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Print certain lines based on condition

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# 1  
Old 08-15-2013
Print certain lines based on condition
Hi All,
I have following listing

Code:
[172.29.38.40]
Filesystem    GB blocks      Free Used    Iused Iused Mounted on
/dev/hd2           4.00      0.31   93    63080    43 /usr

[172.29.38.41]
Filesystem    GB blocks      Free Used    Iused Iused Mounted on

[172.29.38.42]
Filesystem    GB blocks      Free Used    Iused Iused Mounted on

[172.29.38.43]
Filesystem    GB blocks      Free Used    Iused Iused Mounted on
/dev/appvglv05     14.00      0.59   96      316     1 /OUTBOUND

[172.29.38.45]
Filesystem    GB blocks      Free Used    Iused Iused Mounted on
/dev/hd2           4.00      0.29   93    63070    44 /usr
/dev/hd3          10.00      0.90   91    24577    10 /tmp


As you can see I use the script to capture those filesystem more than 90% and output to a logfile. But I just want those only server that exceeded to display out. Pls share thanks.
Last edited by Don Cragun; 08-15-2013 at 11:41 PM.. Reason: Add CODE tags.
# 2  
Old 08-16-2013
The following awk script seems to do what you want:
Code:
awk '
function printlist(     i) {
        for(i = 1; i <= c; i++)
                printf("%s\n", l[i])
        print "" 
}
/^[[]/ {if(c > 2) printlist()
        c = 0 
}
/./ {   l[++c] = $0
}
END {   if(c > 2) printlist()
}' listing

If you are running this on a Solaris/SunOS system, use /usr/xpg4/bin/awk, /usr/xpg6/bin/awk, or nawk instead of awk.

With your sample input, this produces the following output:
Code:
[172.29.38.40]
Filesystem    GB blocks      Free Used    Iused Iused Mounted on
/dev/hd2           4.00      0.31   93    63080    43 /usr

[172.29.38.43]
Filesystem    GB blocks      Free Used    Iused Iused Mounted on
/dev/appvglv05     14.00      0.59   96      316     1 /OUTBOUND

[172.29.38.45]
Filesystem    GB blocks      Free Used    Iused Iused Mounted on
/dev/hd2           4.00      0.29   93    63070    44 /usr
/dev/hd3          10.00      0.90   91    24577    10 /tmp

# 3  
Old 08-16-2013
awesome. Very powerful awk script command. Thanks for your sharing.

---------- Post updated at 02:39 AM ---------- Previous update was at 12:51 AM ----------

Thanks Don for your kind sharing. Btw, can you explain a bit on the highlighted part that I not understand.

Code:
awk '
function printlist(     i) {
        for(i = 1; i <= c; i++)
                printf("%s\n", l[i])
        print "" 
}
/^[[]/ {if(c > 2) printlist()   <<= if found first character is [ and c > 2 initialize the var c
        c = 0 
}
/./ {   l[++c] = $0              <<= can elaborate more on this?
}
END {   if(c > 2) printlist()   
}' listing

Thank You Very Much.
# 4  
Old 08-16-2013
Alternatively:
Code:
awk 'NF>10' RS= ORS='\n\n' file

# 5  
Old 08-16-2013
Hi Scrutinizer, the script you posted it print out all rows. The one Don did, yes I had tested it is working. Thanks.
# 6  
Old 08-16-2013
It works fine for me. Again it show the power of record selector.

Code:
awk 'NF>10' RS= ORS='\n\n' file
[172.29.38.40]
Filesystem    GB blocks      Free Used    Iused Iused Mounted on
/dev/hd2           4.00      0.31   93    63080    43 /usr

[172.29.38.43]
Filesystem    GB blocks      Free Used    Iused Iused Mounted on
/dev/appvglv05     14.00      0.59   96      316     1 /OUTBOUND

[172.29.38.45]
Filesystem    GB blocks      Free Used    Iused Iused Mounted on
/dev/hd2           4.00      0.29   93    63070    44 /usr
/dev/hd3          10.00      0.90   91    24577    10 /tmp

@ckwan
What is your output of:
Code:
awk '{$1=$1}1' RS= ORS='\n\n' file

# 7  
Old 08-16-2013
@ckwan
My suggestion would only work if the format is like in your sample (10 fields in the header) and if there are no additional space characters on empty lines (i.e. two consecutive newlines), which is usually the case with output from utilities. If that is not the case then a solution like Don's would work more reliably...
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