I see an line in my JavaScript code like this:
var regex = /[^\w\s]/gi;
What's the meaning of this /gi
in the regex?
Other part I can understand as it accepts a group of word and spaces, but not /gi
.
The gi modifier is used to do a case insensitive search of all occurrences of a regular expression in a string.
Basically (0+1)* mathes any sequence of ones and zeroes. So, in your example (0+1)*1(0+1)* should match any sequence that has 1. It would not match 000 , but it would match 010 , 1 , 111 etc. (0+1) means 0 OR 1.
The difference between ?: and ?= is that ?= excludes the expression from the entire match while ?: just doesn't create a capturing group. So for example a(?:b) will match the "ab" in "abc", while a(?=
(debug) Inserting the `g' flag tells the compiler to insert more information about the source code into the executable than it normally would. This makes use of a debugger such as gdb much easier, since it will be able to refer to variable names that occur in the source code.
g modifier: global. All matches (don't return on first match) i modifier: insensitive. Case insensitive match (ignores case of [a-zA-Z])
In your case though i
is immaterial as you dont capture [a-zA-Z]
.
For input like !@#$
if g
modifier is not there regex will return first match !
See here.
If g
is there it will return the whole or whatever it can match.See here
The beginning and ending /
are called delimiters. They tell the interpreter where the regex begins and ends. Anything after the closing delimiter is called a "modifier," in this case g
and i
.
The g
and i
modifiers have these meanings:
g
= global, match all instances of the pattern in a string, not just onei
= case-insensitive (so, for example, /a/i
will match the string "a"
or "A"
.In the context you gave (/[^\w\s]/gi
), the i
is meaningless, because there are no case-specific portions of the regex.
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