Related
Obviously, you can use the | (pipe?) to represent OR, but is there a way to represent AND as well?
Specifically, I'd like to match paragraphs of text that contain ALL of a certain phrase, but in no particular order.
Use a non-consuming regular expression.
The typical (i.e. Perl/Java) notation is:
(?=expr)
This means "match expr but after that continue matching at the original match-point."
You can do as many of these as you want, and this will be an "and." Example:
(?=match this expression)(?=match this too)(?=oh, and this)
You can even add capture groups inside the non-consuming expressions if you need to save some of the data therein.
You need to use lookahead as some of the other responders have said, but the lookahead has to account for other characters between its target word and the current match position. For example:
(?=.*word1)(?=.*word2)(?=.*word3)
The .* in the first lookahead lets it match however many characters it needs to before it gets to "word1". Then the match position is reset and the second lookahead seeks out "word2". Reset again, and the final part matches "word3"; since it's the last word you're checking for, it isn't necessary that it be in a lookahead, but it doesn't hurt.
In order to match a whole paragraph, you need to anchor the regex at both ends and add a final .* to consume the remaining characters. Using Perl-style notation, that would be:
/^(?=.*word1)(?=.*word2)(?=.*word3).*$/m
The 'm' modifier is for multline mode; it lets the ^ and $ match at paragraph boundaries ("line boundaries" in regex-speak). It's essential in this case that you not use the 's' modifier, which lets the dot metacharacter match newlines as well as all other characters.
Finally, you want to make sure you're matching whole words and not just fragments of longer words, so you need to add word boundaries:
/^(?=.*\bword1\b)(?=.*\bword2\b)(?=.*\bword3\b).*$/m
Look at this example:
We have 2 regexps A and B and we want to match both of them, so in pseudo-code it looks like this:
pattern = "/A AND B/"
It can be written without using the AND operator like this:
pattern = "/NOT (NOT A OR NOT B)/"
in PCRE:
"/(^(^A|^B))/"
regexp_match(pattern,data)
The AND operator is implicit in the RegExp syntax.
The OR operator has instead to be specified with a pipe.
The following RegExp:
var re = /ab/;
means the letter a AND the letter b.
It also works with groups:
var re = /(co)(de)/;
it means the group co AND the group de.
Replacing the (implicit) AND with an OR would require the following lines:
var re = /a|b/;
var re = /(co)|(de)/;
You can do that with a regular expression but probably you'll want to some else. For example use several regexp and combine them in a if clause.
You can enumerate all possible permutations with a standard regexp, like this (matches a, b and c in any order):
(abc)|(bca)|(acb)|(bac)|(cab)|(cba)
However, this makes a very long and probably inefficient regexp, if you have more than couple terms.
If you are using some extended regexp version, like Perl's or Java's, they have better ways to do this. Other answers have suggested using positive lookahead operation.
Is it not possible in your case to do the AND on several matching results? in pseudocode
regexp_match(pattern1, data) && regexp_match(pattern2, data) && ...
Why not use awk?
with awk regex AND, OR matters is so simple
awk '/WORD1/ && /WORD2/ && /WORD3/' myfile
The order is always implied in the structure of the regular expression. To accomplish what you want, you'll have to match the input string multiple times against different expressions.
What you want to do is not possible with a single regexp.
If you use Perl regular expressions, you can use positive lookahead:
For example
(?=[1-9][0-9]{2})[0-9]*[05]\b
would be numbers greater than 100 and divisible by 5
In addition to the accepted answer
I will provide you with some practical examples that will get things more clear to some of You. For example lets say we have those three lines of text:
[12/Oct/2015:00:37:29 +0200] // only this + will get selected
[12/Oct/2015:00:37:x9 +0200]
[12/Oct/2015:00:37:29 +020x]
See demo here DEMO
What we want to do here is to select the + sign but only if it's after two numbers with a space and if it's before four numbers. Those are the only constraints. We would use this regular expression to achieve it:
'~(?<=\d{2} )\+(?=\d{4})~g'
Note if you separate the expression it will give you different results.
