unix-timestamp

问题 How Can I get the date time in unix time as byte array which should fill 4 bytes space in Java? Something like that: byte[] productionDate = new byte[] { (byte) 0xC8, (byte) 0x34, (byte) 0x94, 0x54 }; 回答1: First: Unix time is a number of seconds since 01-01-1970 00:00:00 UTC. Java's System.currentTimeMillis() returns milliseconds since 01-01-1970 00:00:00 UTC. So you will have to divide by 1000 to get Unix time: int unixTime = (int)(System.currentTimeMillis() / 1000); Then you'll have to get

问题 I require the current date and time to be logged for my application. I have written the code in C. I have attached the code #include <stdio.h> #include <time.h> int main() { time_t t; while(1) { time(&t); printf("Today's date and time : %s",ctime(&t)); } } The output is Today's date and time : Wed Dec 31 23:59:59 1969 Today's date and time : Wed Dec 31 23:59:59 1969 Today's date and time : Wed Dec 31 23:59:59 1969 Today's date and time : Wed Dec 31 23:59:59 1969 The time is not getting

问题 I need to insert data to a given external table which should be partitioned by the inserted date. My question is how is Hive handling the timestamp generation? When I select a timestamp for all inserted records like this: WITH delta_insert AS ( SELECT trg.*, from_unixtime(unix_timestamp()) AS generic_timestamp FROM target_table trg ) SELECT * FROM delta_insert; Will the timestamp always be identical for all records, even if the query takes a lot of time to un? Or should I alternatively only

问题 I need to insert data to a given external table which should be partitioned by the inserted date. My question is how is Hive handling the timestamp generation? When I select a timestamp for all inserted records like this: WITH delta_insert AS ( SELECT trg.*, from_unixtime(unix_timestamp()) AS generic_timestamp FROM target_table trg ) SELECT * FROM delta_insert; Will the timestamp always be identical for all records, even if the query takes a lot of time to un? Or should I alternatively only

问题 I need to insert data to a given external table which should be partitioned by the inserted date. My question is how is Hive handling the timestamp generation? When I select a timestamp for all inserted records like this: WITH delta_insert AS ( SELECT trg.*, from_unixtime(unix_timestamp()) AS generic_timestamp FROM target_table trg ) SELECT * FROM delta_insert; Will the timestamp always be identical for all records, even if the query takes a lot of time to un? Or should I alternatively only

问题 How to get the date of Monday and Sunday in a week for a date? This gives date for 'last' monday: date -dlast-monday +%Y%m%d I want to pass a date as parameter to find the Monday and Sunday for that week. Basically, I want to get Sunday and Monday for a week, for ANY date, NOT only for last monday . 回答1: Try this: export day=2013-10-01 date -d "$day -$(date -d $day +%w) days" This will always print the Sunday before the given date (or the date itself). date -d "$day -$(date -d $day +%u) days"

问题 I want to change Datetime (2014-12-23 00:00:00) into unixtime. I tried it with the Datetime function but it didn´t work. I got the Datetime stamps in an array. Zeit =np.array(Jahresgang1.ix[ :,'Zeitstempel']) t = pd.to_datetime(Zeit, unit='s') unixtime = pd.DataFrame(t) print unixtime Thanks a lot 回答1: I think you can subtract the date 1970-1-1 to create a timedelta and then access the attribute total_seconds: In [130]: s = pd.Series(pd.datetime(2012,1,1)) s Out[130]: 0 2012-01-01 dtype:

问题 This question already has answers here : Convert a date format in epoch (5 answers) How to convert timestamp string to epoch time? (4 answers) Closed 2 years ago . I am pulling data from an API, and one of the pieces of data I get is a date and time. The date and time is returned in a string format and I need to get that in to a Unix Timestamp for my mySQL database. The String looks like: "2018-10-23 18:00:00" Does anyone know how I would go about getting that from a String to a Unix

问题 Hi I have a column with number datatype the data like 1310112000 this is a date, but I don't know how to make it in an understandable format: ex: 10-mar-2013 12:00:00 pm Can any one please help me. 回答1: That is EPOCH time: number of seconds since Epoch(1970-01-01). Use this: SELECT CAST(DATE '1970-01-01' + ( 1 / 24 / 60 / 60 ) * '1310112003' AS TIMESTAMP) FROM DUAL; Result: 08-JUL-11 08.00.03.000000000 AM 回答2: Please try select from_unixtime(floor(EPOCH_TIMESTAMP/1000)) from table; This will

问题 I have timestamps looking like this: 2015-03-21T11:08:14.859831 2015-03-21T11:07:22.956087 I read a Wiki article on ISO 8601, but did not get the meaning of the last 6 digits here. I tried getting it down to milliseconds using "yyyy-MM-dd'T'HH:mm:ss.sss" or "yyyy-MM-dd'T'HH:mm:ss.ssssss" . Is it just more precise than milliseconds - up to microseconds? 回答1: Is it just more precise than milliseconds? Yes, it's microseconds in this case. ISO-8601 doesn't actually specify a maximum precision. It