Asked  9 Months ago    Answers:  5   Viewed   57 times

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?



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 with leading zeros 00 through 23

G should be the same, but without leading zeroes though. I suspect that your PHP is set to a different timezone than +0800. Can you confirm your timezone (date_default_timezone_get())?


OP confirmed that his timezone was set to UTC, in which case it maskes perfect sense that it shows 7 in the morning, as date uses PHPs default timezone.

If you want to "inherit" the Timezone, while getting more flexibility, you should switch to DateTime:

echo (new DateTime($test))->format("Y-m-d H:i:s");
Wednesday, March 31, 2021
answered 9 Months ago

I ended up with this script:

set hour=%time:~0,2%
if "%hour:~0,1%" == " " set hour=0%hour:~1,1%
echo hour=%hour%
set min=%time:~3,2%
if "%min:~0,1%" == " " set min=0%min:~1,1%
echo min=%min%
set secs=%time:~6,2%
if "%secs:~0,1%" == " " set secs=0%secs:~1,1%
echo secs=%secs%

set year=%date:~-4%
echo year=%year%

:: On WIN2008R2 e.g. I needed to make your 'set month=%date:~3,2%' like below ::otherwise 00 appears for MONTH

set month=%date:~4,2%
if "%month:~0,1%" == " " set month=0%month:~1,1%
echo month=%month%
set day=%date:~0,2%
if "%day:~0,1%" == " " set day=0%day:~1,1%
echo day=%day%

set datetimef=%year%%month%%day%_%hour%%min%%secs%

echo datetimef=%datetimef%
Saturday, June 5, 2021
answered 6 Months ago

I've tested this with JDK 1.8.0_131 for Mac OS X and JDK 1.8.0111 for Windows (both worked).

I've created a DateTimeFormatter with optional sections (delimited by []), to parse both cases (MM/dd/yyyy and yyyy-MM-dd'T'HH:mm:ss).

The same formatter worked for your case (LocalDate), but there are some considerations below.

// parse both formats (use optional section, delimited by [])
DateTimeFormatter parser = DateTimeFormatter.ofPattern("[MM/dd/yyyy][yyyy-MM-dd'T'HH:mm:ss]");

// parse MM/dd/yyyy
LocalDate d1 = LocalDate.parse("10/16/2016", parser);
// parse yyyy-MM-dd'T'HH:mm:ss
LocalDate d2 = LocalDate.parse("2016-10-16T10:20:30", parser);

// parser.format(d1) is the same as d1.format(parser)

The output is:


PS: this works only with LocalDate. If I try to format an object with time fields (like LocalDateTime), both formats are used:


This prints:


Note that it formatted with both patterns. My guess is that the formatter checks if the object has the fields in each optional section. As the LocalDate has no time fields (hour/minute/second), the second pattern fails and it prints only the first one (MM/dd/yyyy). But the LocalDateTime has all the time fields, and both patterns are valid, so both are used to format.

My conclusion is: this isn't a general solution (like the Joda-Time's version), it's more like a "lucky" case where the patterns involved created the desired situation. But I wouldn't rely on that for all cases.

Anyway, if you are only using LocalDate, you can try to use this code. But if you're working with another types, then you'll probably have to use another formatter for the output, like this:

// parser/formatter for month/day/year
DateTimeFormatter mdy = DateTimeFormatter.ofPattern("MM/dd/yyyy");
// parser for both patterns
DateTimeFormatter parser = new DateTimeFormatterBuilder()
    // optional MM/dd/yyyy
    // optional yyyy-MM-dd'T'HH:mm:ss (use built-in formatter)
    // create formatter

// parse MM/dd/yyyy
LocalDate d1 = LocalDate.parse("10/16/2016", parser);
// parse yyyy-MM-dd'T'HH:mm:ss
LocalDate d2 = LocalDate.parse("2016-10-16T10:20:30", parser);

// use mdy to format

// format object with time fields: using mdy formatter to avoid multiple pattern problem

The output is:


Friday, July 2, 2021
answered 5 Months ago

I had the same issue. My approach to solve this is to modify the date format within Google Data Studio by creating a new dimension, reformatting the mySQL Datetime into the desired format.

In your example you use DATE_FORMAT. I wonder if you apply this in Data Studio (which does not know DATE_FORMAT) or in mySQL?. If you do it in data studio and leave your mySQL/phpmyadmin untouched you can use this:

TODATE(your-date-field, 'DEFAULT_DASH', '%Y%m%d')

This will take the date format YYYY-MM-DD with optional time in HH:ii:ss and reformat it into YYYYMMDD which works with data studio.

Sunday, September 19, 2021
answered 3 Months ago

You can use ISOdatetime, which is just a simple wrapper to as.POSIXct. Make sure to specify the sec argument as zero.

Data$timestamp <- with(Data, ISOdatetime(YY,MM,DD,hh,mm,0))
Friday, November 5, 2021
answered 4 Weeks ago
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