5

libfaketime and Mac OS, LD_PRELOAD

view full story
linux-howto

http://unix.stackexchange.com – Running Mac OS 10.7.2 and trying to run libfaketime and nothing happen. I installed and ran exactly as directed only the time and date stay exactly as are set in the system? Is anyone else able to successfully run this on OSX? $ date Mon Jan 30 21:05:09 CST 2012 $ LD_PRELOAD=./libfaketime.so.1 FAKETIME="-15d" date Mon Jan 30 21:05:32 CST 2012 (HowTos)

Danial's picture
Submitted by Danial on 01/31/2012 – Made popular on 01/31/2012

Stories similar to libfaketime and Mac OS, LD_PRELOAD

date time stamps in bash 35 weeks 3 days ago

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.

I have an SQL fact table which is used for looking up a date and returning another date.

The table schema is as follows:

TABLE date_lookup ( pk_date DATE, plus1_months DATE, plus2_months DATE, plus3_months DATE );

UNIQUE INDEX on date_lookup(pk_date);

I have a load file (pipe delimited) containing dates from 01-28-2012 to 03-31-2014.

The following is an example of the load file:

01-2

Get day from date variable 34 weeks 1 day ago

Ok...this is really bumming me out because it seems like it should be simple, but for some reason, I just can't get it...I've also googled and searched these forums, and haven't found exactly what I'm looking for...

I have a date in mm/dd/yyyy format in a variable...I need to extract the abbreviated name of the day of week from that variable...so I start with this: 10/18/2012

and end up with thi

Given a date and time in a format that is not recognized by date, how can I get date to recognize the date and time?

For example:

$ date -d "09SEP2012:23:58:46" date: invalid date `09SEP2012:23:58:46' $ date -d "09SEP2012:23:58:46" --magic-option "ddMMMYYY:hh:mm:ss" Sun Sep 9 23:58:46 MDT 2012

Does --magic-option exist?

I have a CSV file with date time formatted like this:

8/2/2012 1:34:01 PM

I need to find all the occurrences (probably regex) of such date and convert it (in file) to mysql default datetime format (2012-08-02 13:34:01).

I found that

date --date="8/2/2012 1:34:01 PM" +"%F %T"

is working properly and I'm getting correct result.

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?

Convert any Date format in unix 21 weeks 4 hours ago

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).

Convert any Date format in unix 21 weeks 4 hours ago

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

I’d 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