I have made a shell script for getting the list of mac address using awk and arp-scan command. I want to strip the mac address to only last 4 digits i.e (i want to print only the letters yy)
我使用awk和arp-scan命令创建了一个shell脚本来获取mac地址列表。我想剥去mac地址只有4位数，即（我想只打印字母yy）
ac:1e:04:0e:yy:yy
ax:8d:5c:27:yy:yy
ax:ee:fb:55:yy:yy
dx:37:42:c9:yy:yy
cx:bf:9c:a4:yy:yy
5 个解决方案
#1
2
Try cut -d: -f5-
尝试切-d：-f5-
(Options meaning: delimiter : and fields 5 and up.)
（选项含义：分隔符：和字段5及以上。）
EDIT: Or in awk, as you requested: awk -F: '{ print $5 ":" $6 }'
编辑：或者在awk中，按照您的要求：awk -F：'{print $ 5“：”$ 6}“
#2
0
Your expected output is NOT clear but based on your example output shown.
您的预期输出不明确，但基于您显示的示例输出。
your_command | awk -F":" '{print $NF}'
I am setting : as field separator so you could get any field which you want to print, so to see field numbers following may help you in same.
我正在设置：作为字段分隔符，因此您可以获得要打印的任何字段，因此查看以下字段编号可能会对您有所帮助。
your_command | awk -F":" '{for(i=1;i<=NF;i++){print "field number",i,"field value",$i}}'
#3
0
here are a few
这里有几个
line=cx:bf:9c:a4:yy:yy
echo ${line:(-5)}
line=cx:bf:9c:a4:yy:yy
echo $line | cut -d":" -f5-
#4
0
I imagine you want to strip the trailing spaces, but it isn't clear whether you want yy:yy or yyyy.
我想你想要去除尾随空格，但是你不清楚你是否想要yy：yy或yyyy。
Anyhow, there are multiple ways to it but you already are running AWK and have the MAC in $2.
无论如何，有多种方法，但你已经运行AWK并且MAC为2美元。
In the first case it would be:
在第一种情况下，它将是：
awk '{match($2,/([^:]{2}:[^:]{2}) *$/,m); print m[0]}'
yy:yy
In the second (no colon :):
在第二个（没有冒号:)：
awk 'match($2,/([^:]{2}):([^:]{2}) *$/,m); print m[1]m[2]}'
yyyy
In case you don't have match available in your AWK, you'd need to resort to gensub.
如果您的AWK中没有可用的匹配项，则需要使用gensub。
awk '{print gensub(/.*([^:]{2}:[^:]{2}) *$/,"\\1","g",$2)}'
yy:yy
or:
要么：
awk '{print gensub(/.*([^:]{2}):([^:]{2}) *$/,"\\1\\2","g",$0)}'
yyyy
Edit:
编辑：
I now realized the trailing spaces were added by anubhava in his edit; they were not present in the original question! You can then simply keep the last n characters:
我现在意识到在他的编辑中anubhava添加了尾随空格;他们没有出现在原始问题中！然后，您可以简单地保留最后n个字符：
awk '{print substr($2,13,5)}'
yy:yy
or:
要么：
awk '{print substr($2,13,2)substr($2,16,2)}'
yyyy
#5
0
Taking into account that the mac address always is 6 octets, you probably could just do something like this to get the last 2 octets:
考虑到mac地址总是6个八位字节，你可能只需要做这样的事情来得到最后2个八位字节：
awk '{print substr($0,13)}' input.txt
While testing on the fly by using arp -an I notice that the output was not always printing the mac addresses in some cases it was returning something like:
在使用arp -an进行即时测试时，我注意到输出并不总是打印mac地址，在某些情况下，它返回的内容如下：
(169.254.113.54) at (incomplete) on en4 [ethernet]
Therefore probably is better to filter the input to guarantee a mac address, this can be done by applying this regex:
因此，最好过滤输入以保证mac地址，这可以通过应用此正则表达式来完成：
^([0-9A-Fa-f]{2}[:-]){5}([0-9A-Fa-f]{2})$
Applying the regex in awk and only printing the 2 last octecs:
在awk中应用正则表达式并仅打印最后两个octec：
arp -an | awk '{if ($4 ~ /^([0-9A-Fa-f]{2}[:-]){5}([0-9A-Fa-f]{2})$/) print substr($4,13)}'
This will filter the column $4 and verify that is a valid MAC address, then it uses substr to just return the last "letters"
这将过滤$ 4列并验证它是否是有效的MAC地址，然后它使用substr返回最后的“字母”
You could also split by : and print the output in multiple ways, for example:
您还可以拆分：并以多种方式打印输出，例如：
awk '{if ($4 ~ /^([0-9A-Fa-f]{2}[:-]){5}([0-9A-Fa-f]{2})$/) split($4,a,":"); print a[5] ":" a[6]}
Notice the exp ~ /regexp/
注意exp~ / regexp /
This is true if the expression exp (taken as a string) is matched by regexp.
