Bash - 使用循环打印具有特定列的所有行

Question

I've stucked on this problem for few hours, so I decided to ask to this community.

I have a log file like this:

gzz kre 1
mnt ttt 1
ddr ppp 2
ret typ 2
epo sst 1
plt ewr 3

I want to divide this file in three different files, each one having the same value in the third column.

I am able to do this statically, using awk:

awk '$3 == 1'  dataTX.txt > dataTX_2_$i.txt

Nevertheless, I can't do this using a loop. I'm trying with this:

for i in `seq 1 3`;
do
    awk '$3 == $i'  dataTX.txt > dataTX_2_$i.txt

done

But nothing appears in the three output files.

Can anyone help me?

Thanks a lot :-)

Answer 1

你可以用单个awk中的所有内容：

awk '{print > ("dataTX_2_" $3 ".txt")}' dataTX.txt

Answer 2

There is a better answer, see @anubhava's.

Approach

Your approach (and my original answer, below) have several problems:

• Reading the input file several times instead of once
• Hardcoding of the values instead of picking up dynamically from the input

@anubhava's solution takes care of these, by redirecting the output inside a single awk process, making a single pass over the input, and dynamically picking up the values to use for the output filenames. As an added bonus, there are no more conditions necessary.

Original answer, band-aiding the wrong approach

You need to use double-quotes to embed a shell variable, and then in that case escape the \\$ in $3 , like this:

for i in `seq 1 3`;
do
    awk "\$3 == $i"  dataTX.txt > dataTX_2_$i.txt    
done

Btw, avoid seq if possible. This will do the job just as well, and more portable:

for i in {1..3};
do
    awk "\$3 == $i"  dataTX.txt > dataTX_2_$i.txt    
done

Instead of messing with the quoting, another option is to inject a value into an Awk variable with the -v flag:

for i in {1..3};
do
    awk -v i=$i '$3 == i'  dataTX.txt > dataTX_2_$i.txt    
done