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grep return more than one line

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# 1  
Old 09-16-2010
grep return more than one line
Hi,

I have a lot of log files which contain lines in the format of (date info) or (info), and when I use grep to search for "date" I was able to get the (date info) line, but some info lines are a separate line after the (date info) line... for example like:

(date info info info)
(date info info)
(info info)
(date info) (or end of file)
I want grep to return the date info line and any further info lines containing the date parameter i enter.

any help would be greatly appreciated, thanks.
# 2  
Old 09-16-2010
Post real sample data.
# 3  
Old 09-16-2010
2010-07-19 10:29:00,813 ERROR [WebContainer : 10]...
2010-07-19 10:29:15,813 ERROR [WebContainer : 10].........
2010-07-19 10:29:30,813 ERROR [WebContainer : 10].......
at com..*****......
at com..*****.........

so if i look for the "2010-07-19 10:29:30"

I want the last 3 lines of this log.

thanks
# 4  
Old 09-16-2010
Try:
Code:
awk -vs="2010-07-19 10:29:30" '$0~"^"s{p=1}!($0~"^"s) && /^2010/{p=0}p' logfile

This User Gave Thanks to bartus11 For This Post:
a27wang
# 5  
Old 09-16-2010
wow that worked very well!! thanks, just a little more to ask though: the returned format was more of a paragraph instead of lines, for example:
i have several lines with the same date and time, they got returned in a big chunk, is there anyway to separate them by lines like originally shown in the log files?
# 6  
Old 09-16-2010
Like this?
Code:
awk -vs="2010-07-19" '$0~"^"s{p=1}!($0~"^"s) && /^2010/{if(p==1){print "\n"}p=0}p' logfile

# 7  
Old 09-16-2010
I don't think it worked, it returned in the same format, thanks
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