kmod.attr: fix string.gsub patterns and return values

basename(): return one value, not two (principle of least confusion).
strip_compress_sfx(): ditto

strip_compress_sfx(): fix it, it had several problems:
    (a) incorrect escaping of . (period) ... masked by incorrect use of backslash
    (b) lua has no | "or" pattern

Use simpler gsub patterns / replacements, presumably faster (did not measure).

Signed-off-by: Denys Vlasenko <dvlasenk@redhat.com>

kmod.attr

@@ -1,10 +1,31 @@
 %__kmod_path           ^/lib/modules/.*/(modules.builtin|.*\.ko|.*\.ko\.gz|.*\.ko\.bz2|.*\.ko\.xz|.*\.ko\.zst)$
+
+# Notes on Lua:
+# The backslash in strings (like "\n" newline) needs to be doubled
+# because we are inside rpm macro. Single backslashes before most chars
+# disappear (removed by rpm's parser), so "\n" turns into just "n".
+# In string.gsub patterns, unlike regexps, backslash has no special meaning.
+# It can't escape . and such. (Use one-character set [.] to represent
+# literal period, or lua's percent escape: %.)
+# Pipe (|) has no special meaning too.
+
 %__kmod_provides() %{lua:
   function basename(fn)
-      return string.gsub(fn, "(.*/)(.*)", "%2")
+      local b = string.gsub(fn, ".*/", "")
+      -- the above adjusts gsub() result to 1 value
+      -- "return f()" construct would return _all_ values, two in case of gsub()
+      return b
   end
   function strip_compress_sfx(fn)
-      return string.gsub(fn, "(.*)(\.gz|\.bz2|\.xz|\.zst)?$", "%1")
+      local cnt
+      fn, cnt = string.gsub(fn, "%.gz$", "")
+      if cnt == 1 then return fn; end
+      fn, cnt = string.gsub(fn, "%.bz2$", "")
+      if cnt == 1 then return fn; end
+      fn, cnt = string.gsub(fn, "%.xz$", "")
+      if cnt == 1 then return fn; end
+      fn, cnt = string.gsub(fn, "%.zst$", "")
+      return fn
   end
   function printdep(mod)
       print("kmod("..mod..") ")