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On Jun 03, 2005, at 12:25, Premshree Pillai wrote:

Is there a simple XML parser (preferably DOM) for Lua?

XPP rather than DOM:

http://dev.alt.textdrive.com/file/lu/LUXMLInputStream.lua

Usage example:

local anInputStream = LUXMLInputStream.new( aContent )

while ( true ) do
local aType, aContent, aName, someAttributes = anInputStream.read()


if ( aType == nil ) then
break
elseif ( aType == LUXMLInputStream.Text ) then
print( aType, aContent )
elseif ( someAttributes ~= nil ) then
print( aType, aName, someAttributes.keys().sort().toString() )
else
print( aType, aName )
end
end


Cheers

--
PA, Onnay Equitursay
http://alt.textdrive.com/