mjhjmtu

I'm trying to match a pattern and then print out two lines after it. I'm successfully able to do it with the following code.

awk '/'202404'/c=3c&&c--' consolelog.log > testing.txt

I'm trying to replace the 202404 with something I can enter in the terminal. So far I have tried

echo "Enter peak value"
read peak
awk '/$peak/c=3c&&c--' consolelog.log > testing.txt

But it does not work.

I don't necessarily need to use awk, grep or sed or anything will do. I just need to match a user inputted pattern and print 2 lines after it.

***UPDATE*****
I was playing around with it and I got it to work. It was something I totally missed. Here's what I did.

awk 'c&&!--c;/'$peak'/c=2' consolelog.log > testing.txt

Your code:

echo "Enter peak value"
read peak
awk '/$peak/c=3c&&c--' consolelog.log > testing.txt

This does not work because strong quotes prevent the shell from performing variable expansion. This is good because awk uses a lot of syntax that weak quotes would make very difficult to use.

There are, however, ways to use environment variables within an awk script.

You can declare your own variable:

read peak
awk -vpeak="$peak" '$0 ~ peak c=3 c&&c--' consolelog.log > testing.txt

Or you can just tell awk to look at the one you just used read to fill:

read peak
export peak
awk '$0 ~ ENVIRON["peak"] c=3 c&&c--' consolelog.log > testing.txt

echo "Enter peak value"
read peak
awk '/'"$peak"'/c=3c&&c--' consolelog.log > testing.txt

Try this.

You need to close single quotes, and then you can use your Bash variable. I recommend placing it in a double-quote environment because otherwise it would cause trouble with other arguments to awk.