AWK捕获正则表达式

Question

我已经使用这个小脚本几个月了，现在已经成功了。今天，我意识到有一个输出似乎无法捕捉，屏幕空白并显示新的提示符：

user@computer ~]$ myscan ipsFile 23

user@computer ~]$

以下是代码

#!/bin/bash

sudo nmap -v -Pn  -p T:$2 -reason -i $1 | awk ' {

        if (/syn-ack/) {
                print "Yes"
                c++
        }

        else if (/no-response|reset|host-unreach/) {
                print "No"
                c++
        }
}

END { print c} '

如果我对其中一个it运行nmap，则它将返回

Starting Nmap 5.51 ( http://nmap.org ) at 2017-09-26 11:44 CDT
Initiating Parallel DNS resolution of 1 host. at 11:44
Completed Parallel DNS resolution of 1 host. at 11:44, 0.00s elapsed
Initiating Connect Scan at 11:44
Scanning 1.1.1.1 [1 port]
Completed Connect Scan at 11:44, 0.20s elapsed (1 total ports)
Nmap scan report for 1.1.1.1
Host is up, received user-set (0.20s latency).
PORT   STATE    SERVICE REASON
23/tcp filtered telnet  host-unreach

Read data files from: /usr/share/nmap
Nmap done: 1 IP address (1 host up) scanned in 0.26 seconds

如何捕获“主机未到达”部分？

Answer 1

让我们试着调试一下。执行以下命令：

nmap -v -Pn -p T:23 -reason -i ipsFile | awk '{print $0}/syn-ack/{print "Yes";c++}/no-response|reset|host-unreach/{print "No";c++}END {print c}' > out.txt

这里唯一的区别是awk脚本将$0 (即nmap调用的输出)打印到文件out.txt。尝试grep您的unreach值。

我自己试了试，发现我得到的不是host-unreach，而是net-unreach。你的情况可能也是一样。

Answer 2

您是否尝试过将stderr转换为stdout，如下所示

#!/bin/bash

sudo nmap -v -Pn  -p T:$2 -reason -i $1 2>&1 | awk ' {

    if (/syn-ack/) {
            print "Yes"
            c++
    }

    else if (/no-response|reset|host-unreach/) {
            print "No"
            c++
    }
}

END { print c} '