通过regexp过滤linux bash输出的每一行
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filter each line of linux bash output by regexp
提问by T. Webster
I want to filter the output of arbitrary output e.g. cator objdumpto only display lines which contain "pattern".
我想过滤任意输出的输出，例如cat或objdump仅显示包含“模式”的行。
Is there a one-liner UNIX/Linux command to do this?
是否有单行 UNIX/Linux 命令来执行此操作？
e.g. cat filepath | xargs grep 'pattern' -lis not working for me
例如cat filepath | xargs grep 'pattern' -l不适合我
采纳答案by Explosion Pills
cat file | grep pattern
You could also just use grep pattern fileif it's a static file.
grep pattern file如果它是静态文件，您也可以使用。
回答by Paul Calabro
Better to use grep -eor egrep(this allows for extended regular expressions).
Then you can do more robust things with regex:
最好使用grep -eor egrep（这允许扩展正则表达式）。然后你可以用正则表达式做更强大的事情：
cat my_phonebook | egrep "[0-9]{10}"
To show all 10 digit phone numbers in a file.
显示文件中的所有 10 位电话号码。
If you toss in a -o, only the numbers get returned (instead of the before and after content on the line).
如果您投入 a -o，则只会返回数字（而不是行上的前后内容）。