Or perhaps you want to select some text between tags... but not the tags! Then you could use:
'~(?<=<p>).*?(?=<\/p>)~g'
for this text:
<p>Hello !</p> <p>I wont select tags! Only text with in</p>
See demo here DEMO
You could pipe your output to another regex. Using grep, you could do this:
grep A | grep B
((yes).*(no))|((no).*(yes))
Will match sentence having both yes and no at the same time, regardless the order in which they appear:
Do i like cookies? **Yes**, i do. But milk - **no**, definitely no.
**No**, you may not have my phone. **Yes**, you may go f yourself.
Will both match, ignoring case.
Use AND outside the regular expression. In PHP lookahead operator did not not seem to work for me, instead I used this
if( preg_match("/^.{3,}$/",$pass1) && !preg_match("/\s{1}/",$pass1))
return true;
else
return false;
The above regex will match if the password length is 3 characters or more and there are no spaces in the password.
Here is a possible "form" for "and" operator:
Take the following regex for an example:
If we want to match words without the "e" character, we could do this:
/\b[^\We]+\b/g
\W means NOT a "word" character.
^\W means a "word" character.
[^\We] means a "word" character, but not an "e".
see it in action: word without e
"and" Operator for Regular Expressions
I think this pattern can be used as an "and" operator for regular expressions.
In general, if:
A = not a
B = not b
then:
[^AB] = not(A or B)
= not(A) and not(B)
= a and b
Difference Set
So, if we want to implement the concept of difference set in regular expressions, we could do this:
a - b = a and not(b)
= a and B
= [^Ab]
Obviously, you can use the | (pipe?) to represent OR, but is there a way to represent AND as well?
Specifically, I'd like to match paragraphs of text that contain ALL of a certain phrase, but in no particular order.
Use a non-consuming regular expression.
The typical (i.e. Perl/Java) notation is:
(?=expr)
This means "match expr but after that continue matching at the original match-point."
You can do as many of these as you want, and this will be an "and." Example:
(?=match this expression)(?=match this too)(?=oh, and this)
You can even add capture groups inside the non-consuming expressions if you need to save some of the data therein.
You need to use lookahead as some of the other responders have said, but the lookahead has to account for other characters between its target word and the current match position. For example:
(?=.*word1)(?=.*word2)(?=.*word3)
The .* in the first lookahead lets it match however many characters it needs to before it gets to "word1". Then the match position is reset and the second lookahead seeks out "word2". Reset again, and the final part matches "word3"; since it's the last word you're checking for, it isn't necessary that it be in a lookahead, but it doesn't hurt.
In order to match a whole paragraph, you need to anchor the regex at both ends and add a final .* to consume the remaining characters. Using Perl-style notation, that would be:
/^(?=.*word1)(?=.*word2)(?=.*word3).*$/m
The 'm' modifier is for multline mode; it lets the ^ and $ match at paragraph boundaries ("line boundaries" in regex-speak). It's essential in this case that you not use the 's' modifier, which lets the dot metacharacter match newlines as well as all other characters.
Finally, you want to make sure you're matching whole words and not just fragments of longer words, so you need to add word boundaries:
/^(?=.*\bword1\b)(?=.*\bword2\b)(?=.*\bword3\b).*$/m
Look at this example:
We have 2 regexps A and B and we want to match both of them, so in pseudo-code it looks like this:
pattern = "/A AND B/"
It can be written without using the AND operator like this:
pattern = "/NOT (NOT A OR NOT B)/"
in PCRE:
"/(^(^A|^B))/"
regexp_match(pattern,data)
The AND operator is implicit in the RegExp syntax.
The OR operator has instead to be specified with a pipe.
The following RegExp:
var re = /ab/;
means the letter a AND the letter b.
It also works with groups:
var re = /(co)(de)/;
it means the group co AND the group de.
Replacing the (implicit) AND with an OR would require the following lines:
var re = /a|b/;
var re = /(co)|(de)/;
You can do that with a regular expression but probably you'll want to some else. For example use several regexp and combine them in a if clause.
You can enumerate all possible permutations with a standard regexp, like this (matches a, b and c in any order):
(abc)|(bca)|(acb)|(bac)|(cab)|(cba)
However, this makes a very long and probably inefficient regexp, if you have more than couple terms.
If you are using some extended regexp version, like Perl's or Java's, they have better ways to do this. Other answers have suggested using positive lookahead operation.
Is it not possible in your case to do the AND on several matching results? in pseudocode
regexp_match(pattern1, data) && regexp_match(pattern2, data) && ...
Why not use awk?
with awk regex AND, OR matters is so simple
awk '/WORD1/ && /WORD2/ && /WORD3/' myfile
The order is always implied in the structure of the regular expression. To accomplish what you want, you'll have to match the input string multiple times against different expressions.
What you want to do is not possible with a single regexp.
If you use Perl regular expressions, you can use positive lookahead:
For example
(?=[1-9][0-9]{2})[0-9]*[05]\b
would be numbers greater than 100 and divisible by 5
In addition to the accepted answer
I will provide you with some practical examples that will get things more clear to some of You. For example lets say we have those three lines of text:
[12/Oct/2015:00:37:29 +0200] // only this + will get selected
[12/Oct/2015:00:37:x9 +0200]
[12/Oct/2015:00:37:29 +020x]
See demo here DEMO
What we want to do here is to select the + sign but only if it's after two numbers with a space and if it's before four numbers. Those are the only constraints. We would use this regular expression to achieve it:
'~(?<=\d{2} )\+(?=\d{4})~g'
Note if you separate the expression it will give you different results.
Or perhaps you want to select some text between tags... but not the tags! Then you could use:
'~(?<=<p>).*?(?=<\/p>)~g'
for this text:
<p>Hello !</p> <p>I wont select tags! Only text with in</p>
See demo here DEMO
You could pipe your output to another regex. Using grep, you could do this:
grep A | grep B
((yes).*(no))|((no).*(yes))
Will match sentence having both yes and no at the same time, regardless the order in which they appear:
Do i like cookies? **Yes**, i do. But milk - **no**, definitely no.
**No**, you may not have my phone. **Yes**, you may go f yourself.
Will both match, ignoring case.
Use AND outside the regular expression. In PHP lookahead operator did not not seem to work for me, instead I used this
if( preg_match("/^.{3,}$/",$pass1) && !preg_match("/\s{1}/",$pass1))
return true;
else
return false;
The above regex will match if the password length is 3 characters or more and there are no spaces in the password.
Here is a possible "form" for "and" operator:
Take the following regex for an example:
If we want to match words without the "e" character, we could do this:
/\b[^\We]+\b/g
\W means NOT a "word" character.
^\W means a "word" character.
[^\We] means a "word" character, but not an "e".
see it in action: word without e
"and" Operator for Regular Expressions
I think this pattern can be used as an "and" operator for regular expressions.
In general, if:
A = not a
B = not b
then:
[^AB] = not(A or B)
= not(A) and not(B)
= a and b
Difference Set
So, if we want to implement the concept of difference set in regular expressions, we could do this:
a - b = a and not(b)
= a and B
= [^Ab]
consider following scenario
input string = "WIPR.NS"
i have to replace this with "WIPR2.NS"
i am using following logic.
match pattern = "(.*)\.NS$" \\ any string that ends with .NS
replace pattern = "$12.NS"
In above case, since there is no group with index 12, i get result $12.NS
But what i want is "WIPR2.NS".
If i don't have digit 2 to replace, it works in all other cases but not working for 2.
How to resolve this case?
Thanks in advance,
Alok
Usually depends entirely on your regex engine (I'm not familiar with those that use $1 to represent a capture group, I'm more used to \1 but you'd have the same problem with that).
Some will provide a delimiter that you can use, like:
replace pattern = "${1}2.NS"
which clearly indicates that you want capture group 1 followed by the literal 2.NS.
In fact, by looking at this page, it appears that's exactly the way to do it (assuming .NET):
To replace with the first backreference immediately followed by the digit 9, use ${1}9. If you type $19, and there are less than 19 backreferences, the $19 will be interpreted as literal text, and appear in the result string as such.
Also keep in mind that Jay provides an excellent answer for this specific use case that doesn't require capture groups at all (by just replacing .NS with 2.NS).
You may want to look into that as a possibility - I'll leave this answer here since:
it's the accepted answer; and
it probably better for the more complex cases, like changing X([A-Z])4([A-Z]) with X${1}5${2}, where you have variable text on either side of the bit you wish to modify.
