bash 使用 grep 一次搜索两个表达式
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Search with grep for two expressions at once
提问by sup
This is basic but I am unable to google it. Can I use on invokation of grep to do
这是基本的，但我无法谷歌它。我可以在调用 grep 时使用它吗
grep expr1 | grep expr2
so that it prints lines including both expr1 and expr2?
以便它打印包含 expr1 和 expr2 的行？
采纳答案by twm
Try this:
尝试这个：
grep 'expr1.*expr2\|expr2.*expr1'
grep 'expr1.*expr2\|expr2.*expr1'
That's a little more complicated than it needs to be if you know that "expr2" will always come after "expr1". In that case, you could simplify it to:
如果您知道“expr2”总是在“expr1”之后，那么这比实际需要的要复杂一些。在这种情况下，您可以将其简化为：
grep 'expr1.*expr2'
grep 'expr1.*expr2'
回答by rounin
回答by NPE
What you have should work.
你所拥有的应该工作。
The following is an alternative way to achieve the same effect:
以下是实现相同效果的另一种方法：
grep -E 'expr1.*expr2|expr2.*expr1'
回答by user3728877
Actually your own suggestion is nearly correct if you want to get the lines that contain both expressions. Use:
实际上，如果您想获得包含两个表达式的行，您自己的建议几乎是正确的。用：
grep expr1 somefile | grep expr2
