[SOLVED] Bash: Extract variable value from string

Issue

I'd need to extract the value of a variable "error" from a log file. Here's a sample line:

WARN (Periodic Recovery) IJ000906: error=15 check server.log

I'd need to capture the value of "error". Looking on similar answers, I've come up with:

echo "WARN (Periodic Recovery) IJ000906: error=15 check server.log" |  grep -P '\d+ (error=?)' -o

However it does not produce any value. Can you recommend a working solution for this case?

Solution

Using sed

$ echo "WARN (Periodic Recovery) IJ000906: error=15 check server.log" | sed 's/.*error=\([^ ]*\).*/\1/'
15

Answered By - HatLess

Answer Checked By - Senaida (WPSolving Volunteer)