2
简而言之，写了一些功能来保存输入我试图设置一个vmap [ping]，这将允许我选择我输入的内容，并将此选项传递给一个函数（因为在命令行上键入函数调用，键入参数（带引号）和转义反斜杠等......抵消通过调用函数节省的大部分时间）Vim vmap，发送选定的文本作为参数功能
对于（一个简单的）示例，假设我有以下功能
func Test(iStr)
if a:iStr[0] =~ [a-zA-z]
echo "hello"
else
echo "hello world"
endif
endfunc
我希望能够可视化选择一些文本，然后用一些按键映射，F2说，它会调用测试（ISTR）与选择作为参数iStr
我相信，与更多的工作（即一些方法来指定我所选择的应该是内部测试（）），以下
vmap <F2> :call Test()
是我后。事情是我已经尝试了一些变体（猜测加上一些有趣的推理：help map），我没有得到任何有用的东西。更新，我一直在使用一个辅助函数的Test2（）只
call Test(<C-W>)
为它的身体试图...我总觉得我需要我的光标（以某种方式）下抢字的保持，那么我做了 - 因为我可以从内部的Test2
传递到测试（......）至于功能我要打电话，以下（但不完全）函数（和辅助功能）的实际例子允许我转换表格的形式，比方说，
f_k^{(j)}g
到
f_1^{(j)}g, f_2^{(j)}g, \dots, f_{n-1}^{(j)}g, f_n^{(j)}g
在我想要一个程序方面
a) type the repeated term in vim
b) visually select it
c) hit some mapping key that will call SumOrSequence(iExpression, iIndex)
d) provide "k" as a parameter
e) press enter
f) see the change made by SumOrSequence(...)
为SumOrSequence（...）的代码如下：
func SumOrSequence(iExpression, iIndex)
"need to check validity of these - maybe set a default
let default = Interrogate("do with defaults? yes [y] (2,1,n,0,\",\"), yes but specify last term [d[a-Z]], no [n]")
if default == "y"
let leftTerms = 2
let rightTerms = 1
let lastTermIndex = "n"
let firstTermIndex = 0
let operator = ","
let dotType = "\\dots"
elseif default =~ 'd[a-zA-Z]'
let leftTerms = 2
let rightTerms = 1
let lastTermIndex = default[1]
let firstTermIndex = 0
let operator = Interrogate("what separates terms? add [+], subtract [-], times [*], comma [,], ampersand [&]?")
let dotType = "\\cdots"
else "so n or anything else
let leftTerms = InterrogateNumber("how many terms before dots? ")
let rightTerms = InterrogateNumber("how many terms after dots? ")
let lastTermIndex = Interrogate("what is last term index?")
let firstTermIndex = Interrogate("what is first term index?")
let operator = Interrogate("what separates terms? add [+], subtract [-], times [*], comma [,], ampersand [&]?") "need to check only any of these provided
let dotType = "\\cdots"
endif
if operator == ","
let dotType = "\\dots"
endif
if operator == "*"
let operator = "\\times"
endif
let leftCount = 0
let oExpression = ""
while leftCount < leftTerms
if leftCount > 0
let oExpression .= operator . " "
endif
let oExpression .= ReplaceIndex(a:iExpression, a:iIndex, leftCount,1)
let leftCount += 1
endwhile
let oExpression .= operator . " " . dotType . " "
let rightCount = rightTerms-1
while rightCount > 0
"here we are going to be counting backwards from some number denoting number of terms - may need to know if we actually have a number!
