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>
> The quickest code for doing this is to write a loop through
> the string,
I see from this and your post on xsl-list that you're a man who likes
low-level coding.
What's wrong with:
(char)Integer.parseInt(Pattern.compile("&#([0-9]+);").matcher(bunch).gro
up(1));
Also untested.
OK, add a couple more lines to do hex as well (but the OP didn't ask for
that).
Michael Kay
> with states for '&', '#', digit and ';'. Should be about
> twenty lines in Java at most. I believe it is also the
> quickest way to write the code.
>
> static String bunchToString(String bunch) {
> StringBuffer buf=new StringBuffer();
> int cc=0; char ch; int state=0; int i;
> for(i=0;i!=bunch.length();++i) {
> switch(ch=bunch.charAt(i)) {
> case '&': state=1; break;
> case '#': state=state==1?2:0; break;
> case 'x': state=state==2?3:0; break;
> case '0': case '1': case '2': case '3': case '4':
> case '5': case '6': case '7': case '8': case '9':
> if(state==4||state==5) break;
> state=state==2?4:state==3?5:0; break;
> case 'A': case 'B': case 'C': case 'D': case 'E': case 'F':
> case 'a': case 'b': case 'c': case 'd': case 'e': case 'f':
> if(state==5) break;
> state=state==3?5:0; break;
> case ';': state=(state==4||state==5)?6:0; break;
> }
> switch(state) {
> case 0: buf.append(ch); break;
> case 1: case 2: case 3: break;
> case 4: cc=cc*10+ch-'0'; break;
> case 5:
> cc=cc*16+('0'<=ch&&ch<='9'?ch-'0':
> 'A'<=ch&&ch<='F'?ch+10-'A':ch+10-'a'); break;
> case 6: buf.append((char)cc); cc=0; break;
> default: throw new RuntimeException("should not happen");
> }
> }
> return buf.toString();
> }
>
> Untested, but the idea should be clear.
>
> David Tolpin
> http://davidashen.net/
>
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