Mixing
How can I place newlines in my regex to make it more readable? [duplicate]
This question already has an answer here:
Commenting Regular Expressions
4 answers
I have a really long email-matching regex in JavasScript that I would like to break into multiple lines without changing the regex functionality. I know some regex engines offer a way to insert newlines for readability, is there a way to do that in JS?
javascript regex
asked Nov 21 '18 at 23:59
temporary_user_nametemporary_user_name
16.8k2997163
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Nov 22 '18 at 9:30
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
This question already has an answer here:
Commenting Regular Expressions
4 answers
I have a really long email-matching regex in JavasScript that I would like to break into multiple lines without changing the regex functionality. I know some regex engines offer a way to insert newlines for readability, is there a way to do that in JS?
javascript regex
asked Nov 21 '18 at 23:59
temporary_user_nametemporary_user_name
16.8k2997163
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Nov 22 '18 at 9:30
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
Can you provide the regex?
– ibrahim mahrir
Nov 22 '18 at 0:01
1
I mean I could but it shouldn't make any difference to the question, it's arbitrary. It applies equally to all regexes. But it's the email-matching regex from the MDN docs. And yeah that's true I could do it that way, but it's not really exactly what I'm looking for here and more of a work-around.
– temporary_user_name
Nov 22 '18 at 0:06
add a comment |
This question already has an answer here:
Commenting Regular Expressions
4 answers
I have a really long email-matching regex in JavasScript that I would like to break into multiple lines without changing the regex functionality. I know some regex engines offer a way to insert newlines for readability, is there a way to do that in JS?
javascript regex
asked Nov 21 '18 at 23:59
temporary_user_nametemporary_user_name
16.8k2997163
This question already has an answer here:
Commenting Regular Expressions
4 answers
I have a really long email-matching regex in JavasScript that I would like to break into multiple lines without changing the regex functionality. I know some regex engines offer a way to insert newlines for readability, is there a way to do that in JS?
This question already has an answer here:
Commenting Regular Expressions
4 answers
javascript regex
javascript regex
asked Nov 21 '18 at 23:59
temporary_user_nametemporary_user_name
16.8k2997163
asked Nov 21 '18 at 23:59
temporary_user_nametemporary_user_name
16.8k2997163
asked Nov 21 '18 at 23:59
temporary_user_nametemporary_user_name
16.8k2997163
asked Nov 21 '18 at 23:59
temporary_user_nametemporary_user_name
16.8k2997163
asked Nov 21 '18 at 23:59
temporary_user_nametemporary_user_name
16.8k2997163
16.8k2997163
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Nov 22 '18 at 9:30
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Nov 22 '18 at 9:30
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
Can you provide the regex?
– ibrahim mahrir
Nov 22 '18 at 0:01
1
I mean I could but it shouldn't make any difference to the question, it's arbitrary. It applies equally to all regexes. But it's the email-matching regex from the MDN docs. And yeah that's true I could do it that way, but it's not really exactly what I'm looking for here and more of a work-around.
– temporary_user_name
Nov 22 '18 at 0:06
add a comment |
Can you provide the regex?
– ibrahim mahrir
Nov 22 '18 at 0:01
1
I mean I could but it shouldn't make any difference to the question, it's arbitrary. It applies equally to all regexes. But it's the email-matching regex from the MDN docs. And yeah that's true I could do it that way, but it's not really exactly what I'm looking for here and more of a work-around.
– temporary_user_name
Nov 22 '18 at 0:06
Can you provide the regex?
– ibrahim mahrir
Nov 22 '18 at 0:01
Can you provide the regex?
– ibrahim mahrir
Nov 22 '18 at 0:01
1
1
I mean I could but it shouldn't make any difference to the question, it's arbitrary. It applies equally to all regexes. But it's the email-matching regex from the MDN docs. And yeah that's true I could do it that way, but it's not really exactly what I'm looking for here and more of a work-around.
