find greater than value


 
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# 1  
Old 10-12-2011
find greater than value

Hi

I want to find greater than and min value.

Code:
dategrep()
{
 varlinenum=$1
 varSESSTRANS_CL="$(egrep -n "<\/SESSTRANSFORMATIONINST>" tmpsess9580.txt | cut -d":" -f1)"
 echo $varSESSTRANS_CL
}

dategrep 8

Output of the above command is:

Quote:
5
7
10
13
18
21
I want to find out greater than 8 and min of the above result.
Answer should be : 10

Can anyone help me on above requirement.

Thanks,
Mallik.
# 2  
Old 10-12-2011
Code:
$ cat outfile
5
7
10
13
18
21
$ nawk '{if($0>8){print $0;exit}}' outfile
10
$

# 3  
Old 10-12-2011
i want to pass 8 value as parameter. When i am passing parameter it not printing anything.

Code:
$ln_num=8
$awk '{if($0>$ln_num){print $0;exit}}' outfile

# 4  
Old 10-12-2011
Quote:
Originally Posted by tmalik79
Hi

I want to find greater than and min value.

Code:
dategrep()
{
 varlinenum=$1
 varSESSTRANS_CL="$(egrep -n "<\/SESSTRANSFORMATIONINST>" tmpsess9580.txt | cut -d":" -f1)"
 echo $varSESSTRANS_CL
}

dategrep 8

Output of the above command is:



I want to find out greater than 8 and min of the above result.
Answer should be : 10

Can anyone help me on above requirement.

Thanks,
Mallik.
what is contents of tmpsess9580.txt ?
# 5  
Old 10-12-2011
if as a parameter ..
Code:
$ nawk -v var=8 '{if($0>var){print $0;exit}}' outfile
10
$

# 6  
Old 10-12-2011
Quote:
Originally Posted by tmalik79
i want to pass 8 value as parameter. When i am passing parameter it not printing anything.

Code:
$ln_num=8
$awk '{if($0>$ln_num){print $0;exit}}' outfile

use like this
Code:
awk '{if($0>'$ln_num'){print $0;exit}}' outfile

or
Code:
awk -v ln_num=8 '{if($0>ln_num){print $0;exit}}' outfile

# 7  
Old 10-12-2011
If the file is not sorted and still you wish to get the "min" value

Code:
 
TESTBOX>awk -v m=9999999999999 '$0>8 && $0<m { m=$0 } END { print m}'  outfile

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