Re: Challenge - Creative Way To Detect Alpha Characters

Given that your language is limited and '||' concatanates stings I would guess at SQL.

CASE WHEN LOWER(foo) = UPPER(foo) THEN 'No Alpha' ELSE 'Alpha' END
[download]

However in some SQL varients you can use translate (has an effect like tr///)

CASE WHEN INSTR(TRANSLATE(foo, 
  'abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ', 
  'XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX'),    
'X')>0
 THEN 'Alpha' ELSE 'No Alpha' END
[download]

Or Even

CASE WHEN LENGTH(TRANSLATE(foo, 
  '#abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ', 
  '#'))<>LENGTH(foo)
 THEN 'Alpha' ELSE 'No Alpha' END
[download]

Hope it Helps UnderMine

Comment on Re: Challenge - Creative Way To Detect Alpha Characters Select or Download Code

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Re^2: Challenge - Creative Way To Detect Alpha Characters by jdalbec (Deacon) on Sep 14, 2004 at 02:22 UTC
My guess would be REXX.	[reply]
Re^3: Challenge - Creative Way To Detect Alpha Characters by ambrus (Abbot) on Sep 14, 2004 at 07:15 UTC
I first thought it could be maple. Maple uses \|\| for string catenation (in version 5 and 7, version 3 uses `.`). It does not have regexps. However, maple (version 7 at least) has string functions to do this: with(StringTools): str:="2-h/": if ""<>Select(IsAlpha, str) then `has alphas` else `no alphas` fi; str:="2-/": if ""<>Select(IsAlpha, str) then `has alphas` else `no alphas` fi; [download]	[reply] [d/l] [select]
Re^3: Challenge - Creative Way To Detect Alpha Characters by parv (Parson) on Sep 14, 2004 at 21:30 UTC
... in that case ... `/* rexx */ strings.1 = 85865487878 strings.2 = 'oewiopeoewirpo iep ' strings.3 = '4889jfkjdk' strings.4 = 'hfhjh 767484' strings.5 = '<&jZ>(){}' strings.6 = '<&>(){}' do i = 1 to 6 say has_alpha(strings.i) '-> alpha? ' "'"strings.i"'" end exit has_alpha: procedure parse upper arg string if (length(string) = 0 \| datatype(string) = 'NUM') then return 0 start = c2d( 'A' ) stop = c2d( 'Z' ) do while start <= stop if ( pos(d2c(start) , string) \= 0 ) then return 1 start = start +1 end return 0` [download] ...which gives... 0 -> alpha? '85865487878' 1 -> alpha? 'oewiopeoewirpo iep ' 1 -> alpha? '4889jfkjdk' 1 -> alpha? 'hfhjh 767484' 1 -> alpha? '<&jZ>(){}' 0 -> alpha? '<&>(){}'	[reply] [d/l]