digitalmars.D.learn - String Manipulation

okibi (5/5) Jul 10 2007 I have a question for you all.

okibi (3/12) Jul 10 2007 Well, I ended up just doing some splits to get what the location of the ...

Gilles G. (15/31) Jul 10 2007 Hello,

Gilles G. (10/43) Jul 10 2007 Oups, sorry!

Frits van Bommel (35/46) Jul 10 2007 Did you try regexes (regular expressions)? (see

okibi (3/56) Jul 10 2007 That's exactly what I wanted! I tried using regex, but I've never really...

okibi <okibi ratedo.com> writes:

I have a question for you all.

If I have the following string or char[], how would I get the xsl filename out
of it?

char[] myStr = "...<?xml-stylesheet type=\"text/xsl\"
href=\"example.xsl\"?>...";

Is there a way to get it to return just example.xsl?

Thanks!

Jul 10 2007

okibi <okibi ratedo.com> writes:

Well, I ended up just doing some splits to get what the location of the xsl
file will be. Still, I'd like someone to tell me if there is an easier way.

Thanks!

okibi Wrote:

 I have a question for you all.
 
 If I have the following string or char[], how would I get the xsl filename out
of it?
 
 char[] myStr = "...<?xml-stylesheet type=\"text/xsl\"
href=\"example.xsl\"?>...";
 
 Is there a way to get it to return just example.xsl?
 
 Thanks!

Jul 10 2007

Gilles G. <schaouette free.fr> writes:

Hello,
maybe you can just use a regular expression...
In D, this will give something like:
import std.regexp;
void main()
{
    char[] myStr = "...<?xml-stylesheet
type=\"text/xsl\"href=\"example.xsl\"?>..."; 
    // this is the first RegExp I found, it may be flawed... It assumes the xsl
    // name has the extension ".xsl"
    auto nameOfStyleSheetRe = new RegExp(^.*href=\\\"(.*\.xsl).*$);
    auto m = nameOfStyleSheetRe.match(myStr);
    // m[1] should now contain the string "example.xsl"
    ... do what you want
}
okibi Wrote:

 Well, I ended up just doing some splits to get what the location of the xsl
file will be. Still, I'd like someone to tell me if there is an easier way.
 
 Thanks!
 
 okibi Wrote:
 
 I have a question for you all.
 
 If I have the following string or char[], how would I get the xsl filename out
of it?
 
 char[] myStr = "...<?xml-stylesheet type=\"text/xsl\"
href=\"example.xsl\"?>...";
 
 Is there a way to get it to return just example.xsl?
 
 Thanks!

Jul 10 2007

Gilles G. <schaouette free.fr> writes:

Oups, sorry!
Of course you should use
RegExp(r"^.*href=\\\"(.*\.xsl).*$");
instead of
RegExp(^.*href=\\\"(.*\.xsl).*$);

(I Forgot the quotes _and_ the r in front of the RegExp to indicate that
backslash must not be treated as an escape character...)

Regards.
--
Gilles

Gilles G. Wrote:

 Hello,
 maybe you can just use a regular expression...
 In D, this will give something like:
 import std.regexp;
 void main()
 {
     char[] myStr = "...<?xml-stylesheet
type=\"text/xsl\"href=\"example.xsl\"?>..."; 
     // this is the first RegExp I found, it may be flawed... It assumes the xsl
     // name has the extension ".xsl"
     auto nameOfStyleSheetRe = new RegExp(^.*href=\\\"(.*\.xsl).*$);
     auto m = nameOfStyleSheetRe.match(myStr);
     // m[1] should now contain the string "example.xsl"
     ... do what you want
 }
 okibi Wrote:
 
 Well, I ended up just doing some splits to get what the location of the xsl
file will be. Still, I'd like someone to tell me if there is an easier way.
 
 Thanks!
 
 okibi Wrote:
 
 I have a question for you all.
 
 If I have the following string or char[], how would I get the xsl filename out
of it?
 
 char[] myStr = "...<?xml-stylesheet type=\"text/xsl\"
href=\"example.xsl\"?>...";
 
 Is there a way to get it to return just example.xsl?
 
 Thanks!

Jul 10 2007

Frits van Bommel <fvbommel REMwOVExCAPSs.nl> writes:

[fixed upside-down reply]
okibi wrote:
 okibi Wrote:
 
 I have a question for you all.

