find greater than value

Shell Programming and Scripting
1

Hi

I want to find greater than and min value.

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.

2 
$ cat outfile
5
7
10
13
18
21
$ nawk '{if($0>8){print $0;exit}}' outfile
10
$
3

i want to pass 8 value as parameter. When i am passing parameter it not printing anything.

$ln_num=8
$awk '{if($0>$ln_num){print $0;exit}}' outfile
4

what is contents of tmpsess9580.txt ?

5

if as a parameter ..

$ nawk -v var=8 '{if($0>var){print $0;exit}}' outfile
10
$
6

use like this

awk '{if($0>'$ln_num'){print $0;exit}}' outfile

or

awk -v ln_num=8 '{if($0>ln_num){print $0;exit}}' outfile
7

If the file is not sorted and still you wish to get the "min" value

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

My output is in variable and i want to read that variable and print the required number.

The above solutions are perfectly working with outfile. But I dot want to create file to store values and process that.

This is what i am trying here.

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

dategrep 8

Output of the echo $varSESSTRANS_CL is..

5
7
10
13
18
21

echo $varSESSTRANS_CL | awk '{if($1>'$varlinenum'){print $1;exit}}'

this stmt is not working or not printing anything..

9 
echo "$varSESSTRANS_CL" | awk '{if($1>'$varlinenum'){print $1;exit}}'
1 Like
10

Thank you very much.