如果满足条件，则打印上一行 [英] Print previous line if condition is met

问题描述

我想grep一个单词，然后在该行中找到第二列，并检查它是否大于一个值.是的，我要打印上一行.

I would like to grep a word and then find the second column in the line and check if it is bigger than a value. Is yes, I want to print the previous line.

例如:

输入文件

AAAAAAAAAAAAA
BB  2
CCCCCCCCCCCCC
BB 0.1

输出

AAAAAAAAAAAAA

现在，我要搜索BB，并且如果该行中的第二列(2或0.1)大于1，我想打印前一行.

Now, I want to search for BB and if the second column (2 or 0.1) in that line is bigger than 1, I want to print the previous line.

有人可以帮我grep和awk吗?谢谢.也欢迎任何其他建议.谢谢.

Can somebody help me with grep and awk? Thanks. Any other suggestions are also welcome. Thanks.

推荐答案

这可以是一种方法:

$ awk '$1=="BB" && $2>1 {print f} {f=$1}' file
AAAAAAAAAAAAA

说明

• $1=="BB" && $2>1 {print f}如果第一个字段恰好是BB并且第二个字段大于1，则打印f，即存储的值.
• {f=$1}将当前行存储在f中，以便在读取下一行时可以访问.

Explanation

• $1=="BB" && $2>1 {print f} if the 1st field is exactly BB and 2nd field is bigger than 1, then print f, a stored value.
• {f=$1} store the current line in f, so that it is accessible when reading the next line.

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