I have a CSV file with a date format like this;
11/19/2012 17:37:00,1.372,121.6
11/19/2012 17:38:00,0.743,121.6
Want to change the time stamp to seconds after 1970 so I can get the data in rrdtool.
I am a newbie to shell scripting
I have a large log file , i need to work on the part of the log file for a particular date.
Is there a way to find the first occurance of the date string and last occurance of the next day date date string and move this section to a new file.
to explain it further.
There is a file like this
-------------------
10/30/2012 some old data
10/31/2012 some dat
I'm looking for a way to have the "date" command output the date in a specific format.
I'm not familiar with the different ways to use the date command at all.
My web-scraping scripts have long enjoyed using date -d to read in human-formatted time and date stamps like "March 11, 1999" and convert to any other format I need via the -s parameter.
How might I get it to understand dates printed in other locales like 27 Kwi, 13:54 in Polish?
I have a CSV file with the below data minor changes are required.
while generating CSV file from excel sheet ,date(Format :Mon 8/28/2012)
got converted into the below format with numbers 41148,41149 so on.
Could anyone please let me know how to the convert the numbers(41148,41149 so on.) to its actual date(Format :Mon 8/28/2012).
Code:
abc.net1,BW: 1.07 M,,,,,,,,,,,,,,,,,,,,,,,
Hrly Avg (IN
How can I convert any user inputted date into yyyy/mm/dd?
For example, the user may input date in one of the following formats:
20120121
2012-01-21
01/21/2012
01/21/2012
...
But I need to convert any of the date entered by the user into yyyy/mm/dd (2012/01/2012).
How can I convert any user inputted date into yyyy/mm/dd?
For example user can input date in one of the following formats 20120121 , 2012-01-21, 01/21/2012, 01/21/2012 etc.
But I need to convert any of the date entered by user into yyyy/mm/dd (2012/01/2012). How?
This is the script I was using , but it is not working.
echo "Please enter the date: "
read X
a=$X+"%y/%m/%d"
echo $a
Hi,
When I use the `date' command on RHEL5, I can get the time shown in 24 hr format. I wonder if the time can be shown in 12 hr format. But I don't want to use the `date +FORMAT' to do that, neither `alias date='date +FORMAT''. I just want a simple `date' to show time in 12hr format.
Is there any configuration file about the 12/24 hr format?
Id like to convert a date string in the form of sun aug 19 09:03:10 EDT 2012, to unixtime timestamp using awk.
I tried
This is how each line of the file looks like, different date and time in this format
Sun Aug 19 08:33:45 EDT 2012, user1(108.6.217.236) all: test on the 17th
Through some help here, I was able to add delimiter and make it the way I want
Sun Aug 19 08:33:45 EDT 20