I am trying to convert DateTime object to string using formatting but due to some strange reason, it ignores AM/PM. It would just take the magnitude of the Hour in the object.
I am using the following format:
StartDate.ToString("yyyy-MM-dd hh:mm:ss.fff");
and
String.Format("0:yyyy-MM-dd hh:mm:ss.fff", StartDate);
I don't think that there is a difference between the two, but just wanted to give it a try. If I pass a value 4/28/2012 6:00:00 AM or 4/28/2012 6:00:00 PM, the result is the same "2012-04-27 06:00:00.000"
You've used hh which uses the 12-hour clock. You want HH which uses the 24-hour clock.
See MSDN for more details about custom format strings.
Note that you may wish to specify the invariant culture, unless you really want the time separator to depend on the current culture:
string formatted = StartDate.ToString("yyyy-MM-dd HH:mm:ss.fff",
CultureInfo.InvariantCulture);
(Note that you shouldn't have the braces if you're passing the format string to ToString. I'll assume this was just a typo in the question.)
If you want to use the 12-hour clock, use tt in the format string to produce the AM/PM designator.
Related
I am doing an F# course by Richard Broida. In one lecture I am supposed to parse a CSV document. Everything works fine except for parsing datetime. My local culture settings are Slovak (sk-SK) and I am trying to parse the following datetime string:
9/14/2016 19:31
I have read through the CSV file and found out that months, days and hours can all have both one or two digits. Years always have 4 digits and minutes always have 2 digits. So, the datetime 10/7/2016 9:01 is also a valid one.
I have created an adapter function:
let dateTimeParseAdapter format provider date =
DateTime.ParseExact(date, format, provider)
and then called
dateTimeParseAdapter "M/d/yyyy h:mm" CultureInfo.InvariantCulture "9/14/2016 19:31"
but I got an exception message saying
System.FormatException: String was not recognized as a valid DateTime.
I tried adjusting the format string to exactly match the number of digits in my single date. However, even the following code resulted in the same error message.
dateTimeParseAdapter "M/dd/yyyy hh:mm" CultureInfo.InvariantCulture "9/14/2016 19:31"
Why am I getting the exception message? Should I change the CultureInfo settings? Or is there a pre-defined datetime format to parse my datetimes?
When working with 24 hour format you should use H (HH) not h hh (which is for 12 hour format) pattern:
format = #"M/d/yyyy H:mm";
since you are using 24 hour in your value you must put capitals in hours. Change
"M/dd/yyyy hh:mm"
to
"M/dd/yyyy HH:mm"
I get this value from my backend service: 171054. It represents hh:mm:ss. But when I use the formatting options from the docs it gives me back 00:00:00.
Things I've tried:
moment('171054').format('hh-mm-ss')
moment('171054').format('HH-mm-ss')
moment('171054').format('HH-MM-SS')
You are confusing format option with parsing option. Since your input string is not in ISO 8601 format, you have to specify format when parsing.
Here a working example for your use case:
var mom = moment('171054', 'HHmmss');
console.log(mom.format());
console.log(mom.format('HH:mm:ss'));
<script src="https://cdnjs.cloudflare.com/ajax/libs/moment.js/2.15.1/moment.min.js"></script>
I am not sure if momentjs can read that as date since there is no identifier. I would suggest changing the value for example you have 171054, get each 2 digits since you sure that this is represent as hh:mm:ss then add identifier between then like ":" or "-" then try us momentjs formatting again.
I insert my date/time data into a CHAR column in the format: '6/4/2015 2:08:00 PM'.
I want that this should get automatically converted to format:
'2015-06-04 14:08:00' so that it can be used in a query because the format of DATETIME is YYYY-MM-DD hh:mm:ss.fffff.
How to convert it?
Given that you've stored the data in a string format (CHAR or VARCHAR), you have to decide how to make it work as a DATETIME YEAR TO SECOND value. For computational efficiency, and for storage efficiency, it would be better to store the value as a DATETIME YEAR TO SECOND value, converting it on input and (if necessary) reconverting on output. However, if you will frequently display the value without doing computations (including comparisons or sorting) it, then maybe a rococo locale-dependent string notation is OK.
The key function for converting the string to a DATETIME value is TO_DATE. You also need to look at the TO_CHAR function because that documents the format codes that you need to use, and because you'll use that to convert a DATETIME value to your original format.
Assuming the column name is time_string, then you need to use:
TO_DATE(time_string, '%m/%d/%Y %I:%M %x') -- What goes in place of x?
to convert to a DATETIME YEAR TO SECOND — or maybe DATETIME YEAR TO MINUTE — value (which will be further manipulated as if by EXTEND as necessary).
I would personally almost certainly convert the database column to DATETIME YEAR TO SECOND and, when necessary, convert to the string format on output with TO_CHAR. The column name would now be time_value (for sake of concreteness):
TO_CHAR(time_value, '%m/%d/%Y %I:%M %x') -- What goes in place of x?
