read lines between search pattern


 
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# 1  
Old 12-08-2010
read lines between search pattern

I have a file split something like

01/11/2010:
No of users 100

02/11/2010:
No of users 102

03/11/2010:
No of users 99

...

I want to search the file for a particular date and then extract the following line with the date, something like

02/11/2010 No of users 102

I can grep the date or read the file but now sure how to read the line(s) before the next date?
# 2  
Old 12-08-2010
Code:
grep -A1 "02/11/2010" file

# 3  
Old 12-08-2010
Code:
grep -A 1 "02/11/2010" your_file | tr -s ":\n" " "

tyler_durden
# 4  
Old 12-08-2010
thanks for the reply, unfortunately grep -A is recognised, I'm using AIX
# 5  
Old 12-08-2010
Quote:
Originally Posted by gefa
...unfortunately grep -A is recognised, I'm using AIX
I guess you mean it is not recognized.

If that's the case then you could use awk to join the pair of lines and then grep just the date of interest.

Code:
$
$ cat f45
01/11/2010:
No of users 100
 
02/11/2010:
No of users 102
 
03/11/2010:
No of users 99
 
$
$ awk '/:$/{printf $0 " "} !/:$/{print}' f45 | grep "02/11/2010"
02/11/2010: No of users 102
$
$

Or you could use Perl -

Code:
$
$ perl -lne 'BEGIN {$d="02/11/2010"} if (/$d/){printf("%s ",$_); $x=1} elsif ($x){print; $x=0}' f45
02/11/2010: No of users 102
$
$

tyler_durden
# 6  
Old 12-08-2010
MySQL

Yes both solutions work, many thanks for your help and speedy resonse.Smilie
# 7  
Old 12-08-2010
Code:
nawk '$1 ~ str {printf $0 OFS;n=1;next} n--==1 {print}' str='02/11/2010' myFile

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