You don't need to do anything with what precedes the .NS, since only what is being matched is subject to replacement.
match pattern = "\.NS$" (any string that ends with .NS -- don't forget to escape the .)
replace pattern = "2.NS"
You can further refine this with lookaround zero-width assertions, but that depends on your regex engine, and you have not specified the environment/programming language in which you are working.
Ok i got this regex:
^[\w\s]+=["']\w+['"]
Now the regex will match:
a href='google'
a href="google"
and also
a href='google"
How can i enforce regex to match its quote?
If first quote is single quote, how can i make the last quote also a single quote not a double quote
Read about backreferences.
^[\w\s]+=(["'])\w+?\1
Note that you want to put a ? after the second + or else it will be greedy. However, in general this is not the right way to parse HTML. Use Beautiful Soup.
I am afraid you will have to do it the long way:
^[\w\s]+=("\w+"|'\w+')
More technically, ensuring correct matching / nesting of quotes is not a problem for a regular grammar so for more complex problems you would have to use a proper parser (or perl6 style extended regular expression but they technically do not class as regular expressions).
Replace the ['"] with \1 to use a back reference (capture group)
^[\w\s]+=["']\w+\1
What exactly do you want to match? It sounds you want to match:
word (tagname)
mandatory whitespace
word (attr name)
optional whitespace
=
optional whitespace
either single quoted or double quoted anything (attr value)
That would be: ^(\w+)\s+(\w+)\s*=\s*(?:'([^']*)'|"([^"]*)")
This will allow matches like:
a href='' - empty attr
a href='Hello world' - spaces and other non-word characters in quoted part
a href="one 'n two" - quotes of different kind in quoted part
a href = 'google' - spaces on both sides of =
And disallow things like these that your original regexp allows:
a b c href='google' - extra words
='google' - only spaces on the left
href='google' - only attr on the left
It still doesn't sound exactly right - you're trying to match a tag with exactly one attribute?
With this regexp, tag name will be in $1, attr name in $2, and attr value in either $3 or $4 (the other being nil - most languages distinguish group not taken with nil vs group taken but empty with "" if you need it).
Regexp that would ensure attr value gets in the same group would be messier if you wanted to allow single quotes in doubly quoted attr value and vice verse - something like ^(\w+)\s+(\w+)\s*=\s*(['"])((?:(?!\3).)*)\3 ((?!) is zero-width negative look-ahead - (?:(?!\3).) means something like [^\3] except the latter isn't supported).
If you don't care about this ^(\w+)\s+(\w+)\s*=\s*(['"])(['"]*)\3 will do just fine (for both $3 will be quote type, and $4 attr value).
By the way re (["'])\w+?\1 above - \w doesn't match quotes, so this ? doesn't change anything.
Having said all that, use a real HTML parser ;-)
These regexps will work in Perl and Ruby. Other languages usually copy Perl's regexp system, but often introduce minor changes so some adjustments might be necessary. Especially the one with negative look-aheads might be unsupported.
Try this:
^[\w\s]+="\w+"|^[\w\s]+='\w+'
Obviously, you can use the | (pipe?) to represent OR, but is there a way to represent AND as well?
Specifically, I'd like to match paragraphs of text that contain ALL of a certain phrase, but in no particular order.
Use a non-consuming regular expression.
The typical (i.e. Perl/Java) notation is:
(?=expr)
This means "match expr but after that continue matching at the original match-point."
You can do as many of these as you want, and this will be an "and." Example:
(?=match this expression)(?=match this too)(?=oh, and this)
You can even add capture groups inside the non-consuming expressions if you need to save some of the data therein.
You need to use lookahead as some of the other responders have said, but the lookahead has to account for other characters between its target word and the current match position. For example:
(?=.*word1)(?=.*word2)(?=.*word3)
The .* in the first lookahead lets it match however many characters it needs to before it gets to "word1". Then the match position is reset and the second lookahead seeks out "word2". Reset again, and the final part matches "word3"; since it's the last word you're checking for, it isn't necessary that it be in a lookahead, but it doesn't hurt.
In order to match a whole paragraph, you need to anchor the regex at both ends and add a final .* to consume the remaining characters. Using Perl-style notation, that would be:
/^(?=.*word1)(?=.*word2)(?=.*word3).*$/m
The 'm' modifier is for multline mode; it lets the ^ and $ match at paragraph boundaries ("line boundaries" in regex-speak). It's essential in this case that you not use the 's' modifier, which lets the dot metacharacter match newlines as well as all other characters.
