Change known words 1 and 3 of a string whilst keeping unknown word 2 constant
I have a text file and I'm trying to find all instances of Word1 Word2 Word3 in a text file and replace it with Word4 Word2 Word5. Word2 is unknown string but the rest of the words are known.
Here is what I have tried so far
I have a string (...) foobarfoo (...) and I want to replace it with (...) hatbarcar (...)
sed -i 's/foo.*foo/hat.*car/g' data.txt
but the result I get is
(...) hat.*car (...)
So the wildcard is finding the word string I want but I then want to use this same wildcard to write the string which replaces the old one.
Is this possible/anyone have any suggestions?
sed wildcards string
edited Dec 16 at 11:49
Rui F Ribeiro
38.9k1479129
asked Sep 12 '17 at 9:01
JHS
31
add a comment |
I have a text file and I'm trying to find all instances of Word1 Word2 Word3 in a text file and replace it with Word4 Word2 Word5. Word2 is unknown string but the rest of the words are known.
Here is what I have tried so far
I have a string (...) foobarfoo (...) and I want to replace it with (...) hatbarcar (...)
sed -i 's/foo.*foo/hat.*car/g' data.txt
but the result I get is
(...) hat.*car (...)
So the wildcard is finding the word string I want but I then want to use this same wildcard to write the string which replaces the old one.
Is this possible/anyone have any suggestions?
sed wildcards string
edited Dec 16 at 11:49
Rui F Ribeiro
38.9k1479129
asked Sep 12 '17 at 9:01
JHS
31
add a comment |
I have a text file and I'm trying to find all instances of Word1 Word2 Word3 in a text file and replace it with Word4 Word2 Word5. Word2 is unknown string but the rest of the words are known.
Here is what I have tried so far
I have a string (...) foobarfoo (...) and I want to replace it with (...) hatbarcar (...)
sed -i 's/foo.*foo/hat.*car/g' data.txt
but the result I get is
(...) hat.*car (...)
So the wildcard is finding the word string I want but I then want to use this same wildcard to write the string which replaces the old one.
Is this possible/anyone have any suggestions?
sed wildcards string
edited Dec 16 at 11:49
Rui F Ribeiro
38.9k1479129
asked Sep 12 '17 at 9:01
JHS
31
I have a text file and I'm trying to find all instances of Word1 Word2 Word3 in a text file and replace it with Word4 Word2 Word5. Word2 is unknown string but the rest of the words are known.
Here is what I have tried so far
I have a string (...) foobarfoo (...) and I want to replace it with (...) hatbarcar (...)
sed -i 's/foo.*foo/hat.*car/g' data.txt
but the result I get is
(...) hat.*car (...)
So the wildcard is finding the word string I want but I then want to use this same wildcard to write the string which replaces the old one.
Is this possible/anyone have any suggestions?
sed wildcards string
sed wildcards string
edited Dec 16 at 11:49
Rui F Ribeiro
38.9k1479129
asked Sep 12 '17 at 9:01
JHS
31
edited Dec 16 at 11:49
Rui F Ribeiro
38.9k1479129
asked Sep 12 '17 at 9:01
JHS
31
edited Dec 16 at 11:49
Rui F Ribeiro
38.9k1479129
edited Dec 16 at 11:49
Rui F Ribeiro
38.9k1479129
edited Dec 16 at 11:49
Rui F Ribeiro
38.9k1479129
38.9k1479129
asked Sep 12 '17 at 9:01
JHS
31
asked Sep 12 '17 at 9:01
JHS
31
asked Sep 12 '17 at 9:01
JHS
31
31
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
The problem with a
sed -i 's/foo(.*)foo/hat1car/g'
approach is that it would change fooxfoo fooyfoo to hatxfoo fooycar as that .* is greedy.
You can use perl instead with its non-greedy .*? operator.
perl -i -pe 's/foo(.*?)foo/hat$1car/g'
(which also has the advantage of being more portable. That -i comes from perl and is not available in many sed implementations (and when it is, is not interpreted the same by all)).
With GNU sed, and provided $POSIXLY_CORRECT is not in the environment, you could do:
sed -i 's/foo/n/g;s/n([^n]*)n/hat1car/g;s/n/foo/g'
That is, replace foo with a character (n, the line delimiter) that cannot occur in the line so we can use [^n]* to achieve the non-greedy equivalent.
If POSIXLY_CORRECT is in the environment, then [^n] would match any character but and n as POSIX requires instead of any character but newline. You can always do:
(unset -v POSIXLY_CORRECT; exec sed...)
if you want your script to still work in environments where POSIXLY_CORRECT is set.
answered Sep 12 '17 at 9:22
Stéphane Chazelas
299k54563913
add a comment |
The replacement string in s/PATTERN/REPLACEMENT/ is not a regular expression.
You will be able to capture what is matched by a bit of the pattern and use it in the replacement if you wish:
sed -r 's/foo(.*)foo/hat1car/g' file
This will capture anything between two occurrences of foo on the same line, and insert that bit between hat and car. The 1 says "insert anything that was captured by the first parentheses".
Note that .* is "greedy" so if you have foobarfoofoobarfoo, 1 will be barfoofoobar, not bar.
edited Sep 12 '17 at 10:12
Stéphane Chazelas
299k54563913
answered Sep 12 '17 at 9:23
Kusalananda
121k16229372
add a comment |
With sed, you can use ( and ) to make a capture group which can be referenced as 1 in the replacement part of the substitution:
sed 's/foo(.*)foo/hat1car/g'
answered Sep 12 '17 at 9:17
Anthony Geoghegan
7,54543954
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
The problem with a
sed -i 's/foo(.*)foo/hat1car/g'
approach is that it would change fooxfoo fooyfoo to hatxfoo fooycar as that .* is greedy.
You can use perl instead with its non-greedy .*? operator.
perl -i -pe 's/foo(.*?)foo/hat$1car/g'
(which also has the advantage of being more portable. That -i comes from perl and is not available in many sed implementations (and when it is, is not interpreted the same by all)).
With GNU sed, and provided $POSIXLY_CORRECT is not in the environment, you could do:
sed -i 's/foo/n/g;s/n([^n]*)n/hat1car/g;s/n/foo/g'
That is, replace foo with a character (n, the line delimiter) that cannot occur in the line so we can use [^n]* to achieve the non-greedy equivalent.
If POSIXLY_CORRECT is in the environment, then [^n] would match any character but and n as POSIX requires instead of any character but newline. You can always do:
(unset -v POSIXLY_CORRECT; exec sed...)
if you want your script to still work in environments where POSIXLY_CORRECT is set.
answered Sep 12 '17 at 9:22
Stéphane Chazelas
299k54563913
add a comment |
The problem with a
sed -i 's/foo(.*)foo/hat1car/g'
approach is that it would change fooxfoo fooyfoo to hatxfoo fooycar as that .* is greedy.
You can use perl instead with its non-greedy .*? operator.
perl -i -pe 's/foo(.*?)foo/hat$1car/g'
(which also has the advantage of being more portable. That -i comes from perl and is not available in many sed implementations (and when it is, is not interpreted the same by all)).
With GNU sed, and provided $POSIXLY_CORRECT is not in the environment, you could do:
sed -i 's/foo/n/g;s/n([^n]*)n/hat1car/g;s/n/foo/g'
That is, replace foo with a character (n, the line delimiter) that cannot occur in the line so we can use [^n]* to achieve the non-greedy equivalent.
If POSIXLY_CORRECT is in the environment, then [^n] would match any character but and n as POSIX requires instead of any character but newline. You can always do:
(unset -v POSIXLY_CORRECT; exec sed...)
if you want your script to still work in environments where POSIXLY_CORRECT is set.
answered Sep 12 '17 at 9:22
Stéphane Chazelas
299k54563913
add a comment |
The problem with a
sed -i 's/foo(.*)foo/hat1car/g'
approach is that it would change fooxfoo fooyfoo to hatxfoo fooycar as that .* is greedy.
You can use perl instead with its non-greedy .*? operator.
perl -i -pe 's/foo(.*?)foo/hat$1car/g'
(which also has the advantage of being more portable. That -i comes from perl and is not available in many sed implementations (and when it is, is not interpreted the same by all)).
With GNU sed, and provided $POSIXLY_CORRECT is not in the environment, you could do:
sed -i 's/foo/n/g;s/n([^n]*)n/hat1car/g;s/n/foo/g'
That is, replace foo with a character (n, the line delimiter) that cannot occur in the line so we can use [^n]* to achieve the non-greedy equivalent.
If POSIXLY_CORRECT is in the environment, then [^n] would match any character but and n as POSIX requires instead of any character but newline. You can always do:
(unset -v POSIXLY_CORRECT; exec sed...)
if you want your script to still work in environments where POSIXLY_CORRECT is set.
answered Sep 12 '17 at 9:22
Stéphane Chazelas
299k54563913
The problem with a
sed -i 's/foo(.*)foo/hat1car/g'
approach is that it would change fooxfoo fooyfoo to hatxfoo fooycar as that .* is greedy.
You can use perl instead with its non-greedy .*? operator.
perl -i -pe 's/foo(.*?)foo/hat$1car/g'
(which also has the advantage of being more portable. That -i comes from perl and is not available in many sed implementations (and when it is, is not interpreted the same by all)).
With GNU sed, and provided $POSIXLY_CORRECT is not in the environment, you could do:
sed -i 's/foo/n/g;s/n([^n]*)n/hat1car/g;s/n/foo/g'
That is, replace foo with a character (n, the line delimiter) that cannot occur in the line so we can use [^n]* to achieve the non-greedy equivalent.
