[SOLVED] egrep move search results all to one line

oliveoyl · 09-23-2016, 11:47 AM

Output of command I'm trying to egrep on:

NodeName=node101 Arch=x86_64 CoresPerSocket=4
CPUAlloc=8 CPUErr=0 CPUTot=8 CPULoad=2.27 Features=core8,mem16gb,gig,harpertown
Gres=(null)
NodeAddr=10.0.1.10 NodeHostName=c0101 Version=15.08
OS=Linux RealMemory=16000 AllocMem=0 FreeMem=2792 Sockets=2 Boards=1
State=ALLOCATED ThreadsPerCore=1 TmpDisk=16000 Weight=1 Owner=N/A
BootTime=2016-05-12T13:52:35 SlurmdStartTime=2016-05-12T13:53:18
CapWatts=n/a
CurrentWatts=0 LowestJoules=0 ConsumedJoules=0
ExtSensorsJoules=n/s ExtSensorsWatts=0 ExtSensorsTemp=n/s

NodeName=node102 Arch=x86_64 CoresPerSocket=4
CPUAlloc=8 CPUErr=0 CPUTot=8 CPULoad=2.14 Features=core8,mem16gb,gig,harpertown
Gres=(null)
NodeAddr=10.0.1.11 NodeHostName=c0102 Version=15.08
OS=Linux RealMemory=16000 AllocMem=0 FreeMem=4779 Sockets=2 Boards=1
State=ALLOCATED ThreadsPerCore=1 TmpDisk=16000 Weight=1 Owner=N/A
BootTime=2016-05-12T13:53:00 SlurmdStartTime=2016-05-12T13:53:42
CapWatts=n/a
CurrentWatts=0 LowestJoules=0 ConsumedJoules=0
ExtSensorsJoules=n/s ExtSensorsWatts=0 ExtSensorsTemp=n/s
...

I found a nifty awk/sed command to get this:
# # mycommand | egrep 'NodeName|State' | sed 'N;s/\n/ /'
OR
# mycommand | egrep 'NodeName|State' | awk 'NR%2{printf "%s ",$0;next;}1'
NodeName=node101 Arch=x86_64 CoresPerSocket=4 State=ALLOCATED ThreadsPerCore=1 TmpDisk=16000 Weight=1 Owner=N/A
NodeName=node102 Arch=x86_64 CoresPerSocket=4 State=ALLOCATED ThreadsPerCore=1 TmpDisk=16000 Weight=1 Owner=N/A
...

What I want to get is just the NodeName and State. What is the proper way to go about getting this result?

NodeName=node101 State=ALLOCATED
NodeName=node102 State=ALLOCATED

pan64 · 09-23-2016, 11:54 AM

actually I used to suggest: do not use both grep and awk, because one of them should be sufficient to solve (such kind of issues).
In your case I would prefer awk, and I would try something like this:

Code:

awk -F'[ =]+' ' /NodeName/ { var=$1 }
                /State/ { print "var=" var "State=" $1 }'

It is not tested and obviously not a complete solution.

oliveoyl · 09-23-2016, 12:25 PM

Quote:

Originally Posted by pan64

actually I used to suggest: do not use both grep and awk, because one of them should be sufficient to solve (such kind of issues).
In your case I would prefer awk, and I would try something like this:

Code:

awk -F'[ =]+' ' /NodeName/ { var=$1 }
                /State/ { print "var=" var "State=" $1 }'

It is not tested and obviously not a complete solution.

Can you please explain that field separtor value?

pan64 · 09-23-2016, 12:30 PM

that is a regexp, it can be any number of = and/or space.
That will make something like this:

Code:

CPUAlloc=8 CPUErr=0 CPUTot=8 CPULoad=2.14 Features=core8,mem16gb,gig,harpertown
   $1    $2  $3   $4  $5   $6  $7    $8     $9       $10

(just an example)

oliveoyl · 09-23-2016, 12:43 PM

This worked:

awk -F'[ =]+' '/NodeName/ { var=$2 } /State/ {print "NodeName=" var " State=" $3}

Trying to understand it. The var is assigning the values from /NodeName/ $2 to it?

Somewhat related but what is the equivalent of egrep 'foo|bar' in awk?

pan64 · 09-23-2016, 12:50 PM

var is a variable, named var. var=$2 is an assignment, this will save the content of the second field in the variable named var (just remember the field separator).
/pattern/ will look for the line containing that pattern.

Code:

egrep 'foo|bar'
awk '/foo|bar/'

or did I misunderstand something?

Turbocapitalist · 09-23-2016, 01:55 PM

Quote:

Originally Posted by oliveoyl

Somewhat related but what is the equivalent of egrep 'foo|bar' in awk?

It would be:

Code:

awk '/foo|bar/'

which is short for

Code:

awk '/foo|bar/ { print }'

which is, in turn, short for

Code:

awk '/foo|bar/ { print $0 }'

"awk" is mostly just a bunch of if-then statements, with the ifs and thens left out. So in pseudo-code the above might be expressed as

Code:

if ( $0 =~ /foo|bar/ ) {
     print $0;
}