使用line.contains查找精确字符串

Question

如果有些信息被认为缺乏知识，那么很抱歉，刚刚开始学习Java。

在这个工作中，用户搜索道路和城镇。问题是，当搜索类似于“Cabramatta”的结果时，结果“Cabramatta West”也会出现在结果中。

正在阅读的信息格式如下：

威廉街^3^3503^柯林斯街

威廉街^3^3503^柯林斯街

 while(fileName.hasNext())
     {
        String line =fileName.nextLine();
        {
        if(line.contains(suburbInput) && line.contains(roadInput))
           {
              String tramDetails[] = line.split("\\^");
              String crossStreet = tramDetails[0];
              String stopNumber = tramDetails[1];
              int stopNumberInt = Integer.parseInt(stopNumber);
              String trackerID = tramDetails[2];
              int trackerIDInt = Integer.parseInt(trackerID);
              String roadName = tramDetails[3];
              String suburbName = tramDetails[4];

System.out.print("'Suburb': " + suburbName + " 'Road': " + roadName + " 'Cross Street': " + crossStreet + " 'Stop': " + stopNumberInt + " 'Tracker ID': " + trackerIDInt + "\n");

我怎样才能在搜索时找到“Cabramatta”的结果，但在搜索“Cabramatta West”时也能找到结果呢？

Answer 1

在检查之前，您必须拆分您的输入，以便可以使用.equals而不是.contains。

while(fileName.hasNext())
 {
    String line =fileName.nextLine();
    String tramDetails[] = line.split("\\^");
    String suburbName = tramDetails[4];
    String roadName = tramDetails[3];

    if(suburbName.equals(suburbInput) && roadName.equals(roadInput))
    {

          String crossStreet = tramDetails[0];
          String stopNumber = tramDetails[1];
          int stopNumberInt = Integer.parseInt(stopNumber);
          String trackerID = tramDetails[2];
          int trackerIDInt = Integer.parseInt(trackerID);

          System.out.print("'Suburb': " + suburbName 
                            + " 'Road': " + roadName 
                            + " 'Cross Street': " + crossStreet   
                            + " 'Stop': " + stopNumberInt 
                            + " 'Tracker ID': " + trackerIDInt + "\n");
    }

Answer 2

先拆分，然后只使用equals方法的String。下面是一个测试代码：

        String line = "William Street^3^3503^Collins Street^Cabramatta West";
        String suburbInput = "Cabramatta";
        String roadInput = "Collins Street";

        String tramDetails[] = line.split("\\^");
        String crossStreet = tramDetails[0];
        String stopNumber = tramDetails[1];
        int stopNumberInt = Integer.parseInt(stopNumber);
        String trackerID = tramDetails[2];
        int trackerIDInt = Integer.parseInt(trackerID);
        String roadName = tramDetails[3];
        String suburbName = tramDetails[4];

        if (suburbInput.equals(suburbName) && roadInput.equals(roadName))
            System.out.print("'Suburb': " + suburbName + " 'Road': " + roadName + " 'Cross Street': " + crossStreet
                    + " 'Stop': " + stopNumberInt + " 'Tracker ID': " + trackerIDInt + "\n");

        suburbInput = "Cabramatta West";

        if (suburbInput.equals(suburbName) && roadInput.equals(roadName))
            System.out.print("'Suburb': " + suburbName + " 'Road': " + roadName + " 'Cross Street': " + crossStreet
                    + " 'Stop': " + stopNumberInt + " 'Tracker ID': " + trackerIDInt + "\n");

和产出：

'Suburb': Cabramatta West 'Road': Collins Street 'Cross Street': William Street 'Stop': 3 'Tracker ID': 3503

希望这能有所帮助！

Answer 3

您可以在郊区使用String#endsWith()而不是String#contains()：

if (line.endsWith(suburbInput) && line.contains(roadInput))

当然，这只是个创可贴。'Cabramatta‘仍将与'West’匹配。问题是，随着if (...)语句的实现，只能找到可能的匹配。您需要将行解析为要匹配的确切字段，然后显式地测试这些字段。

或者(大锤方法)，您可以实现一个正则表达式匹配器，它将在一次执行中完全匹配所有内容。