awk filtering one column log file

Question

I'm trying to use the following command to extract a time period forthe current day date and grep for the specific message as you can see below:

awk -v date="$(date +%Y%m%d')" '$1="date" && $3>"180000" && $3<"192000"' /app/exploit/log/FILE_send.log | grep "End success file transfer /app/reception/FILENAME.dat to HOSTX"

But I'm having trouble to filter by the date column which is the first. Here's the output of the awk command when I remove the variable from the awk command and specify a date on $1=20190405:

1 - 180050 | INFO | FILE_SEND.sh | | FILENAME.dat | 1800498307000 | End success file transfer /app/reception/FILENAME.dat to HOSTX

Here is the line that log has:

20190405 - 180050 | INFO   | FILE_SEND.sh    |          | FILENAME.dat                    | 1800498307000 | End success file transfer /app/reception/FILENAME.dat to HOSTX

I can't see the reason for the column be replaced by 1. How can I filter the first column for the current day.

Answer 1

You can use the command:

awk -F'[-|]'  -v date="20190405" '$1 == date && $2>180000 && $2<192000' file_name | grep "End success file transfer /app/reception/FILENAME.dat to HOSTX"

The output of awk command will be:

 20190405 - 180050 | INFO   | FILE_SEND.sh    |          | FILENAME.dat                    | 1800498307000 | End success file transfer /app/reception/FILENAME.dat to HOSTX

With variable date use:

awk -F'[-|]'  -v date="$(date +%Y%m%d)" '$1 == date && $2>180000 && $2<192000' filename