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java - 如何使用正则表达式解析 ping 结果

转载 作者:太空狗 更新时间:2023-10-29 16:41:13 26 4
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我想要的是逐行解析 ping 的结果。这对我来说有点棘手,尝试了很多东西,但是好吧……我在 Android 上使用 ping。

例如:

PING google.com (173.194.35.9) 56(84) bytes of data.
64 bytes from mil01s16-in-f9.1e100.net (173.194.35.9): icmp_seq=1 ttl=52 time=33.0 ms

--- google.com ping statistics ---
1 packets transmitted, 1 received, 0% packet loss, time 0ms
rtt min/avg/max/mdev = 33.086/33.086/33.086/0.000 ms

在第一行,我想要 IP 地址,即“56(84) 字节的数据”。第二行“64 bytes”,1,52,33.0 ms等

如果直接ping一个IP,会发生一点点变化

PING 192.168.0.12 (192.168.0.12) 56(84) bytes of data.
64 bytes from 192.168.0.12: icmp_seq=1 ttl=64 time=0.134 ms

--- 192.168.0.12 ping statistics ---
1 packets transmitted, 1 received, 0% packet loss, time 0ms
rtt min/avg/max/mdev = 0.134/0.134/0.134/0.000 ms

但也应该有效!

如果我有一些解释和答案,那就太棒了!

非常感谢!

最佳答案

描述

此表达式将捕获 IP、数据字节、字节、ICMP_SEQ、ttl、时间。我找不到 etc

^PING\b # match ping
[^(]*\(([^)]*)\) # capture IP
\s([^.]*)\. # capture the bytes of data
.*?^(\d+\sbytes) # capture bytes
.*?icmp_seq=(\d+) # capture icmp_seq
.*?ttl=(\d+) # capture ttl
.*?time=(.*?ms) # capture time
.*?(\d+)\spackets\stransmitted # the rest of these lines will capture the other portions of the ping result
.*?(\d+)\sreceived
.*?(\d+%)\spacket\sloss
.*?time\s(\d+ms)
.*?=\s([^\/]*)\/([^\/]*)\/([^\/]*)\/(.*)\sms

enter image description here

例子

实例:http://www.rubular.com/r/uEDoEZwY7U

示例文本

PING google.com (173.194.35.9) 56(84) bytes of data.
64 bytes from mil01s16-in-f9.1e100.net (173.194.35.9): icmp_seq=1 ttl=52 time=33.0 ms

--- google.com ping statistics ---
1 packets transmitted, 1 received, 0% packet loss, time 0ms
rtt min/avg/max/mdev = 33.086/33.086/33.086/0.000 ms


PING 192.168.0.12 (192.168.0.12) 56(84) bytes of data.
64 bytes from 192.168.0.12: icmp_seq=1 ttl=64 time=0.134 ms

--- 192.168.0.12 ping statistics ---
1 packets transmitted, 1 received, 0% packet loss, time 0ms
rtt min/avg/max/mdev = 0.134/0.134/0.134/0.000 ms

示例代码

import java.util.regex.Pattern;
import java.util.regex.Matcher;
class Module1{
public static void main(String[] asd){
String sourcestring = "source string to match with pattern";
Pattern re = Pattern.compile("^PING\\b # match ping
[^(]*\\(([^)]*)\\) # capture IP
\\s([^.]*)\\. # capture the bytes of data
.*?^(\\d+\\sbytes) # capture bytes
.*?icmp_seq=(\\d+) # capture icmp_seq
.*?ttl=(\\d+) # capture ttl
.*?time=(.*?ms) # capture time
.*?(\\d+)\\spackets\\stransmitted
.*?(\\d+)\\sreceived
.*?(\\d+%)\\spacket\\sloss
.*?time\\s(\\d+ms)
.*?=\\s([^\\/]*)\\/([^\\/]*)\\/([^\\/]*)\\/(.*?)\\sms


",Pattern.CASE_INSENSITIVE | Pattern.MULTILINE | Pattern.DOTALL);
Matcher m = re.matcher(sourcestring);
int mIdx = 0;
while (m.find()){
for( int groupIdx = 0; groupIdx < m.groupCount()+1; groupIdx++ ){
System.out.println( "[" + mIdx + "][" + groupIdx + "] = " + m.group(groupIdx));
}
mIdx++;
}
}
}

捕获组

[0][0] = PING google.com (173.194.35.9) 56(84) bytes of data.
64 bytes from mil01s16-in-f9.1e100.net (173.194.35.9): icmp_seq=1 ttl=52 time=33.0 ms

--- google.com ping statistics ---
1 packets transmitted, 1 received, 0% packet loss, time 0ms
rtt min/avg/max/mdev = 33.086/33.086/33.086/0.000 ms
[0][2] = 173.194.35.9
[0][2] = 56(84) bytes of data
[0][3] = 64 bytes
[0][4] = 1
[0][5] = 52
[0][6] = 33.0 ms
[0][7] = 1
[0][8] = 1
[0][9] = 0%
[0][10] = 0ms
[0][11] = 33.086
[0][12] = 33.086
[0][13] = 33.086
[0][14] = 0.000


[1][0] = PING 192.168.0.12 (192.168.0.12) 56(84) bytes of data.
64 bytes from 192.168.0.12: icmp_seq=1 ttl=64 time=0.134 ms

--- 192.168.0.12 ping statistics ---
1 packets transmitted, 1 received, 0% packet loss, time 0ms
rtt min/avg/max/mdev = 0.134/0.134/0.134/0.000 ms
[1][3] = 192.168.0.12
[1][2] = 56(84) bytes of data
[1][3] = 64 bytes
[1][4] = 1
[1][5] = 64
[1][6] = 0.134 ms
[1][7] = 1
[1][8] = 1
[1][9] = 0%
[1][10] = 0ms
[1][11] = 0.134
[1][12] = 0.134
[1][13] = 0.134
[1][14] = 0.000

关于java - 如何使用正则表达式解析 ping 结果,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/17592509/

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