Finally, you want to make sure you're matching whole words and not just fragments of longer words, so you need to add word boundaries:
/^(?=.*\bword1\b)(?=.*\bword2\b)(?=.*\bword3\b).*$/m
Look at this example:
We have 2 regexps A and B and we want to match both of them, so in pseudo-code it looks like this:
pattern = "/A AND B/"
It can be written without using the AND operator like this:
pattern = "/NOT (NOT A OR NOT B)/"
in PCRE:
"/(^(^A|^B))/"
regexp_match(pattern,data)
The AND operator is implicit in the RegExp syntax.
The OR operator has instead to be specified with a pipe.
The following RegExp:
var re = /ab/;
means the letter a AND the letter b.
It also works with groups:
var re = /(co)(de)/;
it means the group co AND the group de.
Replacing the (implicit) AND with an OR would require the following lines:
var re = /a|b/;
var re = /(co)|(de)/;
You can do that with a regular expression but probably you'll want to some else. For example use several regexp and combine them in a if clause.
You can enumerate all possible permutations with a standard regexp, like this (matches a, b and c in any order):
(abc)|(bca)|(acb)|(bac)|(cab)|(cba)
However, this makes a very long and probably inefficient regexp, if you have more than couple terms.
If you are using some extended regexp version, like Perl's or Java's, they have better ways to do this. Other answers have suggested using positive lookahead operation.
Is it not possible in your case to do the AND on several matching results? in pseudocode
regexp_match(pattern1, data) && regexp_match(pattern2, data) && ...
Why not use awk?
with awk regex AND, OR matters is so simple
awk '/WORD1/ && /WORD2/ && /WORD3/' myfile
The order is always implied in the structure of the regular expression. To accomplish what you want, you'll have to match the input string multiple times against different expressions.
What you want to do is not possible with a single regexp.
If you use Perl regular expressions, you can use positive lookahead:
For example
(?=[1-9][0-9]{2})[0-9]*[05]\b
would be numbers greater than 100 and divisible by 5
In addition to the accepted answer
I will provide you with some practical examples that will get things more clear to some of You. For example lets say we have those three lines of text:
[12/Oct/2015:00:37:29 +0200] // only this + will get selected
[12/Oct/2015:00:37:x9 +0200]
[12/Oct/2015:00:37:29 +020x]
See demo here DEMO
What we want to do here is to select the + sign but only if it's after two numbers with a space and if it's before four numbers. Those are the only constraints. We would use this regular expression to achieve it:
'~(?<=\d{2} )\+(?=\d{4})~g'
Note if you separate the expression it will give you different results.
Or perhaps you want to select some text between tags... but not the tags! Then you could use:
'~(?<=<p>).*?(?=<\/p>)~g'
for this text:
<p>Hello !</p> <p>I wont select tags! Only text with in</p>
See demo here DEMO
You could pipe your output to another regex. Using grep, you could do this:
grep A | grep B
((yes).*(no))|((no).*(yes))
Will match sentence having both yes and no at the same time, regardless the order in which they appear:
Do i like cookies? **Yes**, i do. But milk - **no**, definitely no.
**No**, you may not have my phone. **Yes**, you may go f yourself.
Will both match, ignoring case.
Use AND outside the regular expression. In PHP lookahead operator did not not seem to work for me, instead I used this
if( preg_match("/^.{3,}$/",$pass1) && !preg_match("/\s{1}/",$pass1))
return true;
else
return false;
The above regex will match if the password length is 3 characters or more and there are no spaces in the password.
Here is a possible "form" for "and" operator:
Take the following regex for an example:
If we want to match words without the "e" character, we could do this:
/\b[^\We]+\b/g
\W means NOT a "word" character.
^\W means a "word" character.
[^\We] means a "word" character, but not an "e".
see it in action: word without e
"and" Operator for Regular Expressions
I think this pattern can be used as an "and" operator for regular expressions.
In general, if:
A = not a
B = not b
then:
[^AB] = not(A or B)
= not(A) and not(B)
= a and b
Difference Set
So, if we want to implement the concept of difference set in regular expressions, we could do this:
a - b = a and not(b)
= a and B
= [^Ab]