Question

我一直困在一个问题，我从文件中逐行读取并将这些行作为字符串存储在arrayList中。然后循环访问该列表以查找在“ring =”之后出现的int我使用的模式是“（？＆lt; = rings =）[0-9] {1}”我在下面的代码中有一个print语句告诉我这个int但它永远不会被使用意味着该方法可能找不到int。从中获取int的示例是。

//Event=ThermostatDay,time=12000
//Event=Bell,time=9000,rings=5
//Event=WaterOn,time=6000

代码是

for (int i = 0; i < fileToArray.size(); i++) {

    try {
        String friskForX = fileToArray.get(i).toString();
        Matcher xTimeSeeker = rinngerPat.matcher(friskForX);
        if (xTimeSeeker.group() != null) {
            System.out.println("will ring more then once ");
            xTimesRing = xTimeSeeker.group();
            int xTimeSeekerInt = Integer.parseInt(xTimesRing);
            System.out.println(xTimeSeekerInt);
        }
    }
    //this catches it but does nothing since some files might not have x value.
    catch (IllegalStateException e) { } 
}

Answer 1

将您的模式更改为：.*rings=(\\d+).*

String pattern=".*rings=(\\d+).*";
Pattern rinngerPat = Pattern.compile(pattern);

然后group(1)会有你的数字

for(String line : fileToArray) {
    Matcher xTimeSeeker = rinngerPat.matcher(line);
    if(xTimeSeeker.find()) {
       xTimesRing = xTimeSeeker.group(1);
       int xTimeSeekerInt = Integer.parseInt(xTimesRing);
       System.out.println(xTimeSeekerInt);
    }
}

自包含的例子：

public static void main(String[] args) {
    List<String> fileToArray = new ArrayList<>();
    fileToArray.add("Event=ThermostatDay,time=12000");
    fileToArray.add("Event=Bell,time=9000,rings=5");
    fileToArray.add("Event=WaterOn,time=6000");
    String pattern=".*rings=(\\d+).*";
    Pattern rinngerPat = Pattern.compile(pattern);
    String xTimesRing;
    for(String line : fileToArray) {
        Matcher xTimeSeeker = rinngerPat.matcher(line);
        if(xTimeSeeker.find()) {
            xTimesRing = xTimeSeeker.group(1);
            int xTimeSeekerInt = Integer.parseInt(xTimesRing);
            System.out.println(xTimeSeekerInt);
        }
    }   
}

从文件中返回一个可以在特定模式之后找到的数字

1 个答案: