Re: Why won't this parse?
Re: Why won't this parse?
- Subject: Re: Why won't this parse?
- From: Paul Berkowitz <email@hidden>
- Date: Thu, 10 May 2001 10:32:52 -0700
On 5/10/01 9:51 AM, "David Graham" <email@hidden> wrote:
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>> set AppleScript's text item delimiters to ASCII number 32 -- space
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> ASCII number 32 is not space. It's
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> 51
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Hmm...
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Then why do I get this?
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ASCII number of " "
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-- 32
Because that's correct.
ASCII number of " " = 32
ASCII character 32 = " "
ASCII number of "3" = 51
ASCII character 51 = "3"
I didn't do it right the last time. Sorry. You can only have ASCII numbers
of single-character strings - individual characters including individual
digits as string. AppleScript will coerce a single digit (3) to its string
equivalent ("3") to come up with the ASCII number of it (51). It appears
that it will also coerce a multi-character string to its first character and
get you the ASCII number, rather than error.
So:
ASCII number "a" = 97
ASCII number "abc" = 97
ASCII number "3" = 51
ASDCII number "32" = 51
ASCII number "325678888997" = 51
ASCII number 3 [coerced to "3"] = 51
ASCII number 32 [coerced to "32"] = 51
ASCII number "325678888997" = 51
The last five are all coerced to
ASCII number "3"
which is the same as
ASCII number of "3"
But you don't want ASCII number of anything. You want the character " ".
That's
ASCII character 32 = " "
Since you wrote 'number' instead of 'character' you got
ASCII number of 32 = ASCII number "3" = 51
and the rest of your script errored, since using {51} as AppleScript's text
item delimiters in a string containing no substring "51" is going to give
you only one text item (the whole string).
--
Paul Berkowitz