# Get the first character from a string?

I’ve got a math function where I’m converting a base36 number to a
base 10 integer and it works wonderfully, except when the first
character is a zero.
I could get this method to work if I could look at the first position
in a string, and if it was a zero, I’d then add a zero to the first
position of the resulting string.
Here’s the equation (and its compliment);

def calc_ovalue(barcode)
barcode.to_i(36)
end

def calc_other(barcode)
onum = calc_ovalue(barcode)
if onum.odd?
(onum + 1).to_s(36).upcase
else
(onum - 1).to_s(36).upcase
end
end

some examples of accepeting the first value and calculating the second
two are;
0N5F 0N5G ovalue = 30003
0V9B 0V9C ovalue = 40511
6JYV 6JYW ovalue = 305815

Thank you,
Kathleen

if you have a string @str, then
@str will give you the charactercode of the first char
that’s already enough to test against the charcode of a zero which is
48

if yo prefer to work with a string as result, use
@str[0,1]
this gives a string result (0 = from position, 1 = length)
or
@str[0…0]
returns a string too, the range is startposition to endposition

Assuming that s is the string; s is the first character. s.chr ==
“1” or s==48 (see the irb session below)ï»¿

a=“1234”
=> “1234”

b=“01234”
=> “01234”

a
=> 49

a.chr
=> “1”

b
=> 48

b.chr
=> “0”

Hope that helps,
Matt

Matt,
I’m seeing how to test the first character of the incoming string
which is (barcode) (see function)
def calc_other(barcode)
onum = calc_ovalue(barcode)
if onum.odd?
(onum + 1).to_s(36).upcase
else
(onum - 1).to_s(36).upcase
end
if barcode == 48
“0” + return
end
end

But then, I don’t see how to attach the preceding zero if the incoming
(barcode) has one. I know this looks stupid, but what I’m attempting
is to add a zero to the first position of the result. I assume result
= string?
Thank you,
Kathleen

will the result always be 4 chars? If so, could you do something like

return_val = my_val.rjust(4).gsub(" ",“0”)

On Sun, Jul 13, 2008 at 1:06 PM, [email protected]
[email protected]

Marlon,
Yes. The problem is that if I submit a barcode like 0123 the other
barcode is calculated as 124. If there is a character other than zero
in the first position it works just fine.
Where are you suggesting that I put the return_val line of code you
show?
Is that gsub example putting a zero at the end of the string?
What / How would you amend the return value of the calc_other
function. Isn’t the result of any method the value inside the ‘return’
keyword?
If so, why can’t the ‘return’ value be modified before the function
closes?
I seem to be having a problem ‘seeing the forest through the trees’.
Kathleen

This is how it could be done:

def calc_ovalue(barcode)
barcode.to_i(36)
end

def calc_other(barcode)
onum = calc_ovalue(barcode)
if onum % 2 == 1
onum+=1
else
onum-=1
end
onum.to_s(36).rjust(4).gsub(" ",“0”).upcase
end

On Sun, Jul 13, 2008 at 3:22 PM, [email protected]
[email protected]

def next_barcode(barcode)
length = barcode.length

val = barcode.to_i(36)
if val.odd?
val += 1
else
val -= 1
end

ret = val.to_s(36).upcase
(‘0’ * (length - ret.length)) + ret
end

tests = [ ‘AAAAAA’, ‘AAAAAZ’, ‘012345’, ‘00000Z’, ‘00ZZZZ’ ]

tests.each do |test|
puts “#{test} --> #{next_barcode(test)}”
end