Reputation: 4629
I am reading an Oracle tutorial on regular expressions. I am on the topic Capturing groups. Though the reference is excellent, but except that a parenthesis represents a group, I am finding many difficulties in understanding the topic. Here are my confusions.
Elaborating with examples would be nice.
Upvotes: 3
Views: 280
Reputation: 11132
Say you have a string, abcabc
, and you want to figure out whether the first part of the string matches the second part. You can do this with a single regex by using capturing groups and backreferences. Here is the regex I would use:
(.+)\1
The way this works is .+
matches any sequence of characters. Because it is in parentheses, it is caught in a group. \1
is a backreference to the 1
st capturing group, so it is the equivalent of the text caught by the capturing group. After a bit of backtracking, the capturing group matches the first part of the string, abc
. The backreference \1
is now the equivalent of abc
, so it matches the second half of the string. The entire string is now matched, so it is confirmed that the first half of the string matches the second half.
Another use of backreferences is in replacing. Say you want to replace all {...}
with [...]
, if the text inside {
and }
is only digits. You can easily do this with capturing groups and backreferences, using the regex
{(\d+)}
And replacing with that with [\1]
.
The regex matches {123}
in the string abc {123} 456
, and captures 123
in the first capturing group. The backreference \1
is now the equivalent of 123
, so replacing {(\d+)}
in abc {123} 456
with [\1]
results in abc [123] 456
.
The reason non-capturing groups exist is because groups in general have more uses that just capturing. The regex (xyz)+
matches a string that consists entirely of the group, xyz
, repeated, such as xyzxyzxyz
. A group is needed because xyz+
only matches xy
and then z
repeated, i.e. xyzzzzz
. The problem with using capturing groups is that they are slightly less efficient compared to non-capturing groups, and they take up an index. If you have a complicated regex with a lot of groups in it, but you only need to reference a single one somewhere in the middle, it's a lot better to just reference \1
rather than trying to count all the groups up to the one you want.
I hope this helps!
Upvotes: 2
Reputation: 6350
([abc])([def](\d+))
has three groups, so I know to refer to them as \1
, \2
and \3
. Note that group 3 is inside 2. They are numbered from the left by where they begin.(foo|bar)
will match "foo"
or "bar"
. If you're not interested in the contents of a group, you can make it non-capturing (e.g: (?:foo|bar)
(varies by dialect)), so as not to "use up" the numbers assigned to groups. But you don't have to, it's just convenient sometimes.\b([a-z])[a-z]*\1\b
The \1
will then be the same as whatever the first group captured. Of course it can be used for much more powerful stuff, but I think you'll get the idea.(Coming up with relevant examples is certainly the hardest part.)
Edit: I answered when the questions were:
- What is the significance of counting groups in an expression?
- There is a special group, called as group-0, which means the entire expression. It is not reported by groupCount() method. Why is that?
- I don't understand what are non-capturing groups?
- Why we need back-references? What is the significance of back-references?
Upvotes: 3
Reputation: 3446
groupCount()
just lets you know how many capture groups you've specified in the expression.(?:)
would be used to, well, not capture a group. Ex. if you need to test if a string contains one of several words and don't want to capture the word in a new group: (?:hello|hi there) world
!== hello|hi there world
. The first matches "hello world" or "hi there world" but the second matches "hello" or "hi there world".^(\d)(?!.*\1)\d+$
would ensure the first digit is unique in a string.Upvotes: 1