Converting grep/sed combination to Java

Hi,

I am using the following grep/sed combination to extract hyperlinks from
a document on Windows 2000.  This is a one-line command:

grep -o -E "href=\"[^&#:;+?/\.0-9A-Za-z_-]*\"" Contents.html|sed -e
"s/href=\"//" -e "s/\"//"

Please could someone tell me how to pass the regular expression part
(the argument to grep), to Pattern.compile in Java.  I have tried
putting \\ before characters which I think may cause problems, but Java
still reports errors within the argument passed to Pattern.compile.

Note: I realised that I had to remove the ^ before the & as this was
only included to make the command valid to the Windows 2000 command
interpreter.

Any help would be much appreciated.

Regards,
Jonny

On Sat, 04 Jun 2005 10:20:50 +0000, Jonny wrote:
> Note: I realised that I had to remove the ^ before the & as this was
> only included to make the command valid to the Windows 2000 command
> interpreter.

man grep. ;)

The ^ inverses the selection, this means you inverted the logic by
removing the ^. Don't do this.

--
In pioneer days they used oxen for heavy pulling, and when one ox
couldn't budge a log, they didn't try to grow a larger ox. We shouldn't
be trying for bigger computers, but for more systems of computers.
--- Rear Admiral Grace Murray Hopper

Stefan Schulz wrote:

> On Sat, 04 Jun 2005 10:20:50 +0000, Jonny wrote: 
>
> man grep. ;)
>
> The ^ inverses the selection, this means you inverted the logic by
> removing the ^. Don't do this.

Thanks for your reply, Stefan.

The problem I am having is not how to use grep, but how to pass the
regular expression to Pattern.compile without getting Java compilation
errors.  Do certain characters need to be escaped?

Regards,
Jonny

In message <6gfoe.7317$%21.3303@newsfe2-gui.ntli.net>, Jonny
<www.mail@ntlworld.com> writes
>Hi,
>
>I am using the following grep/sed combination to extract hyperlinks from
>a document on Windows 2000.  This is a one-line command:
>
>grep -o -E "href=\"[^&#:;+?/\.0-9A-Za-z_-]*\"" Contents.html|sed -e
>"s/href=\"//" -e "s/\"//"
>
>Please could someone tell me how to pass the regular expression part
>(the argument to grep), to Pattern.compile in Java.  I have tried
>putting \\ before characters which I think may cause problems, but Java
>still reports errors within the argument passed to Pattern.compile.

In Java, you can combine the grep and the 2 sed commands.

Try something like:

href=\"(.*?)\"

i.e. look for href=" then collect all characters up to but not including
the first ".

The ? makes the .* non-greedy.

HTH

Jonny wrote:
> Stefan Schulz wrote:
>
> 
>
>
> Thanks for your reply, Stefan.
>
> The problem I am having is not how to use grep, but how to pass the
> regular expression to Pattern.compile without getting Java compilation
> errors.  Do certain characters need to be escaped?

The string to compile is just a string. The only thing that has to be
escaped is the backslash. So every backslash in the pattern must be doubled.

--
Dale King

Nemo wrote:

> In message <6gfoe.7317$%21.3303@newsfe2-gui.ntli.net>, Jonny
> <www.mail@ntlworld.com> writes 
>
> In Java, you can combine the grep and the 2 sed commands.
>
> Try something like:
>
> href=\"(.*?)\"
>
> i.e. look for href=" then collect all characters up to but not including
> the first ".
>
> The ? makes the .* non-greedy.

Thanks Nemo.

That's exactly what I was looking for.  I didn't realise Java used the
advanced regexp syntax.

The statement I finally used, is:

regexp_pattern =
Pattern.compile("href=\\\"(.*?)\\\"",
Pattern.CASE_INSENSITIVE);

Your help is appreciated.

Regards,
Jonny

Dale King wrote:

> Jonny wrote: 
>
> The string to compile is just a string. The only thing that has to be
> escaped is the backslash. So every backslash in the pattern must be doubled.[/colo
r]

There's more to it than that.  Characters also have to be escaped within
the regular expression in order to be compiled by the regexp compiler.
These characters are also escaped by using a backslash.

See my response to Nemo's post, where I have to use three backslashes to
escape a double-quote character.  A regexp compiler would expect a
double-quote to be passed as \", but so would the Java compiler, which
in addition expects the backslash itself to be escaped.  Hence the need
for \\\"

Thanks for your reply.

Regards,
Jonny

Jonny wrote:
> Dale King wrote:
>
> 
>
>
> There's more to it than that.  Characters also have to be escaped within
> the regular expression in order to be compiled by the regexp compiler.
> These characters are also escaped by using a backslash.

I assume he knew that. I was referring to what had to be escaped on top
of the normal regular expression escapes.

> See my response to Nemo's post, where I have to use three backslashes to
> escape a double-quote character.  A regexp compiler would expect a
> double-quote to be passed as \", but so would the Java compiler, which
> in addition expects the backslash itself to be escaped.  Hence the need
> for \\\"

D'oh. Yes forgot to mention the quote. My point was that the thing that
really trips people up is the fact that backslashes must be escaped.

I really wish they would adopt an alternate way to allow you to specify
strings that would let you get around the escaping for things like
regular expressions and path names. Such a thing has been proposed here
before:

<http://groups-beta.google.com/group...ae847810dfcb8e1>

--
Dale King

In message <fazoe.1861$K5.16@newsfe4-win.ntli.net>, Jonny
<www.mail@ntlworld.com> writes
>See my response to Nemo's post, where I have to use three backslashes to
>escape a double-quote character.  A regexp compiler would expect a
>double-quote to be passed as \", but so would the Java compiler, which

Are you sure that you need to quote " in an RE?
However, it won't do any harm, if not required.

I now realise my previous posting was a bit ambiguous/incomplete.
I quoted the " but left off the enclosing " and " - all of which are
needed to turn it into a Java String.

I should have written:

String pattern="href=\"(.*?)\"";

OK?