compare

I need to do a comparaison between an object and NULL. When the object is not NULL I fill it with some data. Here is the code : if (region != null) { .... } This is working but when looping and looping sometime the region object is NOT null (I can see data inside it in debug mode). In step-by-step when debugging, it doesn't go inside the IF statement... When I do a Quick Watch with these following expression : I see the (region == null) return false, AND (region != null) return false too... why and how? Update Someone point out that the object was == and != overloaded: public static bool

This question already has an answer here: How to calculate time elapsed in bash script? 18 answers I'm trying to compare a date given by a user to a date in a file, basically a text file with lots of dates and times listed. for example the user would enter a date such as 22/08/2007 and a time of 1:00, what i need the script to do is count how many dates in the text file are after the date given by the user. I’ve managed to accomplish this by converting each date in the text file to unix timestamp and then comparing the two. Is there no way of simply comparing two dates in bash? Thanks in

I have the following codes: if date in (start, end): print 'in between' else: print 'No!' date, start and end are all variables with the format of 1/1. What should I do to have it print out the right result? i tried date as 10/2, start as 3/14 and end as 11/7 and it's print 'No!', which means it's not running right. I guess have to format them to a date format and then compare them. Thanks for any help! As you are still not satisfied, I have another answer for you. Without using datetime and year. It just uses built-in tuples and comparing them: d1 = (3, 28) d2 = (3, 31) d3 = (4, 2) if d1 < d2

Hi I have two similar files (both with 3 columns). I'd like to check if these two files contains the same elements (but listed in a different orders). First of all I'd like to compare only the 1st columns file1.txt "aba" 0 0 "abc" 0 1 "abd" 1 1 "xxx" 0 0 file2.txt "xyz" 0 0 "aba" 0 0 "xxx" 0 0 "abc" 1 1 How can I do it using awk? I tried to have a look around but I've found only complicate examples. What if I want to include also the other two columns on the comparison? The output should give me the number of matching elements. To print the common elements in both files: $ awk 'NR==FNR{a[$1]

I've got a list of datetime objects, and I want to find the oldest or youngest one. Some of these dates might be in the future. from datetime import datetime datetime_list = [ datetime(2009, 10, 12, 10, 10), datetime(2010, 10, 12, 10, 10), datetime(2010, 10, 12, 10, 10), datetime(2011, 10, 12, 10, 10), #future datetime(2012, 10, 12, 10, 10), #future ] What's the most optimal way to do so? I was thinking of comparing datetime.now() to each one of those. Oldest: oldest = min(datetimes) Youngest before now: now = datetime.datetime.now(pytz.utc) youngest = max(dt for dt in datetimes if dt < now)

I want to compare two dates with time, I want all the results from tbl where date1 > date2 Select * From temp where mydate > '2009-06-29 04:00:44'; but it is just comparing dates not time. it is giving me all the result set of today's date '2009-06-29 11:08:57' '2009-06-29 11:14:35' '2009-06-29 11:12:38' '2009-06-29 11:37:48' '2009-06-29 11:52:17' '2009-06-29 12:12:50' '2009-06-29 12:13:38' '2009-06-29 12:19:24' '2009-06-29 12:27:25' '2009-06-29 12:28:49' '2009-06-29 12:35:54' '2009-06-29 12:36:54' '2009-06-29 12:49:57' '2009-06-29 12:58:04' '2009-06-29 04:13:20' '2009-06-29 04:56:19' '2009-06