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Check if char variable is in an array using a "set" type comparison

Posted on 2015-02-12
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Last Modified: 2015-02-15
Hi all, I'm doing some "set type" comparisons in my app like this:

if h1 in ['0'..'9'] then....

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.. where h1 is just a variable of char. No problem there, but I have a string array that contains a collection of punctuation characters ('!','#','.','?','%' etc) and I'm wondering if there's a comparable way (ie. using the "in" operator) to check if my variable h1 contains any of these characters. As in:

if h1 in PunctuationArray

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... or something like that.

Thanks!
    Shawn
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Question by:shawn857
6 Comments
 
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Accepted Solution

by:
Rgonzo1971 earned 500 total points
ID: 40607394
Hi,

have you tried AnsiMatchStr or MatchStr

source := 'Henry';  
 if AnsiMatchStr(source, ['Brian', 'Jim', 'Henry'])
  then ShowMessage('First match was successful')
  else ShowMessage('First match was not successful');


http://www.delphibasics.co.uk/RTL.asp?Name=AnsiMatchStr

Regards
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Expert Comment

by:mlmcc
ID: 40607899
So if h1 is a string   and your set is like above ['!','#','.','?','%', ',' ] you want to see if any of the punctuation characters are in the string, h1?

So if h1 = "Peter and Paul" the answer is false but if it is "Peter, Paul, and Mary" it is true.

In that case code like this should find it

Haschar := False;
For I := 0  to High(YourArray) 
    If POS(YourArray[I], h1) then
       Haschar := True;

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mlmcc
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Expert Comment

by:Sinisa Vuk
ID: 40607968
Similar to upper example - iterate your array and instead of pos (which can check only one char in time)
use function CheckString from this question: Q_28229176
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Author Closing Comment

by:shawn857
ID: 40609094
Thanks to all who responded.

Cheers
   Shawn
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Expert Comment

by:mlmcc
ID: 40609282
How does that answer the issue or did I not understand your question?

mlmcc
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Author Comment

by:shawn857
ID: 40611738
Hi mlmcc... you were looking at it backwards I think. Variable h1 contains just one character, and I want to see if that character is contained anywhere in the set ['!','#','.','?','%', ',' ]

Thanks for contributing though...

Cheers
   Shawn
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