正则表达式分组

Question

The language I am programming in is Objective C, although that probably has no bearing on solving this problem.

Given 0 to 3 occurrences of 6, 7, 9, 11 and 13 along with their modifiers b or # , how would you write a regex so that you can retrieve the groupings for each value? Is regex even the best solution here? Here's are some examples of some strings that need to be matched:

Example 1: 7#9  //The groupings should be 7 #9

Example 2: 7b9#13 //The groupings should be 7 b9 #13

Example 3: b6 //The grouping should be b6

Example 4: 7#1311 //The groupings should be 7 #13 11

Example 5: 7911 //The groupings should be 7 9 11

To match just one number like #13, I tried the following:

([#b]?13|11|9|7|6)?

It matches for #13, but not for #11. I'm really stuck.

Answer 1

You misplaced the parenthesis. The should be this:

[#b]?(13|11|9|7|6)

Where it looks for the combination of each number with a possible #b

instead of

([#b]?13|11|9|7|6)?

where it only looks for the combination of #13 or 11 or 9 .. etc

Answer 2

The main problem was that the [#|b] set was not being applied to 13, 11, 9, 7 and 6 individually. Also, I duplicated it to handle multiple groupings. Here's the fix:

([b#]?13|[b#]?11|[b#]?9|[b#]?7|[b#]?6)?([b#]?13|[b#]?11|[b#]?9|[b#]?7|[b#]?6)?([b#]?13|[b#]?11|[b#]?9|[b#]?7|[b#]?6)?([b#]?13|[b#]?11|[b#]?9|[b#]?7|[b#]?6)?

For string b9#1113 it now groups as follows:

Group 1: b9
Group 2: #11
Group 3: 13