The following example matches, or selects, all input records with the upper-case letter `J' somewhere in the first field:
$ awk '$1 ~ /J/' inventory-shipped
-| Jan 13 25 15 115
-| Jun 31 42 75 492
-| Jul 24 34 67 436
-| Jan 21 36 64 620
So does this:
awk '{ if ($1 ~ /J/) print }' inventory-shipped
#1
2
Try cut -d: -f5-
尝试切-d：-f5-
(Options meaning: delimiter : and fields 5 and up.)
（选项含义：分隔符：和字段5及以上。）
EDIT: Or in awk, as you requested: awk -F: '{ print $5 ":" $6 }'
编辑：或者在awk中，按照您的要求：awk -F：'{print $ 5“：”$ 6}“
#2
0
Your expected output is NOT clear but based on your example output shown.
您的预期输出不明确，但基于您显示的示例输出。
your_command | awk -F":" '{print $NF}'
I am setting : as field separator so you could get any field which you want to print, so to see field numbers following may help you in same.
我正在设置：作为字段分隔符，因此您可以获得要打印的任何字段，因此查看以下字段编号可能会对您有所帮助。
your_command | awk -F":" '{for(i=1;i<=NF;i++){print "field number",i,"field value",$i}}'
#3
0
here are a few
这里有几个
line=cx:bf:9c:a4:yy:yy
echo ${line:(-5)}
line=cx:bf:9c:a4:yy:yy
echo $line | cut -d":" -f5-
#4
0
I imagine you want to strip the trailing spaces, but it isn't clear whether you want yy:yy or yyyy.
我想你想要去除尾随空格，但是你不清楚你是否想要yy：yy或yyyy。
Anyhow, there are multiple ways to it but you already are running AWK and have the MAC in $2.
无论如何，有多种方法，但你已经运行AWK并且MAC为2美元。
In the first case it would be:
在第一种情况下，它将是：
awk '{match($2,/([^:]{2}:[^:]{2}) *$/,m); print m[0]}'
yy:yy
In the second (no colon :):
在第二个（没有冒号:)：
awk 'match($2,/([^:]{2}):([^:]{2}) *$/,m); print m[1]m[2]}'
yyyy
In case you don't have match available in your AWK, you'd need to resort to gensub.
如果您的AWK中没有可用的匹配项，则需要使用gensub。
awk '{print gensub(/.*([^:]{2}:[^:]{2}) *$/,"\\1","g",$2)}'
yy:yy
or:
要么：
awk '{print gensub(/.*([^:]{2}):([^:]{2}) *$/,"\\1\\2","g",$0)}'
yyyy
Edit:
编辑：
I now realized the trailing spaces were added by anubhava in his edit; they were not present in the original question! You can then simply keep the last n characters:
我现在意识到在他的编辑中anubhava添加了尾随空格;他们没有出现在原始问题中！然后，您可以简单地保留最后n个字符：
awk '{print substr($2,13,5)}'
yy:yy
or:
要么：
awk '{print substr($2,13,2)substr($2,16,2)}'
yyyy
#5
0
Taking into account that the mac address always is 6 octets, you probably could just do something like this to get the last 2 octets:
考虑到mac地址总是6个八位字节，你可能只需要做这样的事情来得到最后2个八位字节：
awk '{print substr($0,13)}' input.txt
While testing on the fly by using arp -an I notice that the output was not always printing the mac addresses in some cases it was returning something like:
在使用arp -an进行即时测试时，我注意到输出并不总是打印mac地址，在某些情况下，它返回的内容如下：
(169.254.113.54) at (incomplete) on en4 [ethernet]
Therefore probably is better to filter the input to guarantee a mac address, this can be done by applying this regex:
因此，最好过滤输入以保证mac地址，这可以通过应用此正则表达式来完成：
^([0-9A-Fa-f]{2}[:-]){5}([0-9A-Fa-f]{2})$
Applying the regex in awk and only printing the 2 last octecs:
在awk中应用正则表达式并仅打印最后两个octec：
arp -an | awk '{if ($4 ~ /^([0-9A-Fa-f]{2}[:-]){5}([0-9A-Fa-f]{2})$/) print substr($4,13)}'
This will filter the column $4 and verify that is a valid MAC address, then it uses substr to just return the last "letters"
这将过滤$ 4列并验证它是否是有效的MAC地址，然后它使用substr返回最后的“字母”
You could also split by : and print the output in multiple ways, for example:
您还可以拆分：并以多种方式打印输出，例如：
awk '{if ($4 ~ /^([0-9A-Fa-f]{2}[:-]){5}([0-9A-Fa-f]{2})$/) split($4,a,":"); print a[5] ":" a[6]}
Notice the exp ~ /regexp/
注意exp~ / regexp /
This is true if the expression exp (taken as a string) is matched by regexp.
The following example matches, or selects, all input records with the upper-case letter `J' somewhere in the first field:
$ awk '$1 ~ /J/' inventory-shipped
-| Jan 13 25 15 115
-| Jun 31 42 75 492
-| Jul 24 34 67 436
-| Jan 21 36 64 620
So does this:
awk '{ if ($1 ~ /J/) print }' inventory-shipped