The manual pages referenced do not immediately lead to a complete specification of the format strings. I think a relevant reference is GL_DATETIME environment variable, but finding that requires more knowledge of the arcana of the Informix product set than is desirable (it is not the first thing that should spring to anyone's mind — not even mine!). If that's correct (it probably is), then one of %p and %r should be used in place of %x in my examples. I have to get Informix (re)configured on my machine to be able to test it.
So I know I can use =FormatDate(MyDate ,"mm/dd/yy") to turn a date into a string. I am then trying to add on a time:
=FormatDate(AdminDate ,"mm/dd/yy") + MyTime
Which works, however, I need to format this back to a datetime field (as I need to compare against a preexisting datetime field). I try using Todate(), but the documentation is very light, and provides very little on what is acceptable in the formatting of the date area, and nothing in the way of time!
I have attempted:
=ToDate(FormatDate(MyDate ,"mm/dd/yy HH:mm:ss") + MyTime ,"mm/dd/yy HH:mm:ss")
but this will only work when there is no time (and it just nulls out the time) and any row with a time will return a #ERROR
Anyone have an insight on formatting datetimes?
Thanks
The correct way to use the FormatDate command to output date and time components together in 12 hour format is:
=FormatDate(AdminDate; "MM/dd/yy hh:mm:ss a")
and in 24 hour format is:
=FormatDate(AdminDate; "MM/dd/yy HH:mm:ss")
Note MM is used in months and mm in minutes.
I am using ColdFusion 10 to make some REST calls and the date returned is using a GMT offset.
Example: 2013-03-25T14:30:40-04:00
I need this formatted for 2 purposes:
Screen Display so it looks something like mm/dd/yyyy hh:mm:ss
To Insert into mySQL.
I have tried a variety of the CF time/date functions but continue to get the "is not a valid date format"
I thought maybe the #ParseDateTime(i.submitted_at,"pop")# would handle it with POP but same issue.
Spent a few hours now trying multiple variations and googling around now just going in circles. Any ideas would be greatly appreciated.
Thanks!
Have a look at the UDF DateConvertISO8601() on CFLib.
DateConvertISO8601(ISO8601dateString, targetZoneOffset) on CFLib
Description:
This function take a string that holds a date in ISO 8601 and converts it to ODBC datetime, but could be adapted to convert to whatever you like. It also will convert to a datetime in a timezone of your choice by specifying the offset, i.e. it could take a datetime in GMT and convert to PT. See http://www.w3.org/TR/NOTE-datetime for description of ISO 8601, the International Standard for the representation of dates and times.
Return Values:
Returns a datetime.
The source code is viewable at the link I provided.
This, 2013-03-25T14:30:40-04:00, is a string. If you run this:
x = left(2013-03-25T14:30:40-04:00, 19);
you get 2013-03-25T14:30:40. You can use the replace function to replace the T with a space. You can then to this
DateTimeVar =parsedatetime('2013-03-25 14:30:40');
Now you have a datetime variable that you can format. If necessary you can do a datediff with the offset from GMT.
This is an informational answer, not a direct answer to the question.
ColdFusion 11 has updated the ParseDateTime() function so that it will correctly convert the ISO-8601 date/time strings to a ColdFusion datetime object.
With a number of remote requests and responses, the date / time values can often be returned in ISO format. In your case, the mask looks like this:
YYYY-MM-DDThh:mm:ssTZD (eg 1997-07-16T19:20:30+01:00)
In this ISO format, the T string is a literal representation of a marker where the time stamp starts in the string (with the offset following directly).
Below is a reusable function that will convert an ISO date format into a useable ColdFusion date time object:
<cffunction name="ISOToDateTime" access="public" returntype="string" output="false"
hint="Converts an ISO 8601 date/time stamp with optional dashes to a ColdFusion
date/time stamp.">
<cfargument name="Date" type="string" required="true" hint="ISO 8601 date/time stamp." />
<cfreturn ARGUMENTS.Date.ReplaceFirst(
"^.*?(\d{4})-?(\d{2})-?(\d{2})T([\d:]+).*$",
"$1-$2-$3 $4"
) />
</cffunction>
You can then call the function like so to output or return a ColdFusion-friendly version of the date time:
ISOToDateTime( "2013-03-25T14:30:40-04:00" )
That function is courtesy of Ben Nadel. The original blog post can be found here:
http://www.bennadel.com/blog/811-Converting-ISO-Date-Time-To-ColdFusion-Date-Time.htm
You can also convert the date time value using the offset, if required. Again, Ben Nadel has a great blog post outlining how to accomplish this:
http://www.bennadel.com/blog/1595-Converting-To-GMT-And-From-GMT-In-ColdFusion-For-Use-With-HTTP-Time-Stamps.htm
CF10 can use this code as stated in the example of the parseDateTime() doc.
<cfset string = "1997-07-16T19:20:30+01:00">
<cfset date = parseDateTime(string, "yyyy-MM-dd'T'HH:mm:ssX")>