Matching an optional substring in a regex

Question

I m developing an algorithm to parse a number out of a series of short-ish strings  These strings are somewhat regular  but there s a few different general forms and several exceptions  I m trying to build a set of regexes that will handle the various forms and exceptions  I ll apply them one after another to see if I get a match    One of these forms goes something like this   X  Y  Z   Where    X is a number I want to capture   Z is static  pre-defined text  it s basically how I determine whether this particular form is applicable or not  Y is a a string of unknown length and content  surrounded by parenthesis     Also  Y is optional  it doesn t always appear in a string with Z and X  So  I want to be able to extract the numbers from all of these strings    10 Z 20  foo  Z 30  bar  Z   Right now  I have a regex that will capture the first one     0-9     Z   My problem is that I don t know how to construct a regex that will match a series of characters if and only if they re enclosed in parenthesis  Can this be done in a single regex

User · Answer

You can do this     0-9                 Z   This will not work with nested parens for Y  however  Nesting requires recursion which isn t strictly regular any more  but context-free   Modern regexp engines can still handle it  albeit with some difficulties  back-references

User · Answer

d   s            s Z   Note the escaped parentheses  and the    zero or once  quantifiers  Any of the groups you don t want to capture can be     non-capture groups    I agree about the spaces   s is a better option there  I also changed the quantifier to insure there are digits at the beginning  As far as newlines  that would depend on context  if the file is parsed line by line it won t be a problem  Another option is to anchor the start and end of the line  add a   at the front and a   at the end

User · Answer

Try this   X    Y     Z

User · Answer

d   s            s Z   Note the escaped parentheses  and the    zero or once  quantifiers  Any of the groups you don t want to capture can be     non-capture groups    I agree about the spaces   s is a better option there  I also changed the quantifier to insure there are digits at the beginning  As far as newlines  that would depend on context  if the file is parsed line by line it won t be a problem  Another option is to anchor the start and end of the line  add a   at the front and a   at the end

User · Answer

d   s            s Z   Note the escaped parentheses  and the    zero or once  quantifiers  Any of the groups you don t want to capture can be     non-capture groups    I agree about the spaces   s is a better option there  I also changed the quantifier to insure there are digits at the beginning  As far as newlines  that would depend on context  if the file is parsed line by line it won t be a problem  Another option is to anchor the start and end of the line  add a   at the front and a   at the end

User · Answer

Try this   X    Y     Z

User · Answer

This ought to work     d  s             s   Z    Haven t tested it though  but let me give you the breakdown  so if there are any bugs left they should be pretty straightforward to find   First the beginning       beginning of string  d    one or more decimal characters  s    one optional whitespace   Then this part               s      Is actually                      Which makes the following contents optional  only if it exists fully             s        an opening bracket          a series of at least one character that is not a closing bracket      followed by a closing bracket  s    followed by one optional whitespace   And the end is made up of  Z    Where  Z   your constant string     the end of the string

User · Answer

You can do this     0-9                 Z   This will not work with nested parens for Y  however  Nesting requires recursion which isn t strictly regular any more  but context-free   Modern regexp engines can still handle it  albeit with some difficulties  back-references

User · Answer

This ought to work     d  s             s   Z    Haven t tested it though  but let me give you the breakdown  so if there are any bugs left they should be pretty straightforward to find   First the beginning       beginning of string  d    one or more decimal characters  s    one optional whitespace   Then this part               s      Is actually                      Which makes the following contents optional  only if it exists fully             s        an opening bracket          a series of at least one character that is not a closing bracket      followed by a closing bracket  s    followed by one optional whitespace   And the end is made up of  Z    Where  Z   your constant string     the end of the string

User · Answer

Try this   X    Y     Z

User · Answer

Try this   X    Y     Z

User · Answer

You can do this     0-9                 Z   This will not work with nested parens for Y  however  Nesting requires recursion which isn t strictly regular any more  but context-free   Modern regexp engines can still handle it  albeit with some difficulties  back-references

User · Answer

You can do this     0-9                 Z   This will not work with nested parens for Y  however  Nesting requires recursion which isn t strictly regular any more  but context-free   Modern regexp engines can still handle it  albeit with some difficulties  back-references

User · Answer

This ought to work     d  s             s   Z    Haven t tested it though  but let me give you the breakdown  so if there are any bugs left they should be pretty straightforward to find   First the beginning       beginning of string  d    one or more decimal characters  s    one optional whitespace   Then this part               s      Is actually                      Which makes the following contents optional  only if it exists fully             s        an opening bracket          a series of at least one character that is not a closing bracket      followed by a closing bracket  s    followed by one optional whitespace   And the end is made up of  Z    Where  Z   your constant string     the end of the string

User · Answer

This ought to work     d  s             s   Z    Haven t tested it though  but let me give you the breakdown  so if there are any bugs left they should be pretty straightforward to find   First the beginning       beginning of string  d    one or more decimal characters  s    one optional whitespace   Then this part               s      Is actually                      Which makes the following contents optional  only if it exists fully             s        an opening bracket          a series of at least one character that is not a closing bracket      followed by a closing bracket  s    followed by one optional whitespace   And the end is made up of  Z    Where  Z   your constant string     the end of the string

User · Answer

d   s            s Z   Note the escaped parentheses  and the    zero or once  quantifiers  Any of the groups you don t want to capture can be     non-capture groups    I agree about the spaces   s is a better option there  I also changed the quantifier to insure there are digits at the beginning  As far as newlines  that would depend on context  if the file is parsed line by line it won t be a problem  Another option is to anchor the start and end of the line  add a   at the front and a   at the end

[regex] Matching an optional substring in a regex

Examples related to regex