datetime-format

问题 I have a date in this format Fri, 15 Jan 2016 15:14:10 +0800 , and I want to display time like this 2016-01-15 15:14:10 . What I tried is: $test = 'Fri, 15 Jan 2016 15:14:10 +0800'; $t = date('Y-m-d G:i:s',strtotime($test)); echo $t; But it is displaying date in this format: 2016-01-15 7:14:10 , it should be 2016-01-15 15:14:10 . How can i do this? 回答1: Use H instead: $test = 'Fri, 15 Jan 2016 15:14:10 +0800'; $t = date('Y-m-d H:i:s',strtotime($test)); echo $t; H : 24-hour format of an hour

问题 Working on a table with a mix of different strings where it's possible to derive a date. period 0 Q2 '20 Base 1 Q3 '20 Base 2 Q1 '21 Base 3 February '20 Base 4 March '20 Peak 5 Summer 22 Base 6 Winter 20 Peak 7 Summer 21 Base 8 Year 2021 9 October '21 Peak I'd like to be able to parse this into a timestamp for analysis in python. First, ideally I want to parse into 4 new columns 1) day 2) month 3) quarter 4) year. Then use these columns to make a datetime (DD-MM-YYYY). period day month

问题 I'm attempting to format a LoacalDate but I didn't find any information. I need to format to another language. The case is simply I want to get the month of the year in Spanish. I'm trying to use: Locale locale = new Locale("es", "ES"); But I don't find a SimpleDateFormat or similar to the format LocalDate. LocalDate.now().getMonth(); Can anyone help me? 回答1: Sorry, I used the older (and still more commonly used) date time classes of Java, as you spoke about SimpleDateFormat which is part of

问题 This question already has answers here : how to convert incoming Json date into java date format? (2 answers) Closed 2 years ago . I am receiving response of Date in JSON format like /Date(1521866513877+0530)/ which I want in format like this 24/03/2018 10:11:53 . 回答1: If there's no native way to parse that, try this regex pattern: \/Date\((\d+)\+(\d+)\)\/ I'm not a java guy but regex is everywhere. I looks like the date is in miliseconds so you need to divide 1521866513877 / 1000 to get the

问题 My date format of "03-01-2017 10:24:48" is getting stored in SQL as "2017-03-01 10:24:48.000" where my 'dd-mm-yyyy' format is getting converted to 'yyyy-mm-dd' format. How can I change the date format of the column to my desired format? My table already contains data. 回答1: SQL Server doesn't store a DateTime in any string format - it's stored as an 8 byte numerical (binary) value. The various settings (language, date format) only influence how the DateTime is shown to you in SQL Server

问题 I use the following code to retrieve the current date and time, then add it to table using following method. Change 2013-07-29 23:20:34.0 to 29th July 2013, 11:20 PM DateTime public Date getCurrentDateTime() { SimpleDateFormat dateFormat = new SimpleDateFormat("yyyy-MM-dd HH:mm:ss"); Date date = new Date(); System.out.println("current:" + dateFormat.format(date)); return date; } Hibernate @Temporal(javax.persistence.TemporalType.TIMESTAMP) public Date getCurrentDateTime() { return date; } 回答1

问题 So when trying to replace some legacy code using SimpleDateFormat and Date, to use java.time.DateTimeFormatter and LocalDate I ran into a problem. The two date formats are not equivalent. At this point I must say I know the two date types are not the same but the scenario I am in means I never care about the time aspect so can ignore it. public Date getDate(String value) { SimpleDateFormat dateFormat = new SimpleDateFormat("dd/MM/yyyy"); try { return dateFormat.parse(value); } catch

问题 So when trying to replace some legacy code using SimpleDateFormat and Date, to use java.time.DateTimeFormatter and LocalDate I ran into a problem. The two date formats are not equivalent. At this point I must say I know the two date types are not the same but the scenario I am in means I never care about the time aspect so can ignore it. public Date getDate(String value) { SimpleDateFormat dateFormat = new SimpleDateFormat("dd/MM/yyyy"); try { return dateFormat.parse(value); } catch

问题 I am looking at the DateTimeFormatter class and I was wondering where are the constants like "M" , "EEE" and "YY" etc are defined. More specifically there should be code like this somewhere given a ChronoField and TextStyle, it should return the DateTimeFormatter string snippet e.g. (ChronoField.MONTH_OF_YEAR, TextStyle.SHORT) should map to the string "MMM" 回答1: After some source digging, I found out that these are pre-defined in a private static final Map , called FIELD_MAP , which is a

问题 I am looking at the DateTimeFormatter class and I was wondering where are the constants like "M" , "EEE" and "YY" etc are defined. More specifically there should be code like this somewhere given a ChronoField and TextStyle, it should return the DateTimeFormatter string snippet e.g. (ChronoField.MONTH_OF_YEAR, TextStyle.SHORT) should map to the string "MMM" 回答1: After some source digging, I found out that these are pre-defined in a private static final Map , called FIELD_MAP , which is a