– temporary_user_name
Nov 22 '18 at 0:06
I mean I could but it shouldn't make any difference to the question, it's arbitrary. It applies equally to all regexes. But it's the email-matching regex from the MDN docs. And yeah that's true I could do it that way, but it's not really exactly what I'm looking for here and more of a work-around.
– temporary_user_name
Nov 22 '18 at 0:06
add a comment |
1 Answer
1
active
oldest
votes
There's no built-in way to accomplish such a thing, but it's not hard to accomplish it yourself. You can use a template literal with String.raw, which will allow you to use newlines in the regex string, without having to double-escape backslashes, and then you can replace all newlines with the empty string before passing it to new RegExp:
const patternStr = String.raw`^
[fg]oo
=
war`;
const pattern = new RegExp(patternStr.replace(/n/g, ''));
console.log(pattern.test('foo=bar'));
console.log(pattern.test('goo=bar'));
console.log(pattern.test('hoo=bar'));You can use a similar technique to allow comments as well:
const patternStr = String.raw`
^ // Match the beginning of the string
[fg]oo // Match either f or g, followed by oo
= // Match an equals sign
war // Match a word character, followed by "ar"
`;
const pattern = new RegExp(
patternStr.replace(/(?: *//.*)?n/g, '')
);
console.log(pattern.test('foo=bar'));
console.log(pattern.test('goo=bar'));
console.log(pattern.test('hoo=bar'));The (?: *//.*)?n pattern means:
(?: *//.*)? - An optional group of zero or more spaces, followed by //, followed by non-newline characters
n - followed by a newline
Of course, this means it'll be impossible to write // as is in the regex, but that's OK, you can just escape forward slashes just like you do with regular expression literals (it'll be parsed by the RegExp constructor as an unnecessary escape character):
const patternStr = String.raw`
^ // Match the beginning of the string
// // Match two literal forward slashes
`;
const pattern = new RegExp(
patternStr.replace(/(?: *//.*)?n/g, '')
);
console.log(pattern.test('//foo'));
console.log(pattern.test('foo'));Another way would be to allow literal //s in the template literal by, when matching the comment // <text> n, make sure that <text> doesn't have any //s in it. This would mean that only the final // on a line would be parsed as a comment, allowing you to use //s earlier on the line, without escaping, without problems, by using (?:(?!//).)* instead of .*:
const patternStr = String.raw`
^ // Match the beginning of the string
// // Match two literal forward slashes
`;
const pattern = new RegExp(
patternStr.replace(/(?: *//(?:(?!//).)*)?n/g, '')
);
console.log(pattern.test('//foo'));
console.log(pattern.test('foo'));Of course, this will mean that //s will only be parsed as actual double-forward-slashes in the regex if there's another // farther right in the line. (If there isn't another // later, you'll have to use // instead)
answered Nov 22 '18 at 0:08
CertainPerformanceCertainPerformance
88.1k154775
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
There's no built-in way to accomplish such a thing, but it's not hard to accomplish it yourself. You can use a template literal with String.raw, which will allow you to use newlines in the regex string, without having to double-escape backslashes, and then you can replace all newlines with the empty string before passing it to new RegExp:
const patternStr = String.raw`^
[fg]oo
=
war`;
const pattern = new RegExp(patternStr.replace(/n/g, ''));
console.log(pattern.test('foo=bar'));
console.log(pattern.test('goo=bar'));
console.log(pattern.test('hoo=bar'));You can use a similar technique to allow comments as well:
const patternStr = String.raw`
^ // Match the beginning of the string
[fg]oo // Match either f or g, followed by oo
= // Match an equals sign
war // Match a word character, followed by "ar"
`;
const pattern = new RegExp(
patternStr.replace(/(?: *//.*)?n/g, '')
);
console.log(pattern.test('foo=bar'));
console.log(pattern.test('goo=bar'));
console.log(pattern.test('hoo=bar'));The (?: *//.*)?n pattern means:
