java - 在java中存储(UniqueString, String)

Question

存储一对值，其中只有一个值是唯一的。

输入:添加(字符串，字符串)

49.205.119.239, hello
14.192.212.57, yollo
49.205.119.239, tello
14.192.212.57, bella

预期输出:

49.205.119.239, hello
14.192.212.57, yollo

我尝试了下面的代码。 uniqueIp.putIfAbsent(line,actualLine); 接受所有值，甚至重复的键(行)。

public static void processLine(String actualLine) throws IOException {
  GetLocationExample obj = new GetLocationExample();
  Map<String, String> uniqueIp = new HashMap<String, String>();
  Pattern pattern = Pattern.compile("([(]100.0)[)][\\]][:]*");
  Matcher matcher = pattern.matcher(actualLine);
  String ip = null;
  String line = null ;
  if (matcher.find())
  {
       ip= "100%";
       Pattern pattern1 = Pattern.compile("\\b(?:(?:25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?)\\.){3}(?:25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?)\\b");
      Matcher matcher1 = pattern1.matcher(actualLine);

      if (matcher1.find())
      { line= (matcher1.group(0));
        uniqueIp.putIfAbsent(line, actualLine);  
      }

      else { System.out.println("Not Present"); }
  }
  else { return; }

  for (String name: uniqueIp.keySet())
  { 
      String key =name.toString();
      String value = uniqueIp.get(name).toString();  
      System.out.println(key + " " + value);  
  } 

尝试了if(!uniqueIp.containsKey(key))。没有成功

Answer 1

不必像大家建议的那样通过调用 containsKey() 进行保护，只需使用 putIfAbsent()方法:

Map<String,String> map = new HashMap<>();
map.putIfAbsent("a1", "hello");
map.putIfAbsent("b1", "yollo");
map.putIfAbsent("a1", "tello");
map.putIfAbsent("b1", "bella");