variable
local= "[ 'service center','New' ]"
I have used the following awk code to remove the square braces and single quotes,
local=gensub(/[[]']+/, "", "g", local);
Expected O/P is
local =" service center,New "
The code I wrote is not working
shell awk regular-expression
asked Feb 1 at 14:01
mittumittu
475
1
Please provide some more context. I think you might better of using a json parser instead of trying to remove characters from your string. And please fix your code as it's not valid.
– RoVo
Feb 1 at 14:53
Can you see my updated question
– mittu
Feb 1 at 14:58
Still no context. And if it's all insideawk, you might want to remove theshelltag from the question.
– RoVo
Feb 1 at 15:08
What exactly is not working. The AWK substitutionlocal=gensub(/[[]']+/, "", "g", local);produces exactly the expected value. The expression/[^,[:punct:][:space:]]/you removed with your edit does not match "([:punct:]or[:space:]) and not," but "not ([:punct:]or[:space:]or,)".
– Bodo
Feb 1 at 15:09
1
@mittu Works for me. I think you should edit your question to include a complete, piece of awk code that exhibits the issue so any readers can repeat it.
– ilkkachu
Feb 1 at 15:36
add a comment |
I have a variable
local= "[ 'service center','New' ]"
I have used the following awk code to remove the square braces and single quotes,
local=gensub(/[[]']+/, "", "g", local);
Expected O/P is
local =" service center,New "
The code I wrote is not working
shell awk regular-expression
asked Feb 1 at 14:01
mittumittu
475
I have a variable
local= "[ 'service center','New' ]"
I have used the following awk code to remove the square braces and single quotes,
local=gensub(/[[]']+/, "", "g", local);
Expected O/P is
local =" service center,New "
The code I wrote is not working
shell awk regular-expression
shell awk regular-expression
asked Feb 1 at 14:01
mittumittu
475
asked Feb 1 at 14:01
mittumittu
475
edited Feb 1 at 14:58
mittu
asked Feb 1 at 14:01
mittumittu
475
asked Feb 1 at 14:01
mittumittu
475
asked Feb 1 at 14:01
mittumittu
475
475
1
Please provide some more context. I think you might better of using a json parser instead of trying to remove characters from your string. And please fix your code as it's not valid.
– RoVo
Feb 1 at 14:53
Can you see my updated question
– mittu
Feb 1 at 14:58
Still no context. And if it's all insideawk, you might want to remove theshelltag from the question.
– RoVo
Feb 1 at 15:08
What exactly is not working. The AWK substitutionlocal=gensub(/[[]']+/, "", "g", local);produces exactly the expected value. The expression/[^,[:punct:][:space:]]/you removed with your edit does not match "([:punct:]or[:space:]) and not," but "not ([:punct:]or[:space:]or,)".
– Bodo
Feb 1 at 15:09
1
@mittu Works for me. I think you should edit your question to include a complete, piece of awk code that exhibits the issue so any readers can repeat it.
– ilkkachu
Feb 1 at 15:36
add a comment |
1
Please provide some more context. I think you might better of using a json parser instead of trying to remove characters from your string. And please fix your code as it's not valid.
– RoVo
Feb 1 at 14:53
Can you see my updated question
– mittu
Feb 1 at 14:58
Still no context. And if it's all insideawk, you might want to remove theshelltag from the question.
– RoVo
Feb 1 at 15:08
What exactly is not working. The AWK substitutionlocal=gensub(/[[]']+/, "", "g", local);produces exactly the expected value. The expression/[^,[:punct:][:space:]]/you removed with your edit does not match "([:punct:]or[:space:]) and not," but "not ([:punct:]or[:space:]or,)".
– Bodo
Feb 1 at 15:09
1
@mittu Works for me. I think you should edit your question to include a complete, piece of awk code that exhibits the issue so any readers can repeat it.
– ilkkachu
Feb 1 at 15:36
1
1
Please provide some more context. I think you might better of using a json parser instead of trying to remove characters from your string. And please fix your code as it's not valid.
– RoVo
Feb 1 at 14:53
Please provide some more context. I think you might better of using a json parser instead of trying to remove characters from your string. And please fix your code as it's not valid.
