Why won't this find work?

methodic · Post by **methodic** » Fri Jul 21, 2006 5:00 am

I have a webpage I am trying to parse... I slurp it with get-url, replace all whitespace with one space (newlines, tabs, multiple spaces, etc)... here is the part I am trying to grab: " Last Login:  7/20/2006 </td>"

here is my find: (find " Last Login:  (.*) " txt 0)

and it returns up to 7/20/2006, but it doesn't stop at the , it keeps going past it to the end of the file... I am using multiple finds on this page, and every of them return fine, are the forward slashes in the date screwing up find?

thanks.

methodic · Post by **methodic** » Fri Jul 21, 2006 5:09 am

Nevermind, I see what I did wrong, sorry. :)

I guess that's a *feature* of find. ;)

cormullion · Post by **cormullion** » Fri Jul 21, 2006 10:38 am

I've found regex useful when working with this stuff:

Code: Select all

(set 'txt "<br> <br> Last Login:&nbsp; 7/20/2006<br> </td>")
(regex "<br> Last Login:&nbsp; (.*)<br>" txt 0) 
;-> ("<br> Last Login:&nbsp; 7/20/2006<br>" 5 36 "7/20/2006" 28 9)

which tells you you want the $1 value rather than the $0:

Code: Select all

(find "<br> Last Login:&nbsp; (.*)<br>" txt 0) 
;-> 5
(println $0)
;-><br> Last Login:&nbsp; 7/20/2006<br> 
(println $1)
;->7/20/2006

I'm still struggling with this myself...! Yellow RegExp-Do belt worn with pride... ;-)

Lutz · Post by **Lutz** » Fri Jul 21, 2006 12:12 pm

In both cases $0 and $1 show the same: $0 the whole area covered by the pattern and $1 the parenthesized subpattern:

Code: Select all

> (regex "<br> Last Login:&nbsp; (.*)<br>" txt 0) 
("<br> Last Login:&nbsp; 7/20/2006<br>" 5 36 "7/20/2006" 28 9)

> $0
"<br> Last Login:&nbsp; 7/20/2006<br>"
> $1
"7/20/2006"

> (find "<br> Last Login:&nbsp; (.*)<br>" txt 0) 
5

> $0
"<br> Last Login:&nbsp; 7/20/2006<br>"
> $1
"7/20/2006"
>

In the list returned by regex imagine 3 members are always grouped together: string, offset and length.

The first group then corresponds to $0 the next to $1 etc.

Lutz

Lutz · Post by **Lutz** » Fri Jul 21, 2006 12:19 pm

to Methodic:

the .* operator will always grab as much as it can and still satify the pattern. You can use the option 512 to invert greediness or put an ? after the star as in .*?

Code: Select all

> (find "a.*c" "abbbbcbbbcd" 0)
0
> $0
"abbbbcbbbc"

> (find "a.*c" "abbbbcbbbcd" 512)
0
> $0
"abbbbc"
> 

> (find "a.*?c" "abbbbcbbbcd" 0)
0
> $0
"abbbbc"

and here the same with parenthesized subexpressions to isolate the subpattern:

Code: Select all

> (find "a(.*)c" "abbbbcbbbcd" 0)
0
> $1
"bbbbcbbb"

> (find "a(.*?)c" "abbbbcbbbcd" 0)
0
> $1
"bbbb"

Lutz

methodic · Post by **methodic** » Sat Jul 22, 2006 10:18 pm

Ah, thats what I was looking for, the (.*?)

thanks so much!