Ping 结果的 Java 正则表达式

Question

我正在努力从 ping 响应中读取数据，如下所示，该响应由我们应用程序中的一个 api 返回。

"PING 2001:0558:4070:0071:021D:D4FF:FEB3:1F0B (2001:558:4070:71:21d:d4ff:feb3:1f0b): 64 data bytes
72 bytes from 2001:558:4070:71:21d:d4ff:feb3:1f0b: icmp_seq=0. time=11.7 ms
72 bytes from 2001:558:4070:71:21d:d4ff:feb3:1f0b: icmp_seq=1. time=12.1 ms
<p />
2001:0558:4070:0071:021D:D4FF:FEB3:1F0B PING Statistics----
2 packets transmitted, 2 packets received, 0% packet loss
round-trip (ms)  min/avg/max/stddev = 11.7/11.9/12.1/0.32"

我需要一个正则表达式模式匹配器来获取

List<Double> times = [11.7, 12.1]
int transmitted = 2
int received = 2
double loss = 0;
double min=11.7
double avg=11.9
double max=12.1
double stddev=0.32

我一直在工作，但可以获得一个可以匹配所有内容的 RE

Pattern re = Pattern.compile( "(.*) = (.+?)/(.+?)/(.+?)/(.*)",
                Pattern.CASE_INSENSITIVE | Pattern.DOTALL);

Answer 1

试试这个。

String s = ""
    + "PING 2001:0558:4070:0071:021D:D4FF:FEB3:1F0B (2001:558:4070:71:21d:d4ff:feb3:1f0b): 64 data bytes\n"
    + "72 bytes from 2001:558:4070:71:21d:d4ff:feb3:1f0b: icmp_seq=0. time=11.7 ms\n"
    + "72 bytes from 2001:558:4070:71:21d:d4ff:feb3:1f0b: icmp_seq=1. time=12.1 ms\n"
    + "<p />\n"
    + "2001:0558:4070:0071:021D:D4FF:FEB3:1F0B PING Statistics----\n"
    + "2 packets transmitted, 2 packets received, 0% packet loss\n"
    + "round-trip (ms)  min/avg/max/stddev = 11.7/11.9/12.1/0.32\n";

Pattern PAT_TIME = Pattern.compile("time=(?<TIME>\\d+.\\d+)\\s+ms");
    Pattern PAT_REST = Pattern.compile(
        "(?<TRANSMITTED>\\d+)\\s+packets transmitted.*"
        + "(?<RECEIVED>\\d+)\\s+(packets )?received.*"
        + "(?<LOSS>\\d+)% packet loss.*" + "min/avg/max/(stddev|mdev)\\s+=\\s+"
        + "(?<MIN>\\d+\\.\\d+)/"
        + "(?<AVG>\\d+\\.\\d+)/"
        + "(?<MAX>\\d+\\.\\d+)/"
        + "(?<STDDEV>\\d+\\.\\d+)",
        Pattern.CASE_INSENSITIVE | Pattern.DOTALL);

List<Double> times = new ArrayList<>();
Matcher m1 = PAT_TIME.matcher(s);
while (m1.find())
    times.add(Double.parseDouble(m1.group("TIME")));
Matcher m2 = PAT_REST.matcher(s);
m2.find();
int transmitted = Integer.parseInt(m2.group("TRANSMITTED"));
int received = Integer.parseInt(m2.group("RECEIVED"));
double loss = Double.parseDouble(m2.group("LOSS"));
double min = Double.parseDouble(m2.group("MIN"));
double avg = Double.parseDouble(m2.group("AVG"));
double max = Double.parseDouble(m2.group("MAX"));
double stddev = Double.parseDouble(m2.group("STDDEV"));
System.out.printf("times=%s transmitted=%d received=%d loss=%f min=%f avg=%f max=%f stddev=%f%n",
    times, transmitted, received, loss, min, avg, max, stddev);

输出

times=[11.7, 12.1] transmitted=2 received=2 loss=0.00000 min=11.7000 avg=11.9000 max=12.1000 stddev=0.320000