VIM 使用 vimrc 更改折叠降价以显示行数和字数

Question

我在 Windows 上使用 GVIM 来编写科幻小说，我将章节折叠起来，以便将它们隐藏起来。折叠时，“行”章节显示如下：

##1 – Spaceman Spoof Encounters the Watterson........ 115 lines [2.4%] +--+--

这很有用，但只有一点点。有些行可能只有几个词，而另一些则可能超过 100 个词。我更喜欢这样的行数和字数：

##1 – Spaceman Spoof Encounters the Watterson........ l:115 w:1032 [4.4%] +--+--

百分比是单词而不是行。任何帮助都将不胜感激。

这是我_vimrc的折叠部分：

function! MarkdownFolds()
    let thisline = getline(v:lnum)
    if match(thisline, '^##') >= 0
        return ">2"
    elseif match(thisline, '^#') >= 0
        return ">1"
    else
        return "="
    endif
endfunction

setlocal foldmethod=expr
setlocal foldexpr=MarkdownFolds()

function! MarkdownFoldText()
    "get first non-blank line
    let fs = v:foldstart
    while getline(fs) =~ '^\s*$' | let fs = nextnonblank(fs + 1)
    endwhile
    if fs > v:foldend
        let line = getline(v:foldstart)
    else
        let line = substitute(getline(fs), '\t', repeat(' ', &tabstop), 'g')
    endif

    let w = winwidth(0) - &foldcolumn - (&number ? 8 : 0)
    let foldSize = 1 + v:foldend - v:foldstart
    let foldWords = v:foldend,v:foldstart!wc -w
    let foldSizeStr = " " . foldSize . " lines "
    let foldWordsStr = " " . foldWords . " w "
    let foldLevelStr = repeat("+--", v:foldlevel)
    let lineCount = line("$")
    let foldPercentage = printf("[%.1f", (foldSize*1.0)/lineCount*100) . "%] "
    " let expansionString = "."
    let expansionString = repeat(".", w - strwidth(foldSizeStr.line.foldLevelStr.foldPercentage))
    return line . expansionString . foldSizeStr . foldPercentage . foldLevelStr
    " return line . expansionString . foldSizeStr . foldPercentage . foldWordsStr . foldLevelStr
    " return line . "......" . foldSizeStr . foldPercentage . foldLevelStr
endfunction

setlocal foldtext=MarkdownFoldText()

Answer 1

这里有一个解决方案：

function! MarkdownFoldText()
    "get first non-blank line
    let fs = v:foldstart
    while getline(fs) =~ '^\s*$' | let fs = nextnonblank(fs + 1)
    endwhile
    if fs > v:foldend
        let line = getline(v:foldstart)
    else
        let line = substitute(getline(fs), '\t', repeat(' ', &tabstop), 'g')
    endif

    let w = winwidth(0) - &foldcolumn - (&number ? 8 : 0)
    let foldSize = 1 + v:foldend - v:foldstart

    let i = 0
    let foldWords=0
    let wordCount=0
    while (i<line('$'))
      let lineWords = len(split(getline(i)))
      if (i<=v:foldend && i>= v:foldstart)
        let foldWords = foldWords + lineWords
      endif
      let wordCount += lineWords
      let i += 1
    endwhile

    let foldWordsStr = " " . foldWords . " w,"
    let foldSizeStr = foldWordsStr . foldSize . " lines "
    let foldLevelStr = repeat("+--", v:foldlevel)
    let foldPercentage = printf("[%.1f", (foldWords*1.0)/wordCount*100) . "%] "
    let expansionString = repeat(".", w - strwidth(foldSizeStr.line.foldLevelStr.foldPercentage))
    return line . expansionString . foldSizeStr . foldPercentage . foldLevelStr
endfunction

这个想法是计算全局字数，以及同时在选定区域中的字数。

然后我们根据要求将百分比替换为字数。

Answer 2

这是另一个解决方案，这次使用 wordcount() 函数：

function! MarkdownFoldText()
    "get first non-blank line
    let fs = v:foldstart
    while getline(fs) =~ '^\s*$' | let fs = nextnonblank(fs + 1)
    endwhile
    if fs > v:foldend
        let line = getline(v:foldstart)
    else
        let line = substitute(getline(fs), '\t', repeat(' ', &tabstop), 'g')
    endif

    let w = winwidth(0) - &foldcolumn - (&number ? 8 : 0)
    let foldSize = 1 + v:foldend - v:foldstart

    let i = v:foldstart
    let foldWords=0
    while (i<v:foldend)
      let lineWords = len(split(getline(i)))
      let foldWords = foldWords + lineWords
      let i += 1
    endwhile
    let wordCount = wordcount()["words"]

    " let foldWords = v:foldend,v:foldstart!wc -w
    let foldWordsStr = " " . foldWords . " w,"
    let foldSizeStr = foldWordsStr . foldSize . " lines "
    let foldLevelStr = repeat("+--", v:foldlevel)
    let foldPercentage = printf("[%.1f", (foldWords*1.0)/wordCount*100) . "%] "
    " let expansionString = "."
    let expansionString = repeat(".", w - strwidth(foldSizeStr.line.foldLevelStr.foldPercentage))
    return line . expansionString . foldSizeStr . foldPercentage . foldLevelStr
    " return line . expansionString . foldSizeStr . foldPercentage . foldWordsStr . foldLevelStr
    " return line . "......" . foldSizeStr . foldPercentage . foldLevelStr
endfunction

应该更快。并不是说这个文件夹与整个缓冲区相比的百分比是多少。使用 wordcount() 函数，我们离终点还有多远将是微不足道的。