– RoVo
Feb 1 at 14:53
Can you see my updated question
– mittu
Feb 1 at 14:58
Can you see my updated question
– mittu
Feb 1 at 14:58
Still no context. And if it's all inside
awk, you might want to remove the shell tag from the question.– RoVo
Feb 1 at 15:08
Still no context. And if it's all inside
awk, you might want to remove the shell tag from the question.– RoVo
Feb 1 at 15:08
What exactly is not working. The AWK substitution
local=gensub(/[[]']+/, "", "g", local); produces exactly the expected value. The expression /[^,[:punct:][:space:]]/ you removed with your edit does not match "([:punct:] or [:space:]) and not ," but "not ([:punct:] or [:space:] or ,)".– Bodo
Feb 1 at 15:09
What exactly is not working. The AWK substitution
local=gensub(/[[]']+/, "", "g", local); produces exactly the expected value. The expression /[^,[:punct:][:space:]]/ you removed with your edit does not match "([:punct:] or [:space:]) and not ," but "not ([:punct:] or [:space:] or ,)".– Bodo
Feb 1 at 15:09
1
1
@mittu Works for me. I think you should edit your question to include a complete, piece of awk code that exhibits the issue so any readers can repeat it.
– ilkkachu
Feb 1 at 15:36
@mittu Works for me. I think you should edit your question to include a complete, piece of awk code that exhibits the issue so any readers can repeat it.
– ilkkachu
Feb 1 at 15:36
add a comment |
2 Answers
2
active
oldest
votes
You can replace commas with a non-punctuation character, remove all remaining puctuation, then restore the commas:
awk -v local="[ 'service center','New' ]" '
BEGIN {
gsub(/,/, SUBSEP, local)
gsub(/[[:punct:]]/, "", local)
gsub(SUBSEP, ",", local)
printf("local="%s"n", local)
}'
local=" service center,New "
Here, I'm using the builtin SUBSEP variable, described in the GNU awk manual thusly:
The default value of
SUBSEPis the string"34", which contains a nonprinting character that is unlikely to appear in an awk program or in most input data
answered Feb 1 at 20:29
glenn jackmanglenn jackman
51.5k572111
Thank you so much
– mittu
Feb 3 at 2:27
add a comment |
Please find below command and it worked fine
cat filename
local= "[ 'service center','New' ]"
sed "s/[^a-zA-Z,= " ]//g" filename
output: local= " service center,New "
Command: sed "s/[^a-zA-Z,= " ]//g" filename
answered Feb 1 at 18:07
Praveen Kumar BSPraveen Kumar BS
1,474138
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
You can replace commas with a non-punctuation character, remove all remaining puctuation, then restore the commas:
awk -v local="[ 'service center','New' ]" '
BEGIN {
gsub(/,/, SUBSEP, local)
gsub(/[[:punct:]]/, "", local)
gsub(SUBSEP, ",", local)
printf("local="%s"n", local)
}'
local=" service center,New "
Here, I'm using the builtin SUBSEP variable, described in the GNU awk manual thusly:
The default value of
SUBSEPis the string"34", which contains a nonprinting character that is unlikely to appear in an awk program or in most input data
answered Feb 1 at 20:29
glenn jackmanglenn jackman
51.5k572111
Thank you so much
– mittu
Feb 3 at 2:27
add a comment |
You can replace commas with a non-punctuation character, remove all remaining puctuation, then restore the commas:
awk -v local="[ 'service center','New' ]" '
BEGIN {
gsub(/,/, SUBSEP, local)
gsub(/[[:punct:]]/, "", local)
gsub(SUBSEP, ",", local)
printf("local="%s"n", local)
}'
local=" service center,New "
Here, I'm using the builtin SUBSEP variable, described in the GNU awk manual thusly:
The default value of
SUBSEPis the string"34", which contains a nonprinting character that is unlikely to appear in an awk program or in most input data
answered Feb 1 at 20:29
glenn jackmanglenn jackman
51.5k572111
Thank you so much
– mittu
Feb 3 at 2:27
add a comment |
You can replace commas with a non-punctuation character, remove all remaining puctuation, then restore the commas:
