how to use Regex?

Posted: 04/14/09 08:40:30

Posted: 04/14/09 20:44:42

As dood pointed out "<" and ">" are operators and you must therefore escape them all which means using something along the lines of

auto r = Regex("\\<title\\>(.*)\\</title\\>");

note that this is now escaping "<" instead of "/" which I don't think needs escaping. You could also try using

auto r = Regex(r"\<title\>(.*)\</title\>");

which becomes easier to read. The "r" before the string removes the need to escape characters within the string, a look at the D language reference on the Digtial Mars website may help you there.

Here's a small program that may make things clearer for you.

import tango.io.Stdout;
import tango.text.Regex;
import tango.net.http.HttpGet;

void main(){
        auto a=new HttpGet("http://www.google.com");
        auto p=cast(char[])a.read();
        auto r=new Regex(r"\<title\>(.*)\</title\>");
        r.test(p);
        Stdout(r.match(0)).newline;
}

Posted: 04/15/09 18:40:09

I have tried running your program and it works fine for me. I am using DMD v1.036 and the tango snapshot from 26/11/2008 both of which are available from the Tango Debian repository. If your versions differ then that may be the cause but I'm not sure.

Posted: 04/15/09 21:49:59

Posted: 04/23/09 19:56:10

a bit late, but I don't have that much time to play with tango now...

anyway, if someone needs conversion from/to latin-1 (based on php 5.2.9)

char[] decode(void[] content)
{
	ubyte[] buf = cast(ubyte[])content;
	ubyte[] result = new ubyte[buf.length * 4];
	uint n = 0;
	uint i = buf.length;
	ubyte* p = buf.ptr;
	while (i > 0)
	{
		uint c = cast(uint)*p;
		if (c < 0x80)
		{
			result[n++] = cast(ubyte)c;
		}
		else if (c < 0x800)
		{
			result[n++] = cast(ubyte)(0xC0 | (c >> 6));
			result[n++] = cast(ubyte)(0x80 | (c & 0x3F));
		}
		else if (c < 0x10000)
		{
			result[n++] = cast(ubyte)(0xE0 | (c >> 12));
			result[n++] = cast(ubyte)(0xC0 | ((c >> 6) & 0x3F));
			result[n++] = cast(ubyte)(0x80 | (c & 0x3F));
		}
		else if (c < 0x200000)
		{
			result[n++] = cast(ubyte)(0xF0 | (c >> 18));
			result[n++] = cast(ubyte)(0xE0 | ((c >> 12) & 0x3F));
			result[n++] = cast(ubyte)(0xC0 | ((c >> 6) & 0x3F));
			result[n++] = cast(ubyte)(0x80 | (c & 0x3F));
		}
		p++;
		i--;
	}
	result.length = n;
	return cast(char[])result;
}

void[] encode(char[] content)
{
	ubyte[] buf = cast(ubyte[])content;
	ubyte[] result = new ubyte[buf.length];
	uint n = 0;
	uint i = buf.length;
	ubyte* p = buf.ptr;
	while (i > 0)
	{
		uint c = *p;
		if (c >= 0xF0)
		{
			// four bytes encoded, 21 bits
			if (i >= 4)
			{
				c = ((p[0] & 0x07) << 18) | ((p[1] & 0x3F) << 12) | ((p[2] & 0x3F) << 6) | (p[3] & 0x3F);
			}
			else
			{
				c = 0x3F;
			}
			p += 4;
			i -= 4;
		}
		else if (c >= 0xE0)
		{
			// three bytes encoded, 16 bits
			if (i >= 3)
			{
				c = ((p[0] & 0x3F) << 12) | ((p[1] & 0x3F) << 6) | (p[2] & 0x3F);
			}
			else
			{
				c = 0x3F;
			}
			p += 3;
			i -= 3;
		}
		else if (c >= 0xC0)
		{
			// two bytes encoded, 11 bits
			if (i >= 2)
			{
				c = ((p[0] & 0x3F) << 6) | (p[1] & 0x3F);
			}
			else
			{
				c = 0x3F;
			}
			p += 2;
			i -= 2;
		}
		else
		{
			p++;
			i--;
		}
		// use '?' (0x3F) if no mapping is possible
		result[n++] = cast(ubyte)((c > 0xFF) ? 0x3F : c);
	}
	result.length = n;
	return cast(void[])result;
}