echo "decrement: " . HandleDecrement(lastTermIndex, rightCount)
let oExpression .= operator . " " . ReplaceIndex(a:iExpression, a:iIndex, HandleDecrement(lastTermIndex, rightCount),1)
let rightCount -= 1
endwhile
let oExpression .= operator . " " . ReplaceIndex(a:iExpression, a:iIndex, lastTermIndex,0)
echo oExpression
endfunc
func ReplaceIndex(iExpression, iIndex, iReplacement, iInsertBraces)
"the game we play here is to search for iIndex in such form that it is not part of any other string
"We should expect this to be the case if the character to the left or right of the index is not in [A-z] (or just to the right if a greek char)
let oExpression = ""
let strEndPosition = strlen(a:iExpression) - 1
let currPosition = 0
let indexLen = strlen(a:iIndex)
while currPosition <= strEndPosition
let indexCounter = 0
let foundIndex = 1
while indexCounter < indexLen
if a:iExpression[currPosition + indexCounter] == a:iIndex[indexCounter]
if a:iExpression[currPosition + indexLen] =~ '[a-zA-Z]'
let foundIndex = 0
let indexCounter = indexLen
elseif a:iExpression[currPosition -1] =~ '[a-zA-Z]' && a:iExpression[currPosition] != "\\"
let foundIndex = 0
let indexCounter = indexLen
else
let indexCounter+=1
endif
else
let indexCounter = indexLen
let foundIndex = 0
endif
endwhile
if foundIndex == 0
let oExpression .= a:iExpression[currPosition]
let currPosition+=1
else
if a:iInsertBraces == 1
let oExpression .= "{" . a:iReplacement . "}"
else
let oExpression .= a:iReplacement
endif
let currPosition+=indexLen
endif
endwhile
echo "oExpression: " . oExpression
return oExpression
endfunc
func HandleIncrement(iExpression, iIncrement)
"and what about negative numbers for iExpression!??? not handling these yet :[
let oExpression = ""
if !(a:iExpression[0] =~ '[0-9]') || a:iExpression < 10 && strlen(a:iExpression) > 1
let oExpression = a:iExpression . " + " . a:iIncrement
else
let oExpression = a:iExpression + a:iIncrement
endif
echo oExpression
return oExpression
endfunc
func HandleDecrement(iExpression, iIncrement)
"TODO and what about negative numbers for iExpression!??? not handling these yet :[
let oExpression = ""
if !(a:iExpression[0] =~ '[0-9]') || a:iExpression < 10 && strlen(a:iExpression) > 1
let oExpression = a:iExpression . " - " . a:iIncrement
else
let oExpression = a:iExpression - a:iIncrement
endif
echo oExpression
return oExpression
endfunc
func Interrogate(iQuestion)
call inputsave()
let answer = input(a:iQuestion)
call inputrestore()
return answer
endfunc
func InterrogateNumber(iQuestion)
call inputsave()
let answer = input(a:iQuestion)
call inputrestore()
"TODO what if negative number??
if !(answer[0] =~ '[0-9]')
let answer = InterrogateNumber(a:iQuestion . " you didn't enter a numerical value ")
endif
return answer
endfunc
至于映射位，我知道看起来我没有做太多的工作，但假设我有更多的东西在挖掘我自己找到答案，任何人都可以帮忙吗？
更新。好吧，我有事情，在一个笨拙的方式排序的工作，也就是说，如果我定义以下helperfunction：
func SumOrSequenceHelper()
let oIndex = Interrogate("index variable? ")
"go to last thing visually selected (I think!), yank it (putting it in the " register), then fetch it via oParam. Then pass this off to SumOrSequence
execute "normal! gvy"
let oExpression = getreg('"')
call SumOrSequence(oExpression, oIndex)
endfunc
vnoremap <F6> :call SumOrSequenceHelper()
那么一切都很好，我可以做一个执行命令来替换我与我的选择从SumOrSequence获得（...）
将是任何改善，但为这一个解决所有意图和目的感激：]
您的解决方案比我的雨衣...感谢的！ （我无法投票这一个尚未）如此确定我怎么会使用com -range的想法，但看看com的帮助和尝试：com我看到该范围不是一个定义的命令 – HexedAgain
有一个例子一个范围命令在这里： http://www.adp-gmbh.ch/vim/user_commands.html（在底部） – zah
这是伟大的 - 我只知道到目前为止的缩写和映射;通过（简短）命令（易于输入参数（无引号））执行正是我想要的！ – HexedAgain