读取每一行并打印成功或失败消息

Question

TBS.log我有一个这样的文件

SYSAUX      70.12
SYSTEM      81.74
UNDOTBS1    5.66
UNDOTBS2    1.93
UNDOTBS3    1.79
USERS       .16

我需要得到如下输出。如果第二列大于 70，则应打印成功消息，否则应打印失败消息例如

SUCCESS: SYSAUX > 70%
FAILURE: UNDOTBS1 < 70%

它应该读取每一行并给出成功或失败消息

Answer 1

awk '{
    if ($2+0 > 70) 
      print "SUCCESS:", $1, "> 70%"; 
    else 
      print "FAILURE:", $1, "<= 70%";
}' TBS.log 
SUCCESS: SYSAUX > 70%
SUCCESS: SYSTEM > 70%
FAILURE: UNDOTBS1 <= 70%
FAILURE: UNDOTBS2 <= 70%
FAILURE: UNDOTBS3 <= 70%
FAILURE: USERS <= 70%