协助Lua功能

如前所述,我对lua相对较新,但同样,我学得很快.我最后一次在这里得到了帮助,这对我帮助很大,而且我能写出更好的剧本.现在我来到另一个问题,我认为这将使我的生活更轻松一些.我不知道我在做什么功能,但我希望有办法做我想做的事情.下面,您将看到我要删除一些不需要的元素的代码示例.是的,我意识到它至少没有效率,所以如果其他人对如何使它更有效率有更好的了解,我会全神贯注.我想做的是用它创建一个函数,这样我就可以通过简单的调用来删除任何变量(比如stripdown(winds)).我感谢所提供的任何帮助,以及任何课程.谢谢！

winds = string.gsub(winds,"%b<>","")  
winds = string.gsub(winds,"%c"," ")  
winds = string.gsub(winds,"  "," ")  
winds = string.gsub(winds,"  "," ")  
winds = string.gsub(winds,"^%s*(.-)%s*$", "%1)")  
winds = string.gsub(winds,"&nbsp;","")  
winds = string.gsub(winds,"/ ", "(")

玩笑

把它变成一个函数是很容易的部分.

function stripdown(winds)
    winds = string.gsub(winds,"%b<>","")  
    winds = string.gsub(winds,"%c"," ")  
    winds = string.gsub(winds,"  "," ")  
    winds = string.gsub(winds,"  "," ")  
    winds = string.gsub(winds,"^%s*(.-)%s*$", "%1)")  
    winds = string.gsub(winds,"&nbsp;","")  
    winds = string.gsub(winds,"/ ", "(")  
    return winds
end

写入的这个函数确实产生并放弃了很多中间字符串结果,这可能是一个相对昂贵的操作.几乎可以肯定值得仔细研究string.gsub()及其pattern language的文档.应该可以在较少的操作中至少完成一些您指定的内容.