awk -v local="[ 'service center','New' ]" '
BEGIN {
gsub(/,/, SUBSEP, local)
gsub(/[[:punct:]]/, "", local)
gsub(SUBSEP, ",", local)
printf("local="%s"n", local)
}'
local=" service center,New "
Here, I'm using the builtin SUBSEP variable, described in the GNU awk manual thusly:
The default value of
SUBSEPis the string"34", which contains a nonprinting character that is unlikely to appear in an awk program or in most input data
answered Feb 1 at 20:29
glenn jackmanglenn jackman
51.5k572111
You can replace commas with a non-punctuation character, remove all remaining puctuation, then restore the commas:
awk -v local="[ 'service center','New' ]" '
BEGIN {
gsub(/,/, SUBSEP, local)
gsub(/[[:punct:]]/, "", local)
gsub(SUBSEP, ",", local)
printf("local="%s"n", local)
}'
local=" service center,New "
Here, I'm using the builtin SUBSEP variable, described in the GNU awk manual thusly:
The default value of
SUBSEPis the string"34", which contains a nonprinting character that is unlikely to appear in an awk program or in most input data
answered Feb 1 at 20:29
glenn jackmanglenn jackman
51.5k572111
answered Feb 1 at 20:29
glenn jackmanglenn jackman
51.5k572111
answered Feb 1 at 20:29
glenn jackmanglenn jackman
51.5k572111
answered Feb 1 at 20:29
glenn jackmanglenn jackman
51.5k572111
51.5k572111
Thank you so much
– mittu
Feb 3 at 2:27
add a comment |
Thank you so much
– mittu
Feb 3 at 2:27
Thank you so much
– mittu
Feb 3 at 2:27
Thank you so much
– mittu
Feb 3 at 2:27
add a comment |
Please find below command and it worked fine
cat filename
local= "[ 'service center','New' ]"
sed "s/[^a-zA-Z,= " ]//g" filename
output: local= " service center,New "
Command: sed "s/[^a-zA-Z,= " ]//g" filename
answered Feb 1 at 18:07
Praveen Kumar BSPraveen Kumar BS
1,474138
add a comment |
Please find below command and it worked fine
cat filename
local= "[ 'service center','New' ]"
sed "s/[^a-zA-Z,= " ]//g" filename
output: local= " service center,New "
Command: sed "s/[^a-zA-Z,= " ]//g" filename
answered Feb 1 at 18:07
Praveen Kumar BSPraveen Kumar BS
1,474138
add a comment |
Please find below command and it worked fine
cat filename
local= "[ 'service center','New' ]"
sed "s/[^a-zA-Z,= " ]//g" filename
output: local= " service center,New "
Command: sed "s/[^a-zA-Z,= " ]//g" filename
answered Feb 1 at 18:07
Praveen Kumar BSPraveen Kumar BS
1,474138
Please find below command and it worked fine
cat filename
local= "[ 'service center','New' ]"
sed "s/[^a-zA-Z,= " ]//g" filename
output: local= " service center,New "
Command: sed "s/[^a-zA-Z,= " ]//g" filename
answered Feb 1 at 18:07
Praveen Kumar BSPraveen Kumar BS
1,474138
answered Feb 1 at 18:07
Praveen Kumar BSPraveen Kumar BS
1,474138
answered Feb 1 at 18:07
Praveen Kumar BSPraveen Kumar BS
1,474138
answered Feb 1 at 18:07
Praveen Kumar BSPraveen Kumar BS
1,474138
1,474138
add a comment |
add a comment |
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1
Please provide some more context. I think you might better of using a json parser instead of trying to remove characters from your string. And please fix your code as it's not valid.
– RoVo
Feb 1 at 14:53
Can you see my updated question
– mittu
Feb 1 at 14:58
Still no context. And if it's all inside
awk, you might want to remove theshelltag from the question.– RoVo
Feb 1 at 15:08
What exactly is not working. The AWK substitution
local=gensub(/[[]']+/, "", "g", local);produces exactly the expected value. The expression/[^,[:punct:][:space:]]/you removed with your edit does not match "([:punct:]or[:space:]) and not," but "not ([:punct:]or[:space:]or,)".– Bodo
Feb 1 at 15:09
1
@mittu Works for me. I think you should edit your question to include a complete, piece of awk code that exhibits the issue so any readers can repeat it.
– ilkkachu
Feb 1 at 15:36