Question

我试图从字符串中删除某个字符，直到该字符串只有x个字符

所以我想要实现的是

c:\folder\new\stuff\all\more\hello\awesome\to.dll

成为

c:\folder\new\

一次删除一个字符，直到该字符串为3 x＆＃39; \

{或者我需要多少反斜杠，因为它每次都不一样。}

到目前为止，我已经尝试过：

function RemoveLastPart(var s: string; I : integer): string;
begin
  repeat
    delete(s, length(s),1)
  until (pos ('\',s) = I) and (s[length(s)]='\');
  result := s;
end;

和

function RemoveStr(var s: string; I : integer): string;
 begin
   while (pos ('\',s) = I)  do
    begin
   delete(s, length(s),1);
   result := s;
 end
 end;

但这不会起作用，因为你的路径仍然可以

c:\folder\new\stuff vs c:\folder\new\

我猜测我错误地使用了pos。我想如果一个字符串拥有＆＃39; \＆＃39;然后做点什么。

由于我从来不知道路径的长度或者它可能包含或不包含的文件夹，因此它有点棘手。

我也试过(pos ('\',s) < I)并将我增加1

所以，如果我希望自己成为3，我会说（pos（＆＃39; \＆＃39;，s）＆lt; 4）

编辑..............

感谢所有回复

我永远不知道它需要多少反斜杠....我有一个源文件夹，它一直在变化

然后我有一个目标文件夹，它也会一直在变化源文件夹（E.G c：\ test）和目标文件夹（E.G D：\ Search Results）

扫描源文件夹以查找文件类型..找到匹配类型后（E.G c：\ test \ somefolder \ anotherfolder.exe）我用目的地替换了源（E.G D：\ Search Results \ somefolder \ anotherfolder.exe）。

然后我使用一个函数来计算源文件夹的反斜杠和搜索结果反斜杠所以

D：\搜索结果\ =两个反斜杠（2）和 D：\搜索结果\ somefolder \ anotherfolder.exe =四个反斜杠（4）

如果我们采取4-2 + 1 = 3

这是我被困的地方......

我现在知道复制整个源目录需要多少反斜杠。

所以我需要3个反斜杠

如果我们向后计算，我们会

D：\ Search Results \ somefolder \ anotherfolder.exe D：\搜索结果\ somefolder \ anotherfolder（这是错误的，这就是为什么我添加了until (pos ('\',s) = I) and (s[length(s)]='\');）

最好是向前计数，以便我们得到

D：\ Search Results \ somefolder \ anotherfolder.exe D：\搜索结果\ somefolder \

再次感谢所有回复。我没有尝试任何建议，但我将调查所有这些，看看我是否可以做一些工作。

Answer 1

由于您只是在寻找特定字符，因此您可以使用另一种方法，只需迭代字符串中的所有字符并计算已经找到特定字符的次数。

以下是实现此目的的代码;

//This function returns desired substring as result and thus does not modiffy input 
//string
//Use this function when it is required for your InputStr to remain intact for further
//processing
function ExtractSubString(InputStr: String; SearchedChar: Char; Count: Integer): String;
var Pos: Integer;
    TimesFound: Integer;
begin
  //Set initial values to local variables
  TimesFound := 0;
  //Set the default result of the method
  //While this is not needed with string results it is still good practice
  //Setting default result is required with most other result types so you don't
  //end up with uninitialized result which can have random value stored in it.
  Result := '';
  //Iterate through all characters in a string
  //NOTE: On moobile platforms strings are Zero-based so modiffy the code acordingly
  //to start with Pos 0
  for Pos := 1 to Length(InputStr) do
  begin
    //Check if specific character matches the one we are looking for
    if InputStr[Pos] = SearchedChar then
    begin
      //Increase the counter which stores how many times have we already found
      //desired character
      TimesFound := TimesFound+1;
      //Check if we found desirecd character enough times
      if TimesFound = Count then
      begin
        //Copy desired part of the input string to result
        Result := Copy(InputStr,0,Pos);
        //Use break command to break out of the loop prematurely
        Break;
      end;
    end;
  end;
end;


//This procedure modifies existing string instead of making a new shortened string
//You need to provide existing string variable as InputStr parameter
//Use only if you need to process string only once othervise use ExtractSubstring
procedure ShortenString(var InputStr: String; SearchedChar: Char; Count: Integer);
var Pos: Integer;
    TimesFound: Integer;
begin
  //Set initial values to local variables
  TimesFound := 0;
  //Iterate through all characters in a string
  //NOTE: On moobile platforms strings are Zero-based so modiffy the code acordingly
  //to start with Pos 0
  for Pos := 1 to Length(InputStr) do
  begin
    //Check if specific character matches the one we are looking for
    if InputStr[Pos] = SearchedChar then
    begin
      //Increase the counter which stores how many times have we already found
      //desired character
      TimesFound := TimesFound+1;
      //Check if we found desirecd character enough times
      if TimesFound = Count then
      begin
        //Adjust the length of the string and thus leaving out all the characters 
        //after desired length
        SetLength(InputStr,Pos);
        //Use break command to break out of the loop prematurely
        Break;
      end;
    end;
  end;
end;

Answer 2

重复使用PosEx次数，以便在结果中出现反斜杠。当您找到 nth 后，使用SetLength()来剪切字符串，或Copy()如果要保留原始字符串并将缩短的字符串作为函数结果返回，例如：

function ChopAtNthBackslash(const s: string; n: integer):string;
var
  i, k: integer;
begin
  k := 0;
  for i := 1 to n do
  begin
    k := PosEx('\', s, k+1);
    if k < 1 then Exit('');
  end;
  Result := Copy(s, 1, k);
end;

尝试从字符串中删除某个字符，直到字符串只有x个字符

2 个答案: