使用awk从文件中打印选定的行

Question

I have a text file with data in the following format.

1 0 0
2 512 6
3 992 12
4 1536 18
5 2016 24
6 2560 29
7 3040 35
8 3552 41
9 4064 47
10 4576 53
11 5088 59
12 5600 65
13 6080 71
14 6592 77
15 7104 83

I want to print all the lines where $1 > 1000.

awk 'BEGIN {$1 > 1000} {print "  " $1 "  "$2 "  "$3}' graph_data_tmp.txt

This doesn't seem to give the output that I am expecting.What am I doing wrong?

Answer 1

You can do this :

awk '$1>1000 {print $0}' graph_data_tmp.txt

print $0 will print all the content of the line

If you want to print the content of the line after the 1000th line/ROW, then you could do the same by replacing $1 with NR . NR represents the number of rows.

awk 'NR>1000 {print $0}' graph_data_tmp.txt

Answer 2

所有你需要的是：

awk '$1>1000' file