[SOLVED] awk + export value to awk not passed

Issue

I am using the following awk (from my previous post - how to get list of folders that older then 30 days ) in order to print the folders that are older than XX days as the following example. Regarding this example we set 30 days as constant in awk syntax

    awk -v d="$(date -d '-30 days' +'%F')" '$1 < d' file
    2023-09-14 /user/hdfs/.sparkStaging/application_1693406696051_0046
    2023-09-14 /user/hdfs/.sparkStaging/application_1693406696051_0049
    2023-09-14 /user/hdfs/.sparkStaging/application_1693406696051_0051
    2023-09-14 /user/hdfs/.sparkStaging/application_1693406696051_0063
    2023-09-14 /user/hdfs/.sparkStaging/application_1693406696051_0064
    2023-09-14 /user/hdfs/.sparkStaging/application_1693406696051_0065
    2023-09-18 /user/hdfs/.sparkStaging/application_1693406696051_0107

Now we want to replace the constant 30 days with a variable, and we changed the awk as the following, so we want to pass the day value to awk, as day=100

awk -v day=100 -v d="$(date -d '-day days' +'%F')" '$1 < d'  file

but we do not see the expected output, and the day setting is not affected at all.

Could you please point out my error?

Solution

The command substitution runs before Awk runs. Also, the single quotes prevent the variable from being interpolated by the shell.

The correct code would look something like

day=100
awk -v d="$(date -d "-$day days" +'%F')" '$1 < d'  file

Make sure you understand the difference between a shell variable and an Awk variable. The shell doesn't know anything about Awk, and vice versa. (Much like you don't know the names or values of variables in the code which is running your computer, unless it exposes an interface to let you inspect its current state.)

Answered By - tripleee

Answer Checked By - Dawn Plyler (WPSolving Volunteer)