(?: *//.*)? - An optional group of zero or more spaces, followed by //, followed by non-newline characters
n - followed by a newline
Of course, this means it'll be impossible to write // as is in the regex, but that's OK, you can just escape forward slashes just like you do with regular expression literals (it'll be parsed by the RegExp constructor as an unnecessary escape character):
const patternStr = String.raw`
^ // Match the beginning of the string
// // Match two literal forward slashes
`;
const pattern = new RegExp(
patternStr.replace(/(?: *//.*)?n/g, '')
);
console.log(pattern.test('//foo'));
console.log(pattern.test('foo'));Another way would be to allow literal //s in the template literal by, when matching the comment // <text> n, make sure that <text> doesn't have any //s in it. This would mean that only the final // on a line would be parsed as a comment, allowing you to use //s earlier on the line, without escaping, without problems, by using (?:(?!//).)* instead of .*:
const patternStr = String.raw`
^ // Match the beginning of the string
// // Match two literal forward slashes
`;
const pattern = new RegExp(
patternStr.replace(/(?: *//(?:(?!//).)*)?n/g, '')
);
console.log(pattern.test('//foo'));
console.log(pattern.test('foo'));Of course, this will mean that //s will only be parsed as actual double-forward-slashes in the regex if there's another // farther right in the line. (If there isn't another // later, you'll have to use // instead)
answered Nov 22 '18 at 0:08
CertainPerformanceCertainPerformance
88.1k154775
add a comment |
There's no built-in way to accomplish such a thing, but it's not hard to accomplish it yourself. You can use a template literal with String.raw, which will allow you to use newlines in the regex string, without having to double-escape backslashes, and then you can replace all newlines with the empty string before passing it to new RegExp:
const patternStr = String.raw`^
[fg]oo
=
war`;
const pattern = new RegExp(patternStr.replace(/n/g, ''));
console.log(pattern.test('foo=bar'));
console.log(pattern.test('goo=bar'));
console.log(pattern.test('hoo=bar'));You can use a similar technique to allow comments as well:
const patternStr = String.raw`
^ // Match the beginning of the string
[fg]oo // Match either f or g, followed by oo
= // Match an equals sign
war // Match a word character, followed by "ar"
`;
const pattern = new RegExp(
patternStr.replace(/(?: *//.*)?n/g, '')
);
console.log(pattern.test('foo=bar'));
console.log(pattern.test('goo=bar'));
console.log(pattern.test('hoo=bar'));The (?: *//.*)?n pattern means:
(?: *//.*)? - An optional group of zero or more spaces, followed by //, followed by non-newline characters
n - followed by a newline
Of course, this means it'll be impossible to write // as is in the regex, but that's OK, you can just escape forward slashes just like you do with regular expression literals (it'll be parsed by the RegExp constructor as an unnecessary escape character):
const patternStr = String.raw`
^ // Match the beginning of the string
// // Match two literal forward slashes
`;
const pattern = new RegExp(
patternStr.replace(/(?: *//.*)?n/g, '')
);
console.log(pattern.test('//foo'));
console.log(pattern.test('foo'));Another way would be to allow literal //s in the template literal by, when matching the comment // <text> n, make sure that <text> doesn't have any //s in it. This would mean that only the final // on a line would be parsed as a comment, allowing you to use //s earlier on the line, without escaping, without problems, by using (?:(?!//).)* instead of .*:
const patternStr = String.raw`
^ // Match the beginning of the string
// // Match two literal forward slashes
`;
const pattern = new RegExp(
patternStr.replace(/(?: *//(?:(?!//).)*)?n/g, '')
);
console.log(pattern.test('//foo'));
console.log(pattern.test('foo'));Of course, this will mean that //s will only be parsed as actual double-forward-slashes in the regex if there's another // farther right in the line. (If there isn't another // later, you'll have to use // instead)