If POSIXLY_CORRECT is in the environment, then [^n] would match any character but and n as POSIX requires instead of any character but newline. You can always do:
(unset -v POSIXLY_CORRECT; exec sed...)
if you want your script to still work in environments where POSIXLY_CORRECT is set.
answered Sep 12 '17 at 9:22
Stéphane Chazelas
299k54563913
edited Sep 12 '17 at 9:33
answered Sep 12 '17 at 9:22
Stéphane Chazelas
299k54563913
answered Sep 12 '17 at 9:22
Stéphane Chazelas
299k54563913
answered Sep 12 '17 at 9:22
Stéphane Chazelas
299k54563913
299k54563913
add a comment |
add a comment |
The replacement string in s/PATTERN/REPLACEMENT/ is not a regular expression.
You will be able to capture what is matched by a bit of the pattern and use it in the replacement if you wish:
sed -r 's/foo(.*)foo/hat1car/g' file
This will capture anything between two occurrences of foo on the same line, and insert that bit between hat and car. The 1 says "insert anything that was captured by the first parentheses".
Note that .* is "greedy" so if you have foobarfoofoobarfoo, 1 will be barfoofoobar, not bar.
edited Sep 12 '17 at 10:12
Stéphane Chazelas
299k54563913
answered Sep 12 '17 at 9:23
Kusalananda
121k16229372
add a comment |
The replacement string in s/PATTERN/REPLACEMENT/ is not a regular expression.
You will be able to capture what is matched by a bit of the pattern and use it in the replacement if you wish:
sed -r 's/foo(.*)foo/hat1car/g' file
This will capture anything between two occurrences of foo on the same line, and insert that bit between hat and car. The 1 says "insert anything that was captured by the first parentheses".
Note that .* is "greedy" so if you have foobarfoofoobarfoo, 1 will be barfoofoobar, not bar.
edited Sep 12 '17 at 10:12
Stéphane Chazelas
299k54563913
answered Sep 12 '17 at 9:23
Kusalananda
121k16229372
add a comment |
The replacement string in s/PATTERN/REPLACEMENT/ is not a regular expression.
You will be able to capture what is matched by a bit of the pattern and use it in the replacement if you wish:
sed -r 's/foo(.*)foo/hat1car/g' file
This will capture anything between two occurrences of foo on the same line, and insert that bit between hat and car. The 1 says "insert anything that was captured by the first parentheses".
Note that .* is "greedy" so if you have foobarfoofoobarfoo, 1 will be barfoofoobar, not bar.
edited Sep 12 '17 at 10:12
Stéphane Chazelas
299k54563913
answered Sep 12 '17 at 9:23
Kusalananda
121k16229372
The replacement string in s/PATTERN/REPLACEMENT/ is not a regular expression.
You will be able to capture what is matched by a bit of the pattern and use it in the replacement if you wish:
sed -r 's/foo(.*)foo/hat1car/g' file
This will capture anything between two occurrences of foo on the same line, and insert that bit between hat and car. The 1 says "insert anything that was captured by the first parentheses".
Note that .* is "greedy" so if you have foobarfoofoobarfoo, 1 will be barfoofoobar, not bar.
edited Sep 12 '17 at 10:12
Stéphane Chazelas
299k54563913
answered Sep 12 '17 at 9:23
Kusalananda
121k16229372
edited Sep 12 '17 at 10:12
Stéphane Chazelas
299k54563913
edited Sep 12 '17 at 10:12
Stéphane Chazelas
299k54563913
edited Sep 12 '17 at 10:12
Stéphane Chazelas
299k54563913
299k54563913
answered Sep 12 '17 at 9:23
Kusalananda
121k16229372
answered Sep 12 '17 at 9:23
Kusalananda
121k16229372
answered Sep 12 '17 at 9:23
Kusalananda
121k16229372
121k16229372
add a comment |
add a comment |
With sed, you can use ( and ) to make a capture group which can be referenced as 1 in the replacement part of the substitution:
sed 's/foo(.*)foo/hat1car/g'
answered Sep 12 '17 at 9:17
Anthony Geoghegan
7,54543954
add a comment |
With sed, you can use ( and ) to make a capture group which can be referenced as 1 in the replacement part of the substitution:
sed 's/foo(.*)foo/hat1car/g'
answered Sep 12 '17 at 9:17
Anthony Geoghegan
7,54543954
add a comment |
With sed, you can use ( and ) to make a capture group which can be referenced as 1 in the replacement part of the substitution:
sed 's/foo(.*)foo/hat1car/g'
answered Sep 12 '17 at 9:17
Anthony Geoghegan
7,54543954
With sed, you can use ( and ) to make a capture group which can be referenced as 1 in the replacement part of the substitution:
sed 's/foo(.*)foo/hat1car/g'
answered Sep 12 '17 at 9:17
Anthony Geoghegan
7,54543954
answered Sep 12 '17 at 9:17
Anthony Geoghegan
7,54543954
answered Sep 12 '17 at 9:17
Anthony Geoghegan
7,54543954
answered Sep 12 '17 at 9:17
Anthony Geoghegan
7,54543954
7,54543954
add a comment |
add a comment |
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