Mansour Al Akeel schrieb:
On Tue Jun 02,2009 10:34 am, Steven Dolg wrote:
Mansour Al Akeel schrieb:
For some reasons {0} is matching only the digits, because 1 is empty
always !
<map:match type="regexp" pattern="(\**)([0-9])$">
......
<map:transform src="transform.xsl">
<map:parameter name="href" value="{1}"/>
<map:parameter name="depth" value="{2}"/>
</map:transform>
My understanding is {0}={1}+{2} but since {1} is empty always, I am
getting {0}={2}. Now the question is how do I capture {1} ?
Since the documentaiton says that regexp works like wildcard matcher,
there should be a way to capture all parts or the regex.
Any one has an idea ?
Your pattern "(\**)([0-9])$" means "any number of asterisks followed by
exactly one digit at the end of the string".
Any number here means 0 or more.
Steven, thank you. I see my mistake. I am looking to match anything ends
with a digit. For example:
hostname/myBlock/news/4
hostname/myBlock/services/2
hostname/myBlock/news/local/3
hostname/myBlock/3
I need the first part and the digit. The first part can be empty as
well.
The pattern "(.*)(\d+)" should work for the cases you mentioned. ("\d+"
will match 1...n digits; if you just want to allow one digit "\d" is the
thing to do)
When using "hostname/myBlock/news/local/4" the groups will be:
- {0} = "news/local/4" ({0} is always the whole expression; and
"hostname/myBlock/" will not be forwarded to the sitemap of your block)
- {1} = "news/local/"
- {2} = "4"
For "hostname/myBlock/3" the groups will be:
- {0} = "3"
- {1} = ""
- {2} = "3"
So {0} == {2}, but this is merely because the first group is empty.
HTH,
Steven
If the string you are using for matching does not contain any asterisks
before the digit the first group will be empty.
So the question is, what do you want to capture with the first group?
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