我使用luaxml将Lua表转换为xml。我使用luaxml时遇到问题,例如,我有一个像这样的lua表
t = {name="John", age=23, contact={email="john@gmail.com", mobile=12345678}}
当我尝试使用LuaXML时,
local x = xml.new("person")
x:append("name")[1] = John
x:append("age")[1] = 23
x1 = xml.new("contact")
x1:append("email")[1] = "john@gmail.com"
x1:append("mobile")[1] = 12345678
x:append("contact")[1] = x1
生成的xml变为:
<person>
<name>John</name>
<age>23</age>
<contact>
<contact>
<email>john@gmail.com</email>
<mobile>12345678</mobile>
</contact>
</contact>
</person>`
xml中有2个联系人。我该怎么做才能使Xml正确?
此外,如何将XML转换回Lua表?
答案 0 :(得分:2)
您的语法有点偏离,您正在为联系人创建一个新表,然后附加一个“联系人”节点并使用此代码同时分配另一个节点:
x1 = xml.new("contact")
x1:append("email")[1] = "john@gmail.com"
x1:append("mobile")[1] = 12345678
x:append("contact")[1] = x1
这应该是这样的:
local contact = xml.new("contact")
contact.email = xml.new("email")
table.insert(contact.email, "john@gmail.com")
contact.mobile = xml.new("mobile")
table.insert(contact.mobile, 12345678)
请记住,每个'node'都是它自己的表值,这是xml.new()返回的值。
以下代码在您致电xml.save(x, "\some\filepath")
时正确创建了xml。要记住的是,无论何时调用xml.new(),你都会得到一个表,我认为可以轻松设置属性...但是使用简单值添加的语法更加冗长
-- generate the root node
local root = xml.new("person")
-- create a new name node to append to the root
local name = xml.new("name")
-- stick the value into the name tag
table.insert(name, "John")
-- create the new age node to append to the root
local age = xml.new("age")
-- stick the value into the age tag
table.insert(age, 23)
-- this actually adds the 'name' and 'age' tags to the root element
root:append(name)
root:append(age)
-- create a new contact node
local contact = xml.new("contact")
-- create a sub tag for the contact named email
contact.email = xml.new("email")
-- insert its value into the email table
table.insert(contact.email, "john@gmail.com")
-- create a sub tag for the contact named mobile
contact.mobile = xml.new("mobile")
table.insert(contact.mobile, 12345678)
-- add the contact node, since it contains all the info for its
-- sub tags, we don't have to worry about adding those explicitly.
root.append(contact)
在这个例子之后,你应该可以很清楚地深入挖掘它。你甚至可以编写函数来轻松创建子标签,使代码更简洁......
答案 1 :(得分:0)
local x = xml.new("person")
x:append("name")[1] = John
x:append("age")[1] = 23
x1 = x:append("contact")
x1:append("email")[1] = "john@gmail.com"
x1:append("mobile")[1] = 12345678
print(x)