bash 具有字段值条件的 linux cut 命令
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linux cut command with field-value condition
提问by Ranjoy Saha
I am trying to figure out a linux shell command which will display portion of a field from a log satisfying condition of other field.
我试图找出一个 linux shell 命令，它将显示满足其他字段条件的日志中的字段的一部分。
Sample log-file:
示例日志文件：
2013-04-15;[email protected];[email protected];outgoing;24;headline
2013-04-15;[email protected];[email protected];incoming;0;chat
2013-04-15;[email protected];[email protected];outgoing;26;headline
2013-04-15;[email protected];333333;incoming;12;chat
2013-04-16;[email protected];555555;incoming;0;chat
I tried to display only the unique numbers from field 2 and 3 separated by delimiter ';' where field 5 != 0
我试图只显示字段 2 和 3 中由分隔符“;”分隔的唯一数字 其中字段 5 != 0
Sample expected output:
示例预期输出：
111111
333333
444444
采纳答案by karakfa
$ awk -F";" '!=0{print , }' fields.txt | grep -o '[0-9]\+' | sort -u
111111
333333
444444
Explanation: extract columns 2 and 3 based on column 5 value; match only the numbers; eliminate duplicates
说明：根据第5列的值提取第2列和第3列；只匹配数字；消除重复
回答by adelphus
grep -Po '\d+(?=([^;]*;)?[^;]*;[^;]+;[^0][^;]*;[a-z]+$)' logfile|sort -u
Should be self-explanatory. lol.
应该是不言自明的。哈哈。
