awk:根据另一列的值打印列

问题描述

我有一个包含六列的文件，我只想在第六列中打印值> 3的行的前两列.

I have a file with six columns, and I only want to print the first two columns of the lines that have a value >3 in the sixth column.

此语句将打印第六行> 3

This statement prints all lines where the sixth column > 3

awk '$6 > 3' file > out

此语句显示前两列:

awk '{print $1,$2}' file > out

任何人都知道如何将这两个命令组合成一个单行吗?

Anyone knows how to combine these two commands into a one-liner?

推荐答案

您就快到了，就像您所说的，将它们合并"！ .试试这个:

you are almost there,just as you said, "combine them"! . try this:

awk '$6>3{print $1,$2}' file >out

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