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Grep result from dd command

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# 1  
Old 05-06-2014
Grep result from dd command
Hi,

I am running following command in a bash script for testing IO and use grep to get throughput number, but it did not work, it displayed everything:

Code:
dd if=/dev/zero of=/dev/null bs=1G count=1 oflag=dsync | grep bytes | awk '{print $7}'

1+0 records in
1+0 records out
536870912 bytes (537 MB) copied, 8.857 s, 60.6 MB/s

How can I extract only the result "60.6 MB/s"?

Also, can I have timeout if it takes too long?

Thank you
# 2  
Old 05-06-2014
Try:
Code:
dd .... 2>&1 | grep ....

This User Gave Thanks to Scrutinizer For This Post:
hce
# 3  
Old 05-06-2014
In addition to what Scrutinizer said, $7 in the last line of output from dd is s,; not 60.6 MB/s. And you don't need both grep and awk. Try:
Code:
dd if=/dev/zero of=/dev/null bs=1G count=1 oflag=dsync 2>&1 | awk '/bytes/{print $(NF-1), $NF}'

Last edited by Don Cragun; 05-06-2014 at 03:27 AM.. Reason: Switch back to original dd parameters.
This User Gave Thanks to Don Cragun For This Post:
hce
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