How to grep a pattern having value greater than 123 in a file?


 
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# 8  
Old 05-23-2011
I have missed one pattern.

Code:
 for ((i=0;i<130;i++)); do echo " $i "|grep -v -e " [0-9] " -e " [0-9][0-9] " -e " 1[0-1][0-9] " -e " 12[0-3] "

 124
 125
 126
 127
 128
 129

# 9  
Old 09-07-2011
how would you do the same thing but only for values before the first "," in each line AND for decimal numbers ie 400.232352,ATK4875739,234,2,33,3?
# 10  
Old 09-07-2011
Quote:
Originally Posted by herot
how would you do the same thing but only for values before the first "," in each line AND for decimal numbers ie 400.232352,ATK4875739,234,2,33,3?
Code:
$
$
$ cat f18
400.232352,ATK4875739,234,2,33,3  # show, because no. from beginning to 1st comma is decimal and > 123
100.23,XYZ,0                      # don't show, because no. from beginning to 1st comma is decimal but < 123
125,ABC,0                         # don't show, because no. from beginning to 1st comma is not a decimal
12PQR,DEF,0                       # don't show, because chars from beginning to 1st comma do not represent a number
123.000001,ATK4875739,234,2,33,3  # show, because no. from beginning to 1st comma is decimal and > 123
-100.23,MNO,0                     # don't show, because no. from beginning to 1st comma is decimal but < 123
$
$
$ perl -lne '/^(.*?),.*/; print if $1 !~ /[a-zA-Z]/ and int($1) != $1 and $1 > 123' f18
400.232352,ATK4875739,234,2,33,3  # show, because no. from beginning to 1st comma is decimal and > 123
123.000001,ATK4875739,234,2,33,3  # show, because no. from beginning to 1st comma is decimal and > 123
$
$
$

tyler_durden
# 11  
Old 09-12-2011
Quote:
Originally Posted by kumaran_5555
I have missed one pattern.
Code:
 for ((i=0;i<130;i++)); do echo " $i "|grep -v -e " [0-9] " -e " [0-9][0-9] " -e " 1[0-1][0-9] " -e " 12[0-3] "

 124
 125
 126
 127
 128
 129

Thanks kumaran_5555 for this. But, why this works only when there is a space before and after the pattern?
# 12  
Old 09-12-2011
Code:
awk -F, '/,/{if($1>123){print}next}{x=$0;gsub(/[^0-9]/,"");if($0+0>123){print x} }' inputFile

Something like this?

Code:
$ cat file
rdate NTPserver login 115 /var/log
rdate NTPserver login 123.4 /var/log
rdate NTPserver login 123 /var/log
rdate NTPserver login 124 /var/log
400.232352,ATK4875739,234,2,33,3
110.232352,ATK4875739,234,2,33,3

$ awk -F, '/,/{if($1>123){print}next}{x=$0;gsub(/[^0-9]/,"");if($0+0>123){print x} }' file
rdate NTPserver login 123.4 /var/log
rdate NTPserver login 124 /var/log
400.232352,ATK4875739,234,2,33,3

--ahamed

Last edited by ahamed101; 09-12-2011 at 03:33 AM.. Reason: Some minor modifications
This User Gave Thanks to ahamed101 For This Post:
# 13  
Old 09-12-2011
hi,

if you have gawk installed, this seems to work :
Code:
echo "rdate NTPserver login 140 /var/log
rdate NTPserver login 121 /var/log" | \
   awk '{var=$0; var=gensub(/[^[:digit:]*]/,"","g",var); if ( var > 123 ) { print $0}}'

This User Gave Thanks to daPeach For This Post:
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