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[awk]: Row begins by random number and field 10 is greater than 10.00%

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# 1  
Old 03-17-2005
[awk]: Row begins by random number and field 10 is greater than 10.00%
Hello!

I wish to extract the pid where CPU is above 10%
Code:
last pid: 22621;  load averages:  4.71,  5.04,  5.13                                                                                                15:08:34
221 processes: 212 sleeping, 2 running, 1 stopped, 6 on cpu
CPU states:     % idle,     % user,     % kernel,     % iowait,     % swap
Memory: 4096M real, 1810M free, 6148M swap free

  PID USERNAME THR PRI NICE  SIZE   RES STATE   TIME    CPU COMMAND
14990 oracle     1   0    0  401M  385M cpu14 316.5H 14.79% oracle
21104 oracle     1  58    0  401M  378M sleep   3:12  4.71% oracle
...

I tried like that
Code:
top | awk '/^[0-9][0-9]*/ && $10 >= "10.00%" {print $1}'

of course it doesn't work as awk is comparing alphabetically...

I wish to compare on value, I can get rid of % sign, I only don't remember how to tell to awk to compare the value not the string.

Thanks
# 2  
Old 03-17-2005
awk can translate string to float automatic.

try:
awk '($10 > 10)'
# 3  
Old 03-17-2005
Maybe use the int() function...
Code:
| awk '(int($1) == $1) && (int($10) > 10) {print $1}'

# 4  
Old 03-18-2005
Quote:
Originally Posted by ZealeS
awk can translate string to float automatic.

try:
awk '($10 > 10)'
yeah! it works fine, thanks! (I sed first to erase '%' signs)
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