Lua string.gsub具有多个模式 [英] Lua string.gsub with Multiple Patterns

问题描述

我正在重命名包含多余字母的电影标题. string.gsub可以将字符串替换为" nil值，但是我大约有200种字符串模式需要替换为".

I am working on renaming the Movie titles that has unwanted letters. The string.gsub can replace a string with "" nil value but I have around 200 string patterns that need to be replaces with "".

现在，对于每个模式，我都必须string.gsub.我当时在想是否有一种方法可以将所有字符串模式都放在单个string.gsub行中.我已经在网上搜索了解决方案，但仍然什么也没得到.

Right now I have to string.gsub for every pattern. I was thinking is there is a way to put all the string patterns in to single string.gsub line. I have searched around the web for the solution but still didn't got anything.

电影标题就是这样B.A.Pass 2013 Hindi 720p DvDRip CROPPED AAC x264 RickyKT 我想删除多余的字符，例如2013，Hindi，720p，DvDRip，CROPPED，AAC，x264，RickyKT.

The movie title is like this B.A.Pass 2013 Hindi 720p DvDRip CROPPED AAC x264 RickyKT and I want to remove the extra characters like 2013, Hindi, 720p, DvDRip, CROPPED, AAC, x264, RickyKT.

推荐答案

您可以将表作为第三个参数传递给string.gsub，如下所示:

You can pass to string.gsub a table as the third argument like this:

local movie = "B.A.Pass 2013 Hindi 720p DvDRip CROPPED AAC x264 RickyKT"
movie = movie:gsub("%S+", {["2013"] = "", ["Hindi"] = "", ["720p"] = "", 
                       ["DvDRip"] = "", ["CROPPED"] = "", ["AAC"] = "", 
                       ["x264"] = "", ["RickyKT"] = ""})

print(movie)

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