awk filter by one col larger than a value

awk -F'\t' '$36 > "2017-10-18 15:00:00" {print ;}' 13224_sample_material_encode_full | sort -r > target.txt

打印第36列大于某一时刻的整行

'$36 > "2017-10-18 15:00:00" {print ;}'

filter: $36 > "2017-10-18 15:00:00"

print whole line: print ;

awk/cut 输出指定列

awk '{for(i=10;i<=20;i++)printf $i""FS;print""}' file

FS就是分隔符

cut -d" " -f 10-20 file

百度知道

逆序

sort reverse: sort -r

按某列排序

sort -k 2 file.txt
按第二列排序
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