answered Nov 22 '18 at 0:08
CertainPerformanceCertainPerformance
88.1k154775
add a comment |
There's no built-in way to accomplish such a thing, but it's not hard to accomplish it yourself. You can use a template literal with String.raw, which will allow you to use newlines in the regex string, without having to double-escape backslashes, and then you can replace all newlines with the empty string before passing it to new RegExp:
const patternStr = String.raw`^
[fg]oo
=
war`;
const pattern = new RegExp(patternStr.replace(/n/g, ''));
console.log(pattern.test('foo=bar'));
console.log(pattern.test('goo=bar'));
console.log(pattern.test('hoo=bar'));You can use a similar technique to allow comments as well:
const patternStr = String.raw`
^ // Match the beginning of the string
[fg]oo // Match either f or g, followed by oo
= // Match an equals sign
war // Match a word character, followed by "ar"
`;
const pattern = new RegExp(
patternStr.replace(/(?: *//.*)?n/g, '')
);
console.log(pattern.test('foo=bar'));
console.log(pattern.test('goo=bar'));
console.log(pattern.test('hoo=bar'));The (?: *//.*)?n pattern means:
(?: *//.*)? - An optional group of zero or more spaces, followed by //, followed by non-newline characters
n - followed by a newline
Of course, this means it'll be impossible to write // as is in the regex, but that's OK, you can just escape forward slashes just like you do with regular expression literals (it'll be parsed by the RegExp constructor as an unnecessary escape character):
const patternStr = String.raw`
^ // Match the beginning of the string
// // Match two literal forward slashes
`;
const pattern = new RegExp(
patternStr.replace(/(?: *//.*)?n/g, '')
);
console.log(pattern.test('//foo'));
console.log(pattern.test('foo'));Another way would be to allow literal //s in the template literal by, when matching the comment // <text> n, make sure that <text> doesn't have any //s in it. This would mean that only the final // on a line would be parsed as a comment, allowing you to use //s earlier on the line, without escaping, without problems, by using (?:(?!//).)* instead of .*:
const patternStr = String.raw`
^ // Match the beginning of the string
// // Match two literal forward slashes
`;
const pattern = new RegExp(
patternStr.replace(/(?: *//(?:(?!//).)*)?n/g, '')
);
console.log(pattern.test('//foo'));
console.log(pattern.test('foo'));Of course, this will mean that //s will only be parsed as actual double-forward-slashes in the regex if there's another // farther right in the line. (If there isn't another // later, you'll have to use // instead)
answered Nov 22 '18 at 0:08
CertainPerformanceCertainPerformance
88.1k154775
There's no built-in way to accomplish such a thing, but it's not hard to accomplish it yourself. You can use a template literal with String.raw, which will allow you to use newlines in the regex string, without having to double-escape backslashes, and then you can replace all newlines with the empty string before passing it to new RegExp:
const patternStr = String.raw`^
[fg]oo
=
war`;
const pattern = new RegExp(patternStr.replace(/n/g, ''));
console.log(pattern.test('foo=bar'));
console.log(pattern.test('goo=bar'));
console.log(pattern.test('hoo=bar'));You can use a similar technique to allow comments as well:
const patternStr = String.raw`
^ // Match the beginning of the string
[fg]oo // Match either f or g, followed by oo
= // Match an equals sign
war // Match a word character, followed by "ar"
`;
const pattern = new RegExp(
patternStr.replace(/(?: *//.*)?n/g, '')
);
console.log(pattern.test('foo=bar'));
console.log(pattern.test('goo=bar'));
console.log(pattern.test('hoo=bar'));The (?: *//.*)?n pattern means:
(?: *//.*)? - An optional group of zero or more spaces, followed by //, followed by non-newline characters
n - followed by a newline
Of course, this means it'll be impossible to write // as is in the regex, but that's OK, you can just escape forward slashes just like you do with regular expression literals (it'll be parsed by the RegExp constructor as an unnecessary escape character):
