Display a few lines before selected word

The following command is correctly returning all the lines with warnings more than 0 from all files.

grep -i warning * | grep -v 'Warnings: 0' | more

I want to see t开发者_JAVA百科he 4 lines above the warnings line where warning is more than 0. The -B4 switch does not work for obvious reasons.

If I understood your question correctly, here is a solution:

grep -v "Warnings: 0" * | grep -B4 -i warning

How about using a little regex instead:

grep -e "Warnings: [1-9][0-9]*" -B4 * | more

The grep should look for 1 or more warnings and print the previous 4 lines.

your are unclear if the warnings message should be include in the output or not. In this version it is.

Try:

awk -F"[: ]" '($1  /Warning/ && $2> 0){a[NR]=$0;for (i=4;i>=0;i--) \
            print a[NR-i]}{a[NR]=$0}' file

HTH Chris