Question

所以如果我有这个：＆＃39;我的消息需要检查＆＃39;。

我有一个字符串：[1][3][4]。这些是我要删除的邮件的元素（例如，＆＃39;我的＆＃39;将是元素1，＆＃39;将是元素3）。

如何循环显示此消息并删除其他字符串中的元素？

示例：

String messageToFilter = "my message that needs checking";
String filter = "[1]-[3]-[4]";

for (String curElement : filter.split("-")) {
    //If I remove element [0], element [3] is then moved to [2]; so not sure what to do!
}

//So at this stage I need the messageToFilter, but with the elements from filter removed.
//In the example above this would output 'message checking'.

Answer 1

向后循环移除项目数组

String[] toFilter = filter.split("-");
for ( int i = toFilter.length() - 1; i >= 0; i-- ){
    ///remove the items
}

Answer 2

首先，您需要将{{1>}中的索引作为整数，然后删除句子中此位置的单词。

filter

输出：

消息检查

Answer 3

我建议您创建第二个字符串（或List）作为结果。

String messageToFilter = "my message that needs checking";
String filter = "[1]-[3]-[4]";
String result = null;
for (String curElement : filter.split("-")) {
   if(curElement.equalsIgnoreCase("[3]")) {
       result = curElement;
       break; // or whatever you need to do with the result.
   }
}

Answer 4

你可以尝试：

public static void main(String[] argss){

   String messageToFilter = "my message that needs checking";
   String filter = "[1]-[3]-[4]";
   ArrayList<Integer> lst=new ArrayList<Integer>();
   String[] fltr = filter.split("-");

   for (String curElement : fltr) {
       // populate lst with indexes to filter !
       lst.add(Character.getNumericValue(curElement.charAt(1)));
   }

   String result="";
   String[] msgSplitted=messageToFilter.split(" ");

   for(int i=0; i<msgSplitted.length;i++){
       if(!lst.contains(i+1))
       {
           //Checks if this index i must be flitered
           //Otherwise, add matching word to result
           result=result+" "+msgSplitted[i];
       }
   }
   System.out.print(result); //your result
}

Answer 5

向后循环并用空格替换索引，然后构造一个没有空格的新String。

    String filterString = "[1]-[3]-[4]";

    String messageToFilter = "my message that needs checking";

    String[] words = messageToFilter.split("\\s");// filterString for white space

    String[] indexes = StringUtils.split(filterString, "]|\\[|-");// filterString the numbers out

    for (int i = words.length - 1; i >= 0; i--) {

        for (int j = indexes.length - 1; j >= 0; j--) {

            if (i > j)
                break;

            if (i == j) {
                int valueAtIndex = Integer.parseInt(indexes[i]);
                words[valueAtIndex-1] = "";
                break;
            }
        }
    }
    StringBuffer bf = new StringBuffer();
    for (String word : words) {
        if(word!="")
            bf.append(word).append(" ");
    }

    System.out.println(bf.toString());

如何循环字符串并删除特定元素？

5 个答案: