Sponsored Content
Top Forums Shell Programming and Scripting I can't seem to pass variables properly into a nawk statement Post 302180144 by DeCoTwc on Sunday 30th of March 2008 05:15:49 AM
Old 03-30-2008
I can't seem to pass variables properly into a nawk statement

Ok, So up front I'm going to say that I'm a very elementary scripter, and I tend to use tools I don't fully understand, but I shotgun at something until I can get it to work...that said, I can't for the life of me understand why I can't get this to go down the way I want it to.

The goal:
-to establish todays date
-to take a look at a log file (in this case the /var/adm/messages log)
-to check if the last line of the log matches todays date and if so ... (eventually it will be to dump all lines of the log with todays date to a text file that will at some point be picked up by the notification server and go out in an email)

My issue is that though I can establish the date, and create variables based upon it, I can't get nawk to accept the variables I've created.

What I have so far:

Code:
#!/usr/bin/bash
#collect todays date (date)
d=`date|awk '{print $3}'`
#collect todays date (month)
m=`date|awk '{print $2}'`
#collect errors for todays date (day and month)
e=`grep ""$d" "$m"" /var/adm/messages`
tail -1 /var/adm/messages|nawk -v m=$m -v d=$d  '{if ($1 == "$m" && $2 == "$d") print "success"}'
#if 
#        [ "$z" = "success" ]; 
#then echo "$e"
#fi

this returns nothing, no errors, just goes straight back to the prompt.

Now as a matter of checking my work I've tried replacing that last nawk statement with
Code:
tail -1 /var/adm/messages|nawk -v m=$m  '{print $m}'

I figured this would at least tell me if my variable is making it into the nawk statement, but this returns:

"nawk: illegal field $(Mar)
input record number 1
source line number 1"

the last thing that I've tried, just to make sure my syntax isn't completely foobar was
Code:
tail -1 /var/adm/messages|nawk -v m=$m -v d=$d  '{if ($1 == "Mar" && $2 == "30") print "success"}'

and that returns success...

So from what I can gather, my overall nawk statement is good, it's just not using the variables from the rest of the script.

So any input would be appreciated. Also (although, I want to learn this way) if anyone has any insight on a better method of keeping track of the messages log (or any log) please chime in.
 

10 More Discussions You Might Find Interesting

1. UNIX for Advanced & Expert Users

pf not working properly even with only "pass in all" and "pass out all" rules

i have two rules in my pf.conf file, "pass in all" and "pass out all" i was having issues with getting pf working to begin with, so i went with starting from nothing and working on up. i have an ultrasparc ultra1 200e, with an added 4-port fast ethernet sbus card, running "3.4 GENERIC#85... (4 Replies)
Discussion started by: xyyz
4 Replies

2. Shell Programming and Scripting

is it possible to pass external variable values to nawk?

Dear friends, please tell me how to pass the external variable values to the nawk command. length=`expr $len2 - $len1` i need to pass $length to following nawk command as mentioned below. nawk '{if((x=index($0,"W/X"))>0){id=substr($0,x, $length);print x;print id;}}' filename1 but I am... (1 Reply)
Discussion started by: swamymns
1 Replies

3. Shell Programming and Scripting

Shell script not processing if statement properly

Hi I am trying to create a shell script that will look for a contracthead file first and if the contract head file does not exist on day1 exit script. Now on day2 if contracthead exists or not run the script uploading files in order such as contract line then contract contact so the... (2 Replies)
Discussion started by: jonathan184
2 Replies

4. Shell Programming and Scripting

how to access values of awk/nawk variables outside the awk/nawk block?

i'm new to shell scripting and have a problem please help me in the script i have a nawk block which has a variable count nawk{ . . . count=count+1 print count } now i want to access the value of the count variable outside the awk block,like.. s=`expr count / m` (m is... (5 Replies)
Discussion started by: saniya
5 Replies

5. AIX

can't parse this nawk statement

hi all i have the following portion in an xml file: </n:AOMessage> <?xml version="1.0" encoding="UTF-8"?> <n:AOMessage xmlns:n="urn:ao:hs:update:shell" xmlns:bo="urn:ao:hs" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="urn:ao:hs:update:shell... (0 Replies)
Discussion started by: chipahoys
0 Replies

6. Shell Programming and Scripting

Can't get shell parameters to pass properly to sqlplus

Gurus, The issue I'm having is that my Shell won't accept SQL parameters properly...... Here's they way I'm running it.... applmgr@ga006hds => sh CW_MigrationDeployScript.sh apps <appspwd> <SID> '01-JAN' '31-MAR' The process just hangs not submitting the SQL job... ... (3 Replies)
Discussion started by: WhoDatWhoDer
3 Replies

7. Shell Programming and Scripting

Pass parameter to nawk from shell script

I need to parse log files using nawk, but I'm not able to pass script input argument (date) to nawk, for example: ------------ #!/bin/ksh read date nawk -F, '{if($1==date) print $4" "$5}' ------------- Is there a way to pass an argument to nawk from shell script. Many thanks... (8 Replies)
Discussion started by: samer.odeh
8 Replies

8. Shell Programming and Scripting

Bash-how to properly READ and PASTE variables.

Recently i made a script for a project at molecular dynamics but am stuck at the last step.The thing i want to do is to ask the user to input the number of particles, then replace the bolded numbers at lines 9 and 17.. code #!/bin/bash #read number of particles echo "insert the number of... (2 Replies)
Discussion started by: arisinhell
2 Replies

9. Shell Programming and Scripting

How to pass nawk variable to shell within the same script?

Hi All, I tried googling but so far no luck, can someone tell me how pass the variable value used inside the nawk command to shell. In the below script i get the value of $c (without color: Total Executed: " c ") but the printf which is outside the nawk command doesn't print the value or it... (4 Replies)
Discussion started by: Optimus81
4 Replies

10. Shell Programming and Scripting

Need to pass shell arguments into Nawk/awk

Hi, I am in critical need of help, Thanks a ton for your help. I need to know how to pass the shell argument into nawk code in AIX. so that my file gets passed into that awk script and it can execute it part. To be detail, i have more than 100 files and in those files a particular field... (6 Replies)
Discussion started by: Selva_2507
6 Replies
sulog(4)							   File Formats 							  sulog(4)

NAME
sulog - su command log file SYNOPSIS
/var/adm/sulog DESCRIPTION
The sulog file is a record of all attempts by users on the system to execute the su(1M) command. Each time su(1M) is executed, an entry is added to the sulog file. Each entry in the sulog file is a single line of the form: SU date time result port user-newuser where date The month and date su(1M) was executed. date is displayed in the form mm/dd where mm is the month number and dd is the day number in the month. time The time su(1M) was executed. time is displayed in the form HH/MM where HH is the hour number (24 hour system) and MM is the minute number. result The result of the su(1M) command. A ` + ' sign is displayed in this field if the su attempt was successful; otherwise a ` - ' sign is displayed. port The name of the terminal device from which su(1M) was executed. user The user id of the user executing the su(1M) command. newuser The user id being switched to with su(1M). EXAMPLES
Example 1 A sample sulog file. Here is a sample sulog file: SU 02/25 09:29 + console root-sys SU 02/25 09:32 + pts/3 user1-root SU 03/02 08:03 + pts/5 user1-root SU 03/03 08:19 + pts/5 user1-root SU 03/09 14:24 - pts/5 guest3-root SU 03/09 14:24 - pts/5 guest3-root SU 03/14 08:31 + pts/4 user1-root FILES
/var/adm/sulog su log file /etc/default/su contains the default location of sulog SEE ALSO
su(1M) SunOS 5.11 6 Jun 1994 sulog(4)
All times are GMT -4. The time now is 04:05 AM.
Unix & Linux Forums Content Copyright 1993-2022. All Rights Reserved.
Privacy Policy