如何在一个线性命令中使用 bash、awk、sed 或 perl 从下一行仅捕获 sdX?
echo ""dfs.datanode.data.dir" : "/rid/sdb/oo/hdfs/data,/rid/sdc/oo/hdfs/data,/rid/sdd/oo/hdfs/data,/rid/sde/oo/hdfs/data,/rid/sdf/oo/hdfs/data","
预期产出
sdb
sdc
sdd
sde
sdf
答案1
您可以使用GREP论点:
-P, --perl-regexp
Interpret the pattern as a Perl-compatible regular expression
(PCRE). This is experimental and grep -P may warn of
unimplemented features.
-o, --only-matching
Print only the matched (non-empty) parts of a matching line,
with each such part on a separate output line.
所以你的命令是:
echo ""dfs.datanode.data.dir" : "/rid/sdb/oo/hdfs/data,/rid/sdc/oo/hdfs/data,/rid/sdd/oo/hdfs/data,/rid/sde/oo/hdfs/data,/rid/sdf/oo/hdfs/data"," | grep -oP "\w*sd\w*"
sdb
sdc
sdd
sde
sdf
答案2
使用
echo ... | grep -Eo "sd[a-z]"
where-E
将模式解释为(扩展)正则表达式并-o
仅打印每行中的匹配部分。
答案3
echo '"dfs.datanode.data.dir" : "/rid/sdb/oo/hdfs/data,/rid/sdc/oo/hdfs/data,/rid/sdd/oo/hdfs/data,/rid/sde/oo/hdfs/data,/rid/sdf/oo/hdfs/data",'
"dfs.datanode.data.dir" : "/rid/sdb/oo/hdfs/data,/rid/sdc/oo/hdfs/data,/rid/sdd/oo/hdfs/data,/rid/sde/oo/hdfs/data,/rid/sdf/oo/hdfs/data",
echo '"dfs.datanode.data.dir" : "/rid/sdb/oo/hdfs/data,/rid/sdc/oo/hdfs/data,/rid/sdd/oo/hdfs/data,/rid/sde/oo/hdfs/data,/rid/sdf/oo/hdfs/data",' | grep -Po 'sd\w'
sdb
sdc
sdd
sde
sdf
答案4
gnu sed:
$ s='"dfs.datanode.data.dir" : "/rid/sdb/oo/hdfs/data,/rid/sdc/oo/hdfs/data,/rid/sdd/oo/hdfs/data,/rid/sde/oo/hdfs/data,/rid/sdf/oo/hdfs/data",'
$ echo $s| sed -E ':b;s~[^,:]+.{,3}/rid/(.+)~\1~;Te;h;s~(\w+)/.*~\1~p;g;tb;:e d'