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之后的预期视图:

{"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"},