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Find and print specific date with awk

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# 1  
Old 06-21-2012
Find and print specific date with awk
hi all
I would like to help me find the problem with this script to find and print to the screen a specific date of a log file that I have on my server, the date it is received as the first argument in the script $ 1
Here I show you a few lines that made ​​the idea of ​​my log file:

Code:
**** run_sql: state_cagendas.sql - 18/Jun/2012 22:59 ****
PL/SQL procedure successfully completed.
**** Completed script /bin/run_sql  - 18/Jun/2012 22:59 ****


**** run_sql: /bin/sql/unid.sql - 19/Jun/2012 22:58 ****
5 rows updated.
****  Completed script /bin/run_sql  - 19/Jun/2012 22:58 ****


**** run_sql: state_cagendas.sql- 19/Jun/2012 22:59 ****
PL/SQL procedure successfully completed.
**** Completed script /bin/run_sql  - 19/Jun/2012 22:59 ****


**** run_sql: c /tarif.sql - 20/Jun/2012 05:15 ****
PL/SQL procedure successfully completed.
**** Termino script /hfa/bin/run_sql  - 20/Jun/2012 05:15 ****


**** run_suspende: opensic- PL_SUSPENDE - 20/Jun/2012 05:30 ***
PL/SQL procedure successfully completed.
**** Termino  /hfa/bin/run_suspende  - 20/Jun/2012 05:30 ***

that is my script called "search_log"
Code:
echo "Enter date to Search (e.g. 19Jun2012): "
read search
day=`echo $search | cut -c1-2`
month=`echo $search | cut -c3-5`
year=`echo $search | cut -c6-9`

echo "BEGIN { count=0 }; /"$day"\/"$month"\/"$year"/ {print;active=\"y\";count=count+1;next}; { if (active == \"y\" && count == 2) print;active=\"n\";next}; { if (active == \"n\") exit}"  > /tmp/prog$$

awk -f /tmp/prog$$  $1
rm /tmp/prog$$

when I run

search_log / logfile

I get the following error in the command line:

awk: syntax error near line 1
awk: bailing out near line 1

please if anyone can give a help to get out of this problem would appreciate it.

thank you very much
# 2  
Old 06-21-2012
You need to see this.
This User Gave Thanks to elixir_sinari For This Post:
gilmore666
# 3  
Old 06-21-2012
Code:
echo "Enter date to Search (e.g. 19-Jun-2012): "
IFS=" /-" read DD MON YYYY

nawk -v D="$DD/$MON/$YYYY" '/\*\*\*\*/ && ($0 ~ D) { P=1 } P ; /\*\*\*\*/ && !($0 ~ D) { P=0 }' inputfile

This User Gave Thanks to Corona688 For This Post:
gilmore666
# 4  
Old 06-21-2012
thank you very much for the speed of responses both solutions were effective.

I changed awk to nawk and it worked.

I tried the second solution of Corona and I also was very useful, the only detail on the last line of the log shows the date of the day "20/Jun/2012" when I look for the log of the day "19/Jun2012" but very good solution anyway. Smilie
# 5  
Old 06-21-2012
I can fix that! Smilie

Code:
nawk -v D=" $DD/$MON/$YYYY " '/\*\*\*\*/ && ($0 ~ D) { P=1 }; /\*\*\*\*/ && !($0 ~ D) { P=0 }; P' inputfile

Also note the spaces around the date. It will match 9/xxx/yyyy without matching 19/xxx/yyyy now.
This User Gave Thanks to Corona688 For This Post:
gilmore666
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