Newbie problem: conversion with unicode

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Francois - 23 Jun 2005 11:08 GMT

Hi all,

I have caracters represented with escape sequence in a file:
\u00B3
\u2074
\u2075

these represents superscript 3, 4, 5 respectively.
Now I read the file and read these as String "\u00B3".
I would convert this representation in a new String object: which would
have the superscrit 3 4 and 5.

What is the correct code for this ? I have compilation errors or
converion errors ...

Thanks a lot for any help !

Francois Rappaz

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Harald - 23 Jun 2005 22:27 GMT

> I have caracters represented with escape sequence in a file:
> \u00B3
[quoted text clipped - 7 lines]
> What is the correct code for this ? I have compilation errors or
> converion errors ...

Strip the backslash, strip the u, convert the resulting four hex digits
to an int (Integer.parseInt() or the likes should help), cast the
resulting int to char, append to a StringBuilder (StringBuffer),
convert the resulting StringBuilder to String.

Harald.

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Java Text Crunching: http://www.ebi.ac.uk/Rebholz-srv/whatizit/software

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Francois - 24 Jun 2005 09:39 GMT

Thanks a lot !

That's is exactly what I needed and it gives something like

String s = rawCode;
byte c = new byte[1]
try {
c[0] = Integer.decode("0x" + rawCode.substring(2)).byteValue();
} catch (NumberFormatExceptione){e.printStackTrace();}
String result = new String(c);

Could I have encoding problems with that code ? or should it worked on
any situation ?
TIA
Francois

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HK - 24 Jun 2005 17:43 GMT

> Thanks a lot !
>
> That's is exactly what I needed and it gives something like
>
> String s = rawCode;

You don't use the s below, but rawCode (should not matter).

> byte c = new byte[1]

This should definitively be char[], not byte[].

> try {
> c[0] = Integer.decode("0x" + rawCode.substring(2)).byteValue();

You certainly want .intValue(), not .byteValue(). Remember that
char --- in Java and very much unlike C/C++ --- is 2 bytes long.
In addition I would rather use

c[0] = Integer.parseInt(rawCode.substring(2), 16);

> } catch (NumberFormatExceptione){e.printStackTrace();}
> String result = new String(c);
>
> Could I have encoding problems with that code ? or should it worked on
> any situation ?

I assume you checked before that rawCode starts with '\\' and 'u'
and that it contains at most 4 hex digits. If it contains 5, digits,
you loose
something in the cast to char.

Harald.

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Francois - 28 Jun 2005 13:52 GMT

Well probably nobody care, but the code above does not work:
the following seems to be alright:

char c[] = new char[n]
s has been read in a file and contain "\u2079" for example
...
c[i]= (char)Integer.parseInt(s.substring(2),16);
...
and for the whole array c
String result = new String(c);

Francois

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Newbie problem: conversion with unicode

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