Dea*_*Iss 7 unix bash scripting grep cut
所以我有一个文件的文件:
puddle2_1557936:/home/rogers.williams/folderz/puddle2
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我想使用grep命令
grep puddle2_1557936
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使用cut命令(或必要时的其他命令)混合显示此部分:
/home/rogers.williams/folderz/puddle2
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到目前为止,我知道如果这样做
grep puddle2_1557936 | cut -d ":" -f1
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然后它会显示
puddle2_1557936
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那么反正的定界符切割命令是什么呢?
注意:解决方案必须从开始grep puddle2_15579636
.
Gil*_*not 24
您无需更改分隔符以显示字符串的右侧部分cut
.
命令的-f
切换cut
是由分隔符:分隔的n-TH元素:
,因此您只需键入:
grep puddle2_1557936 | cut -d ":" -f2
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如果你想要有趣的话,另一个解决方案(适应一点):
使用grep:
grep -oP 'puddle2_1557936:\K.*' <<< 'puddle2_1557936:/home/rogers.williams/folderz/puddle2'
/home/rogers.williams/folderz/puddle2
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grep -oP '(?<=puddle2_1557936:).*' <<< 'puddle2_1557936:/home/rogers.williams/folderz/puddle2'
/home/rogers.williams/folderz/puddle2
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或者使用perl:
perl -lne '/puddle2_1557936:(.*)/ and print $1' <<< 'puddle2_1557936:/home/rogers.williams/folderz/puddle2'
/home/rogers.williams/folderz/puddle2
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ruby -F: -ane '/puddle2_1557936/ and puts $F[1]' <<< 'puddle2_1557936:/home/rogers.williams/folderz/puddle2'
/home/rogers.williams/folderz/puddle2
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或者使用awk:
awk -F'puddle2_1557936:' '{print $2}' <<< 'puddle2_1557936:/home/rogers.williams/folderz/puddle2'
/home/rogers.williams/folderz/puddle2
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或者使用python:
python -c 'import sys; print(sys.argv[1].split("puddle2_1557936:")[1])' 'puddle2_1557936:/home/rogers.williams/folderz/puddle2'
/home/rogers.williams/folderz/puddle2
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或仅使用bash:
IFS=: read _ a <<< "puddle2_1557936:/home/rogers.williams/folderz/puddle2"
echo "$a"
/home/rogers.williams/folderz/puddle2
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js<<EOF
var x = 'puddle2_1557936:/home/rogers.williams/folderz/puddle2'
print(x.substr(x.indexOf(":")+1))
EOF
/home/rogers.williams/folderz/puddle2
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php -r 'preg_match("/puddle2_1557936:(.*)/", $argv[1], $m); echo "$m[1]\n";' 'puddle2_1557936:/home/rogers.williams/folderz/puddle2'
/home/rogers.williams/folderz/puddle2
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