 If I have the following string or char[], how would I get the xsl filename out
of it?

 char[] myStr = "...<?xml-stylesheet type=\"text/xsl\"
href=\"example.xsl\"?>...";

 Is there a way to get it to return just example.xsl?

 Well, I ended up just doing some splits to get what the location of the xsl
file will be. Still, I'd like someone to tell me if there is an easier way.

Did you try regexes (regular expressions)? (see 
http://www.digitalmars.com/d/1.0/phobos/std_regexp.html)

(I see Gilles G. has already suggested regexes since I started this 
post, but I'll post it anyway since I think my suggested regex is better 
:) )

For example:
---
import std.regexp;
import std.stdio;

void main() {
     char[] myStr = "...<?xml-stylesheet type=\"text/xsl\" 
href=\"example.xsl\"?>...";

     if (auto m = search(myStr, `<\?.*href="(.*)".*\?>`)) {
         writefln("Match: '%s'", m.match(1));
     } else {
         writefln("No match found.");
     }
}
---
(Correct for linewrapping before use)
This picks out the text between (double) quotes after 'href=' in a 
'<?'-'?>' block. You'll need to be a bit more tricky if you want to 
handle single quotes as well (or is that a HTML-only thing?). Perhaps 
`<\?.*href=(["'])(.*)\1.*\?>`: The part between the first parentheses 
captures the opening quote after 'href=', the \1 says to match the same 
quote there. The .xsl file name is then m.match(2), since m.match(1) is 
now the opening quote.

By the way: note the use of a backquoted string (``s) to avoid escaping 
the '\'s and '"'s, otherwise the original regexp would be 
"<\\?.*href=\"(.*)\".*\\?>" which is equivalent but uglier (IMHO). You 
could also use this for your XML string to avoid '\"' all over the place.

Also notice the quoting of '?' as '\?' since '?' is a special character 
in regexes.

Jul 10 2007

okibi <okibi ratedo.com> writes:

That's exactly what I wanted! I tried using regex, but I've never really
understood how they work. It's one of those things I think I need someone to
spell it out for me lol.

Thanks!

Frits van Bommel Wrote:

 [fixed upside-down reply]
 okibi wrote:
 okibi Wrote:
 
 I have a question for you all.

 If I have the following string or char[], how would I get the xsl filename out
of it?

 char[] myStr = "...<?xml-stylesheet type=\"text/xsl\"
href=\"example.xsl\"?>...";

 Is there a way to get it to return just example.xsl?

 Well, I ended up just doing some splits to get what the location of the xsl
file will be. Still, I'd like someone to tell me if there is an easier way.

 
 Did you try regexes (regular expressions)? (see 
 http://www.digitalmars.com/d/1.0/phobos/std_regexp.html)
 
 (I see Gilles G. has already suggested regexes since I started this 
 post, but I'll post it anyway since I think my suggested regex is better 
 :) )
 
 For example:
 ---
 import std.regexp;
 import std.stdio;
 
 void main() {
      char[] myStr = "...<?xml-stylesheet type=\"text/xsl\" 
 href=\"example.xsl\"?>...";
 
      if (auto m = search(myStr, `<\?.*href="(.*)".*\?>`)) {
          writefln("Match: '%s'", m.match(1));
      } else {
          writefln("No match found.");
      }
 }
 ---
 (Correct for linewrapping before use)
 This picks out the text between (double) quotes after 'href=' in a 
 '<?'-'?>' block. You'll need to be a bit more tricky if you want to 
 handle single quotes as well (or is that a HTML-only thing?). Perhaps 
 `<\?.*href=(["'])(.*)\1.*\?>`: The part between the first parentheses 
 captures the opening quote after 'href=', the \1 says to match the same 
 quote there. The .xsl file name is then m.match(2), since m.match(1) is 
 now the opening quote.
 
 By the way: note the use of a backquoted string (``s) to avoid escaping 
 the '\'s and '"'s, otherwise the original regexp would be 
 "<\\?.*href=\"(.*)\".*\\?>" which is equivalent but uglier (IMHO). You 
 could also use this for your XML string to avoid '\"' all over the place.
 
 Also notice the quoting of '?' as '\?' since '?' is a special character 
 in regexes.

Jul 10 2007

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digitalmars.D.learn - String Manipulation