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- How do I get the last word in each line with bash 6 answers
如何使用bash 6答案获得每行中的最后一个单词
i have file result.txt which look like this:
我有文件result.txt,如下所示:
acb.xyz.asd
mnt.x.sdr
s.ere.43
I want to write shell script which will extract the last word and print like below:
我想编写shell脚本,它将提取最后一个单词并打印如下:
asd
sdr
43
I tried awk command but it didn't work:
我试过awk命令,但它不起作用:
awk 'NF>1{print $NF}' result.txt
I think i need to use delimiter "." and then parse the result.txt based on parse result. Plz help.
我想我需要使用分隔符“。”然后根据解析结果解析result.txt。 Plz的帮助。
3 个解决方案
#1
I usually use rev | cut | rev to extract the columns from the right.
我通常使用rev |切| rev从右侧提取列。
rev result.txt | cut -d. -f1 | rev
#2
You were indeed missing the -F option with awk
你确实错过了awk的-F选项
awk -F. 'NF>1{print $NF}' result.txt
Alternatively, you can use cut and specify both the delimiter and field that you want
或者,您可以使用剪切并指定所需的分隔符和字段
cut -d. -f3 result.txt
#3
I think sed -r -e 's/[^.]+\.[^.]+.//' will do what you want.
我认为sed -r -e's / [^。] +。[^。] +。//'会做你想要的。
#1
I usually use rev | cut | rev to extract the columns from the right.
我通常使用rev |切| rev从右侧提取列。
rev result.txt | cut -d. -f1 | rev
#2
You were indeed missing the -F option with awk
你确实错过了awk的-F选项
awk -F. 'NF>1{print $NF}' result.txt
Alternatively, you can use cut and specify both the delimiter and field that you want
或者,您可以使用剪切并指定所需的分隔符和字段
cut -d. -f3 result.txt
#3
I think sed -r -e 's/[^.]+\.[^.]+.//' will do what you want.
我认为sed -r -e's / [^。] +。[^。] +。//'会做你想要的。