bash awk seems to get numbers wrong

Asked May 11 '18 at 18:01

Active May 11 '18 at 18:01

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I have a data file fake_data.csv that stores the following data:

ts, value
1527392000, 1
1535392000, 10
1524392000, 200

I ran this command and it doesn't print out anything:

value=$(date +%s) | awk '{ if ($1>=$value) {print}}'  ./fake_data.csv

echo $value:
1526061574

So I think I should get two rows because the 1527392000>$value, and 1535392000>$value.

What went wrong here? Is it because the integers overflows? Does bash support long int type?

asked May 11 '18 at 18:01

daydayup

1

awk is not included with bash, part of bash, etc. It's just awk, not "bash awk". – Charles Duffy May 11 '18 at 18:02
That said, why do you think awk can see your shell variable named `value`? Awk can't see shell variables any more than, say, a Python script can see variables set inside a Java program that started it. – Charles Duffy May 11 '18 at 18:03
`awk -v value="$(date +%s)" '$1 >= value { print }'` – Charles Duffy May 11 '18 at 18:05
Got you! Thank you very much!! – daydayup May 11 '18 at 18:06

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