Help talk:Parser functions in templates

I think there's a logical error here. It's easiest to see in the table at the bottom. In sample 1, is true for both being defined and for being not defined. How can it both be defined and not defined? -Khono
 * I think it is right. If you call, then both and  will return "bar" (in template:foo), why the #if statement treat both as true. --79.136.98.249 23:13, 1 September 2009 (UTC)

White spaces
Is there any parser function that strip leading and tailing white spaces, or one that change them into "%20" or "_"? I try to make a template that includes a link to wikipedia.

--Petter Källström, 79.136.98.249 23:13, 1 September 2009 (UTC)
 * givs: See phoo Wikipedias article - NOK (.../wiki/).
 * givs: See Wikipedias article - OK (.../wiki/phoo).
 * givs: See bar Wikipedias article - NOK (.../wiki/phoo).
 * givs: See Wikipedias article - NOK (../wiki/Parser).

parameter specified or not
One way to test whether a parameter was specified in a template call is thus:  

Could an expert please check whether this is true? I find it returns "v was specified (and may be empty)" only when "|v=" or "|v=text" is in the template call code, otherwise "|v" (without any =) returns "v was not specified ". The latter behaviour is unexpected. -84user 18:20, 27 September 2009 (UTC)
 * I may have misunderstood, but why is the latter behaviour unexpected? When = is missing, v is interpreted as an unnamed parameter, not as parameter v, so parameter v is missing. I believe that is the correct behaviour. Hamilton Abreu 00:09, 24 January 2010 (UTC)