如何替換dart中字串中出現的n個子字串？

我想替換字串中出現的n個子字串。

myString = "I have a mobile. I have a cat.";

我怎樣才能更換第二have的myString

uj5u.com熱心網友回復：

這樣的事情應該作業：

String replaceNthOccurrence(String input, int n, String from, String to) {
  var index = -1;
  
  while (--n >= 0) {
    index = input.indexOf(from,   index);
    if (index == -1) {
      break;
    }
  }
  
  if (index != -1) {
    var result = input.replaceFirst(from, to, index);
    return result;
  }
  
  return input;
} 

void main() {
   var myString = "I have a mobile. I have a cat.";
   var replacedString = replaceNthOccurrence(myString, 2, "have", "had");
   print(replacedString); // prints "I have a mobile. I had a cat."
}

uj5u.com熱心網友回復：

希望這個簡單的功能有幫助。如果您不想要函式，也可以提取函式內容。它只是兩行帶有一些 Dart 魔法的代碼

void main() {
  String myString = 'I have a mobile. I have a cat.';
  String searchFor='have';
  int replaceOn = 2;
  String replaceText = 'newhave';
  String result = customReplace(myString,searchFor,replaceOn,replaceText);
  print(result);
}

String customReplace(String text,String searchText, int replaceOn, String replaceText){
  Match result = searchText.allMatches(text).elementAt(replaceOn - 1);
  return text.replaceRange(result.start,result.end,replaceText);
}

uj5u.com熱心網友回復：

這將是一個更好的解決方案，因為它還會檢查回退。讓我列出所有場景：

1. 如果位置為 0，那么它將替換所有出現的。
2. 如果位置正確，那么它將在同一位置替換。
3. 如果位置錯誤，則它將發回輸入字串。
4. 如果輸入中不存在子字串，則它將發回輸入字串。

void main() {
          String input = "I have a mobile. I have a cat.";
          print(replacenth(input, 'have', 'need', 1));
        }
        
        /// Computes the nth string replace.
        String replacenth(String input, String substr, String replstr,int position) {
          
          if(input.contains(substr))
          {
            var splittedStr = input.split(substr);
            if(splittedStr.length == 0)
              return input;
              
            String finalStr = "";
            
            for(int i = 0; i < splittedStr.length; i  )
            {
              finalStr  = splittedStr[i];
              if(i == (position - 1))
                finalStr  = replstr;
              else if(i < (splittedStr.length - 1))
                finalStr  = substr;
            }
            
            return finalStr;
          }
          
          return input;
        }

uj5u.com熱心網友回復：

讓我們試試這個

void main() {
  var myString = "I have a mobile. I have a cat.I have a cat";
  print(replaceInNthOccurrence(myString, "have", "test", 1));
}

String replaceInNthOccurrence(
    String stringToChange, String searchingWord, String replacingWord, int n) {
  if(n==1){
    return stringToChange.replaceFirst(searchingWord, replacingWord);
  }
  final String separator = "#######";
  
  String splittingString =
      stringToChange.replaceAll(searchingWord, separator   searchingWord);
  var splitArray = splittingString.split(separator);
  print(splitArray);
  String result = "";
  for (int i = 0; i < splitArray.length; i  ) {
    if (i % n == 0) {
      splitArray[i] = splitArray[i].replaceAll(searchingWord, replacingWord);
    }
    result  = splitArray[i];
  }
  return result;
}

這里的正則運算式

void main() {
  var myString = "I have a mobile. I have a cat. I have a cat. I have a cat.";
   final newString =
      myString.replaceAllMapped(new RegExp(r'^(.*?(have.*?){3})have'), (match) {
    return '${match.group(1)}';
  });
  
  print(newString.replaceAll("  "," had "));
}

演示鏈接

標籤：扑 镖

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