I want to filter the output of arbitrary output e.g. cat or objdump to only display lines which contain "pattern".
我想过滤任意输出的输出，例如cat或objdump，只显示包含“模式”的行。
Is there a one-liner UNIX/Linux command to do this?
是否有一行UNIX/Linux命令可以做到这一点?
e.g. cat filepath | xargs grep 'pattern' -l is not working for me
例如:cat filepath | xargs grep 'pattern' -l不适合我
2 个解决方案
#1
13
cat file | grep pattern
You could also just use grep pattern file if it's a static file.
如果是静态文件，也可以使用grep模式文件。
#2
7
Better to use grep -e or egrep(this allows for extended regular expressions). Then you can do more robust things with regex:
最好使用grep -e或鹭(这允许扩展正则表达式)。然后你可以用regex做更健壮的事情:
cat my_phonebook | egrep "[0-9]{10}"
To show all 10 digit phone numbers in a file.
在文件中显示所有10位数的电话号码。
If you toss in a -o, only the numbers get returned (instead of the before and after content on the line).
如果您使用-o，则只有返回的数字(而不是前面和后面的内容)。
#1
13
cat file | grep pattern
You could also just use grep pattern file if it's a static file.
如果是静态文件，也可以使用grep模式文件。
#2
7
Better to use grep -e or egrep(this allows for extended regular expressions). Then you can do more robust things with regex:
最好使用grep -e或鹭(这允许扩展正则表达式)。然后你可以用regex做更健壮的事情:
cat my_phonebook | egrep "[0-9]{10}"
To show all 10 digit phone numbers in a file.
在文件中显示所有10位数的电话号码。
If you toss in a -o, only the numbers get returned (instead of the before and after content on the line).
如果您使用-o，则只有返回的数字(而不是前面和后面的内容)。