[bash/awk] How to increment hour in a loop?

wasim_jd · 12-08-2011, 02:46 AM

Hi All,
Could you please help me to write theis script for every hour,it should increment on hourly basis.

Code:

awk -F  ","  '{ if (($3 ~ /^965/) && ($14 ~ /2011\/12\/07 00:[0-1][0-9]:[0-9][0-9]/)) print $17}' * |sort | uniq -c |sort -nr |head -10
awk -F  ","  '{ if (($3 ~ /^965/) && ($14 ~ /2011\/12\/07 01:[0-1][0-9]:[0-9][0-9]/)) print $17}' * |sort | uniq -c |sort -nr |head -10
awk -F  ","  '{ if (($3 ~ /^965/) && ($14 ~ /2011\/12\/07 02:[0-1][0-9]:[0-9][0-9]/)) print $17}' * |sort | uniq -c |sort -nr |head -10
...
...
...

soplease help me out...

Thanks in advance..

chrism01 · 12-08-2011, 02:57 AM

1. can you explain what you are trying to achieve
2. look into cron http://adminschoice.com/crontab-quic...Crontab%20file

wasim_jd · 12-08-2011, 03:22 AM

If you see here "($14 ~ /2011\/12\/07 02:[0-1][0-9]:[0-9][0-9]/)

Hours should get increased 1 for every hr.
I am trying to take data which matches ($3 ~ /^965/) and print by hour basis...

Thanks
Wasim.

wasim_jd · 12-08-2011, 04:00 AM

Check if this works..

Code:

#!/bin/bash
n==0
while [[ $n -le 23 ]]
do
                  awk -F  ","  '{ if (($3 ~ /^965/) && ($26 ~ /^bulksms/) && ($14 ~ /2011\/12\/07 $n:[0-1][0-9]:[0-9][0-9]/)) print $17}' * |sort | uniq -c |sort -nr |head -10
                  ((n++))
                    done

i executed but i have

./b.sh: line 4: [[: =0: syntax error: operand expected (error token is "=0")

pls do help..

Thanks..

colucix · 12-08-2011, 11:59 AM

The assignment statement in bash (like in any other programming language) requires a single equal signs. You used two of them, so that the second one has been included in the value of the variable n:

Code:

n==0

when the shell performs the variable substitution in line 4, the test statement results in

Code:

while [[ =0 -le 23 ]]

which is obviously wrong.

Regarding your original question, the problem is how to pass a shell variable to an awk program. Your code will not work as expected because you treat $n as a shell variable inside the awk code, whereas in awk $n will be interpreted as the n-th field. Furthermore take in mind that shell variables are not expanded inside single quotes. For an explanation of the correct way to pass a shell variable, please see http://www.gnu.org/s/gawk/manual/gaw...hell-Variables.

Basically you can use the -v option and assign the value $n to the awk variable n. However in this case the "n" inside the regular expression will be treated literally. In other words, you cannot use awk variables inside a regexp. What you need is a dynamic regexp, e.g.

Code:

awk -v n=$n -F,  '{ if (($3 ~ /^965/) && ($26 ~ /^bulksms/) && ($14 ~ "2011/12/07 " n ":[0-1][0-9]:[0-9][0-9]")) print $17}' file

The alternative is to correctly use a sequence of single and double quotes to write down the command:

Code:

awk -F,  '{ if (($3 ~ /^965/) && ($26 ~ /^bulksms/) && ($14 ~ /2011\/12\/07 '"$n"':[0-1][0-9]:[0-9][0-9]/)) print $17}' file

Here you close the initial single quote when you want to let bash do the variable substitution and re-open single quotes to write down the rest of the awk code. Hope this helps.