Question

例如，我有

str = "Beamer-Template!navigation symbols@\\texttt {navigation symbols}"
print(str:gsub('[^!|@%s]+@', ''))

可打印

Beamer-Template!navigation \texttt {navigation symbols}

但应该是

Beamer-Template!\texttt {navigation symbols}

我如何抓住空间？重要的只是foo@bar。该模式适用于像

这样的字符串

str="foo@bar!baz@foobar!nice|crazy"
-> bar!foobar!nice|crazy

但没有额外的空格

str="foo@bar!baz baz@foobar!nice|crazy"
-> bar!baz foobar!nice|crazy

应为bar!foobar!nice|crazy

Answer 1

要匹配makeindex条目，使用LPEG语法可能会很有用。这样，您可以在分隔符处进行拆分，甚至根据匹配的字段执行语义操作。

local lpeg = assert(require"lpeg")
local C, S = lpeg.C, lpeg.S

local sep = S("@!|")
local str = C((1 - sep)^0)

local idx = str * ( "@" * str / function(match) return "@" .. match end
                  + "!" * str / function(match) return "!" .. match end
                  + "|" * str / function(match) return "|" .. match end)^0

print(idx:match("hello!world@foo|bar"))

$ lua test.lua 
hello   !world  @foo    |bar

回答评论：在表中收集匹配项。匹配项将根据其前缀进行收集。

local lpeg = assert(require"lpeg")
local C, Ct, S = lpeg.C, lpeg.Ct, lpeg.S

local sep = S("@!|")
local str = C((1 - sep)^0)

local match = function(expr)
    local prefix = function(prefix)
        return function(match)
            return prefix .. match
        end
    end

    local idx = str * ( "@" * str / prefix("@")
                      + "!" * str / prefix("!")
                      + "|" * str / prefix("|"))^0

    return Ct(idx):match(expr)
end

for _, str in ipairs{
    "hello!world@foo|bar",
    "foo@bar!baz baz@foobar!nice|crazy",
    "foo@bar!baz@foobar!nice|crazy",
    "Beamer-Template!navigation symbols@\\texttt {navigation symbols}"
} do
    local t = match(str)
    print(table.concat(t," "))
end

$ lua test.lua 
hello !world @foo |bar
foo @bar !baz baz @foobar !nice |crazy
foo @bar !baz @foobar !nice |crazy
Beamer-Template !navigation symbols @\texttt {navigation symbols}

Lua：获取一个子字符串

1 个答案: