结肠和逗号之间的Grep数字

Question

我想要查看包含超过70％使用率的所有结果

输出示例：

{"ipaddr":"1.1.1.1","hostname":"host1.test.com","percentage":69,"dir":"/root"},
{"ipaddr":"1.1.1.1","hostname":"host1.test.com","percentage":79,"dir":"/oracle"},
{"ipaddr":"1.1.1.1","hostname":"host1.test.com","percentage":1,"dir":"/oradump"},
{"ipaddr":"1.1.1.1","hostname":"host1.test.com","percentage":90,"dir":"/archive"},

grep后的预期视图：

{"ipaddr":"1.1.1.1","hostname":"host1.test.com","percentage":79,"dir":"/oracle"},
{"ipaddr":"1.1.1.1","hostname":"host1.test.com","percentage":90,"dir":"/archive"},

Answer 1

Awk更适合这里：

$ awk -F'[:,]' '$6>70' file
{"ipaddr":"1.1.1.1","hostname":"host1.test.com","percentage":79,"dir":"/oracle"},
{"ipaddr":"1.1.1.1","hostname":"host1.test.com","percentage":90,"dir":"/archive"},

Answer 2

或者使用Perl：

$ perl -ne'print if /"percentage":([0-9]+),/ and $1 > 70'

（无需麻烦的分离计数）

Answer 3

perl -F'[:,]' -ane 'print if $F[5]>70' file

Answer 4

GNU sed

sed -n '/:[0]\?70,/d;/:[0-1]\?[7-9][0-9],/p' file