awk to skip lines up to and including pattern [duplicate] - awk

Question

This question already has answers here:
Printing with sed or awk a line following a matching pattern
(9 answers)
Closed 5 years ago.
I am trying to use awk to skip all lines including a specific pattern /^#CHROM/ and start processing on the line below. The awk does execute but currently returns all lines in the tab-delimited file. Thank you :).
file
##INFO=<ID=ANN,Number=1,Type=Integer,Description="My custom annotation">
##source_20170530.1=vcf-annotate(r953) -d key=INFO,ID=ANN,Number=1,Type=Integer,Description=My custom annotation -c CHROM,FROM,TO,INFO/ANN
##INFO=<ID=,Number=A,Type=Float,Description="Variant quality">
#CHROM POS ID REF ALT
chr1 948846 . T TA NA NA
chr2 948852 . T TA NA NA
chr3 948888 . T TA NA NA
awk
awk -F'\t' -v OFS="\t" 'NR>/^#CHROM/ {print $1,$2,$3,$4,$5,"ID=1"$6,"ID=2"$7}' file
desiered output
chr1 948846 . T TA ID1=NA ID2=NA
chr2 948852 . T TA ID1=NA ID2=NA
chr3 948888 . T TA ID1=NA ID2=NA

Answer 1

awk 'BEGIN{FS=OFS="\t"} f{print $1,$2,$3,$4,$5,"ID1="$6,"ID2="$7} /^#CHROM/{f=1}' file
See https://stackoverflow.com/a/17914105/1745001 for details on this and other awk search idioms. Yours is a variant of "b" on that page.

Answer 2

Use the following awk approach:
awk -v OFS="\t" '/^#CHROM/{ r=NR }r && NR>r{ $6="ID=1"$6; $7="ID=2"$7; print }' file
The output:
chr1 948846 . T TA ID=1NA ID=2NA
chr2 948852 . T TA ID=1NA ID=2NA
chr3 948888 . T TA ID=1NA ID=2NA
/^#CHROM/{ r=NR } - capturing the pattern line number
The alternative approach would look as below:
awk -v OFS="\t" '/^#CHROM/{ f=1; next }f{ $6="ID=1"$6; $7="ID=2"$7; print }' file