如何从表成员中引用lua表成员？

Question

我在lua有一张桌子:

enUS = {
   LOCALE_STHOUSANDS = ",",  --Thousands separator e.g. comma

   patNumber = "%d+["..LOCALE_STHOUSANDS.."%d]*", --regex to find a number

   ["PreScanPatterns"] = {
      ["^("..patNumber..") Armor$"] = "ARMOR",
   }
}

所以你看到这个表中有一整串自引用:

• LOCAL_STHOUSANDS
  • patNumber
    • ["^("..patNumber..") Armor$"]

如何在lua表中执行自引用？

---

我不想做的是必须硬替换价值观; 有数百个参考:

enUS = {
   LOCALE_STHOUSANDS = ",",  --Thousands separator e.g. comma

   patNumber = "%d+[,%d]*", --regex to find a number

   ["PreScanPatterns"] = {
      ["^(%d+[,%d]*) Armor$"] = "ARMOR",
   }
}

Answer 1

如何在lua表中执行自引用？

你没有.

Lua不是C.在构造表之前,不存在任何表条目.因为表本身尚不存在.因此,您不能在表构造函数中有一个条目引用不存在的表中的另一个条目.

如果你想减少重复输入,那么你应该使用局部变量和do/end块:

do
  local temp_thousands_separator = ","
  local temp_number_pattern = "%d+["..LOCALE_STHOUSANDS.."%d]*"

  enUS = {
   LOCALE_STHOUSANDS = temp_thousands_separator,  --Thousands separator e.g. comma

   patNumber = "%d+["..temp_thousands_separator.."%d]*", --regex to find a number

   ["PreScanPatterns"] = {
      ["^("..temp_number_pattern..") Armor$"] = "ARMOR",
   }
  }
end

该do/end块是存在的,因此临时变量不存在于表创建代码之外.

或者,您可以分阶段进行施工:

  enUS = {}
  enUS.LOCALE_STHOUSANDS = ",",  --Thousands separator e.g. comma

  enUS.patNumber = "%d+["..enUS.LOCALE_STHOUSANDS.."%d]*", --regex to find a number

  enUS["PreScanPatterns"] = {
      ["^("..enUS.patNumber..") Armor$"] = "ARMOR",
   }

Answer 2

无法在构造函数本身内部执行此操作，但您可以在创建表后执行此操作，如下所示：

enUS = {
   LOCALE_STHOUSANDS = ","
}

enUS.patNumber = "%d+["..enUS.LOCALE_STHOUSANDS.."%d]*"
enUS.PreScanPatterns = {
  ["^("..enUS.patNumber..") Armor$"] = "ARMOR",
}

如果你特别需要引用当前表，Lua提供了一个“self”参数，但它只能在函数中访问。

local t = {
  x = 1,
  y = function(self) return self.x end
} 

-- this is functionally identical to t.y
function t:z() return self.x end

-- these are identical and interchangeable
print(t:y(), t.z(t))
-- 1, 1