如何从表成员中引用lua表成员?
发布时间:2020-12-14 21:48:21 所属栏目:大数据 来源:网络整理
导读: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$"] = "ARMOR",}} 所以你看到这个
我在lua有一张桌子:
enUS = { LOCALE_STHOUSANDS = ",",--Thousands separator e.g. comma patNumber = "%d+["..LOCALE_STHOUSANDS.."%d]*",--regex to find a number ["PreScanPatterns"] = { ["^("..patNumber..") Armor$"] = "ARMOR",} } 所以你看到这个表中有一整串自引用: > LOCAL_STHOUSANDS > patNumber > [“^(”.. patNumber ..“)护甲$”] 如何在lua表中执行自引用? 我不想做的是必须硬替换价值观;有数百个参考: enUS = { LOCALE_STHOUSANDS = ",--Thousands separator e.g. comma patNumber = "%d+[,%d]*",--regex to find a number ["PreScanPatterns"] = { ["^(%d+[,%d]*) Armor$"] = "ARMOR",} } 解决方法
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$"] = "ARMOR",} } end do / end块在那里,以便临时变量不存在于表创建代码之外. 或者,您可以分阶段进行施工: enUS = {} enUS.LOCALE_STHOUSANDS = ",--Thousands separator e.g. comma enUS.patNumber = "%d+["..enUS.LOCALE_STHOUSANDS.."%d]*",--regex to find a number enUS["PreScanPatterns"] = { ["^("..enUS.patNumber..") Armor$"] = "ARMOR",} (编辑:李大同) 【声明】本站内容均来自网络,其相关言论仅代表作者个人观点,不代表本站立场。若无意侵犯到您的权利,请及时与联系站长删除相关内容! |