const patternStr = String.raw`
^ // Match the beginning of the string
// // Match two literal forward slashes
`;
const pattern = new RegExp(
patternStr.replace(/(?: *//.*)?n/g, '')
);
console.log(pattern.test('//foo'));
console.log(pattern.test('foo'));Another way would be to allow literal //s in the template literal by, when matching the comment // <text> n, make sure that <text> doesn't have any //s in it. This would mean that only the final // on a line would be parsed as a comment, allowing you to use //s earlier on the line, without escaping, without problems, by using (?:(?!//).)* instead of .*:
const patternStr = String.raw`
^ // Match the beginning of the string
// // Match two literal forward slashes
`;
const pattern = new RegExp(
patternStr.replace(/(?: *//(?:(?!//).)*)?n/g, '')
);
console.log(pattern.test('//foo'));
console.log(pattern.test('foo'));Of course, this will mean that //s will only be parsed as actual double-forward-slashes in the regex if there's another // farther right in the line. (If there isn't another // later, you'll have to use // instead)
const patternStr = String.raw`^
[fg]oo
=
war`;
const pattern = new RegExp(patternStr.replace(/n/g, ''));
console.log(pattern.test('foo=bar'));
console.log(pattern.test('goo=bar'));
console.log(pattern.test('hoo=bar'));const patternStr = String.raw`^
[fg]oo
=
war`;
const pattern = new RegExp(patternStr.replace(/n/g, ''));
console.log(pattern.test('foo=bar'));
console.log(pattern.test('goo=bar'));
console.log(pattern.test('hoo=bar'));const patternStr = String.raw`
^ // Match the beginning of the string
[fg]oo // Match either f or g, followed by oo
= // Match an equals sign
war // Match a word character, followed by "ar"
`;
const pattern = new RegExp(
patternStr.replace(/(?: *//.*)?n/g, '')
);
console.log(pattern.test('foo=bar'));
console.log(pattern.test('goo=bar'));
console.log(pattern.test('hoo=bar'));const patternStr = String.raw`
^ // Match the beginning of the string
[fg]oo // Match either f or g, followed by oo
= // Match an equals sign
war // Match a word character, followed by "ar"
`;
const pattern = new RegExp(
patternStr.replace(/(?: *//.*)?n/g, '')
);
console.log(pattern.test('foo=bar'));
console.log(pattern.test('goo=bar'));
console.log(pattern.test('hoo=bar'));const patternStr = String.raw`
^ // Match the beginning of the string
// // Match two literal forward slashes
`;
const pattern = new RegExp(
patternStr.replace(/(?: *//.*)?n/g, '')
);
console.log(pattern.test('//foo'));
console.log(pattern.test('foo'));const patternStr = String.raw`
^ // Match the beginning of the string
// // Match two literal forward slashes
`;
const pattern = new RegExp(
patternStr.replace(/(?: *//.*)?n/g, '')
);
console.log(pattern.test('//foo'));
console.log(pattern.test('foo'));const patternStr = String.raw`
^ // Match the beginning of the string
// // Match two literal forward slashes
`;
const pattern = new RegExp(
patternStr.replace(/(?: *//(?:(?!//).)*)?n/g, '')
);
console.log(pattern.test('//foo'));
console.log(pattern.test('foo'));const patternStr = String.raw`
^ // Match the beginning of the string
// // Match two literal forward slashes
`;
const pattern = new RegExp(
patternStr.replace(/(?: *//(?:(?!//).)*)?n/g, '')
);
console.log(pattern.test('//foo'));
console.log(pattern.test('foo'));answered Nov 22 '18 at 0:08
CertainPerformanceCertainPerformance
88.1k154775
edited Nov 22 '18 at 0:28
answered Nov 22 '18 at 0:08
CertainPerformanceCertainPerformance
88.1k154775
answered Nov 22 '18 at 0:08
CertainPerformanceCertainPerformance
88.1k154775
answered Nov 22 '18 at 0:08
CertainPerformanceCertainPerformance
88.1k154775
88.1k154775
add a comment |
add a comment |
Can you provide the regex?
– ibrahim mahrir
Nov 22 '18 at 0:01
1
I mean I could but it shouldn't make any difference to the question, it's arbitrary. It applies equally to all regexes. But it's the email-matching regex from the MDN docs. And yeah that's true I could do it that way, but it's not really exactly what I'm looking for here and more of a work-around.
– temporary_user_name
Nov 22 '18 at 0:06