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How to add date as a prefix for each outputline

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# 1  
Old 08-18-2010
How to add date as a prefix for each outputline
What is the best way of printing date-time at the beginning of each output line?

DTTIME=$(date +%Y%m%d%H%M%S)

Code:
myfile.log
    0.0    5.2    0.0   41.6  0.1  0.8   15.0  160.2   8  40 d0
    0.0    0.8    0.0    6.4  0.0  0.0   13.7   53.2   1   4 d2
    3.3  109.4   16.9  721.5  0.9  7.0    7.8   62.1  80 100 d3

How can I get either a static or dynamic datetime printed in front of each line? It would be nice if its a generic solution.

Code:
cat myfile.log | <sometrick>
   20100818124946  0.0    5.2    0.0   41.6  0.1  0.8   15.0  160.2   8  40 d0
   20100818124946  0.0    0.8    0.0    6.4  0.0  0.0   13.7   53.2   1   4 d2
   20100818124946  3.3  109.4   16.9  721.5  0.9  7.0    7.8   62.1  80 100 d3

# 2  
Old 08-18-2010
One way:
Code:
awk -v DTTIME=$(date +%Y%m%d%H%M%S) '{print DTTIME, $0}' myfile.log

# 3  
Old 08-18-2010
Thaks that works,,
On my solaris 8 I have to use /usr/xpg4/bin/awk instead of /bin/awk for inline variable declarations to work i.e . awk -v <var declaration>
# 4  
Old 08-18-2010
sed
Code:
sed "s/^/$(date +%Y%m%d%H%M%S) /" myfile.log

# 5  
Old 08-18-2010
Quote:
Originally Posted by frans
Code:
sed "s/^/$(date +%Y%m%d%H%M%S) /" myfile.log

The date output in sed will keep changing in seconds.

Not sure if this is the original request.
# 6  
Old 08-19-2010
Quote:
Originally Posted by rdcwayx
The date output in sed will keep changing in seconds.
Ooops !
with a fixed date in a variable
Code:
D=$(date +%Y%m%d%H%M%S); sed "s/^/$D /" myfile.log

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