using nested case statements to return value - ruby-on-rails

Question

I'm trying to use case statements to have a value returned, depending on the conditions. Unfortunately I'm getting a string of syntax errors because I'm not completely familiar with how a case statement should be set up.
Basically, a user can belong to a single state (region), so in the user table is where the user's "state_id" is stored (53 could be Oregon for example). I had most of this working using if statements, but it was far too messy, so I was told "case" would be the better way to go. I also have a model "tax" which belongs_to a user and a user can have many of.
Prior to changing to using case, I was doing if user.state_id == 53 for example, which was working, and then the conditions were if self.income <= 10276 which also worked. Now with the case however, although slightly more refactored, I can't get it working.. (many syntax errors due to the evaluators <= and all the others.)
def provincial
case user.state_id
when 53
case income
when <= 10276
return 0
when 10277..37568
return self.income * 0.0506
when 37569..75138
return ((self.income - 37568) * 0.077) + 1901
when 75139..86268
return ((self.income - 75138) * 0.105) + 4794
when 86269..104754
return ((self.income - 86268) * 0.1229) + 5963
when > 104754
return ((self.income - 104754) * 0.147) + 8235
end
when 52
case income
when <= 17593
return 0
when > 17593
return self.income * 0.10
end
end
end

Answer 1

You can't put the variable at the head of the case and then use a "dangling" inequality (an inequality where you only have an expression on one side). You would need the variable with the inequality, or you could use a proc.
So this when clause is fine:
when 10277..37568
return self.income * 0.0506
But this one isn't:
when > 17593
return self.income * 0.10
You can, however, replace it with this:
when proc {|n| n > 17593}
return self.income * 0.10
This gets rid of the "dangling" inequality by using a proc. You might also be able to replace it with this (a little weird, but it works):
Inf = 1.0/0 # Assigns "infinity" to Inf
when 17593..Inf
return self.income * 0.10
You could also move the variable to each when:
def provincial
case user.state_id
when 53
case
when income <= 10276
return 0
when (income >= 10277) and (income <= 37568)
return self.income * 0.0506
when (income >= 37569) and (income <= 75138)
return ((self.income - 37568) * 0.077) + 1901
...
end
when 52
case
when income <= 17593
return 0
when income > 17593
return self.income * 0.10
end
end
end
end
Overall, in this situation, I think I like the proc best. :)
Note you can't mix, in the same case statement, the variable at the head of the case and the variable at the when clauses. So this is not valid:
case x
when x < 3
# do stuff
when 5
# do stuff
end