Zend_Date function returns the value "Feb 10, 2012" . It supposed to return "Oct 2, 2012". If a give the day greater than 12 than it returns the currect output. I don't know what I was missing. Please help me.
Thanks
If your format is YYYY-MM-DD (2012-10-02), try this code:
$str = '2012-10-02';
$date = new Zend_Date($str, Zend_Date::YEAR . '-' . Zend_Date::MONTH . '-' . Zend_Date::DAY);
echo $date->toString();
What is the code you are using to call the function?
Different countries use different date formats. Some use dd/mm/yyyy others use mm/dd/yyyy. The order you are passing in the day and month are different from the way the function expects to be called.
Try reversing the month and day before you call the function.
Related
I am using PowerShell to try and convert a string to a datetime. It should be easy, right?
I am getting the string from a CSV import, and it comes in the format of Jul-16. I have tried multiple ways of getting it into the format I want which is yyyy-MM-dd and I am currently at the following.
$invoice = $object.'Invoice Month'
$invoice = "01-" + $invoice
$invoice = [datetime]::parseexact($invoice, 'yyyy-MM-dd', $null)
But I get the error:
String was not recognized as a valid DateTime.
Am I missing something?
ParseExact is told the format of the date it is expected to parse, not the format you wish to get out.
$invoice = '01-Jul-16'
[datetime]::parseexact($invoice, 'dd-MMM-yy', $null)
If you then wish to output a date string:
[datetime]::parseexact($invoice, 'dd-MMM-yy', $null).ToString('yyyy-MM-dd')
You can simply cast strings to DateTime:
[DateTime]"2020-7-16"
or
[DateTime]"Jul-16"
or
$myDate = [DateTime]"Jul-16";
And you can format the resulting DateTime variable by doing something like this:
'{0:yyyy-MM-dd}' -f [DateTime]'Jul-16'
or
([DateTime]"Jul-16").ToString('yyyy-MM-dd')
or
$myDate = [DateTime]"Jul-16";
'{0:yyyy-MM-dd}' -f $myDate
You need to specify the format it already has, in order to parse it:
$InvoiceDate = [datetime]::ParseExact($invoice, "dd-MMM-yy", $null)
Now you can output it in the format you need:
$InvoiceDate.ToString('yyyy-MM-dd')
or
'{0:yyyy-MM-dd}' -f $InvoiceDate
Chris Dents' answer has already covered the OPs' question but seeing as this was the top search on google for PowerShell format string as date I thought I'd give a different string example.
If like me, you get the time string like this 20190720170000.000000+000
An important thing to note is you need to use ToUniversalTime() when using [System.Management.ManagementDateTimeConverter] otherwise you get offset times against your input.
PS Code
cls
Write-Host "This example is for the 24hr clock with HH"
Write-Host "ToUniversalTime() must be used when using [System.Management.ManagementDateTimeConverter]"
$my_date_24hr_time = "20190720170000.000000+000"
$date_format = "yyyy-MM-dd HH:mm"
[System.Management.ManagementDateTimeConverter]::ToDateTime($my_date_24hr_time).ToUniversalTime();
[System.Management.ManagementDateTimeConverter]::ToDateTime($my_date_24hr_time).ToUniversalTime().ToSTring($date_format)
[datetime]::ParseExact($my_date_24hr_time,"yyyyMMddHHmmss.000000+000",$null).ToSTring($date_format)
Write-Host
Write-Host "-----------------------------"
Write-Host
Write-Host "This example is for the am pm clock with hh"
Write-Host "Again, ToUniversalTime() must be used when using [System.Management.ManagementDateTimeConverter]"
Write-Host
$my_date_ampm_time = "20190720110000.000000+000"
[System.Management.ManagementDateTimeConverter]::ToDateTime($my_date_ampm_time).ToUniversalTime();
[System.Management.ManagementDateTimeConverter]::ToDateTime($my_date_ampm_time).ToUniversalTime().ToSTring($date_format)
[datetime]::ParseExact($my_date_ampm_time,"yyyyMMddhhmmss.000000+000",$null).ToSTring($date_format)
Output
This example is for the 24hr clock with HH
ToUniversalTime() must be used when using [System.Management.ManagementDateTimeConverter]
20 July 2019 17:00:00
2019-07-20 17:00
2019-07-20 17:00
-----------------------------
This example is for the am pm clock with hh
Again, ToUniversalTime() must be used when using [System.Management.ManagementDateTimeConverter]
20 July 2019 11:00:00
2019-07-20 11:00
2019-07-20 11:00
MS doc on [Management.ManagementDateTimeConverter]:
https://learn.microsoft.com/en-us/dotnet/api/system.management.managementdatetimeconverter?view=dotnet-plat-ext-3.1
$invoice = "Jul-16"
[datetime]$newInvoice = "01-" + $invoice
$newInvoice.ToString("yyyy-MM-dd")
There you go, use a type accelerator, but also into a new var, if you want to use it elsewhere, use it like so: $newInvoice.ToString("yyyy-MM-dd")as $newInvoice will always be in the datetime format, unless you cast it as a string afterwards, but will lose the ability to perform datetime functions - adding days etc...
Hope below helps!
PS C:\Users\aameer>$invoice = $object.'Invoice Month'
$invoice = "01-" + $invoice
[datetime]$Format_date =$invoice
Now type is converted. You can use method or can access any property.
Example :$Format_date.AddDays(5)
It's very easy; in my case it works with;
Input:
$Date = '29-07-2022'
DateFormat Convertion:
[datetime]::parseexact($date, 'dd-MM-yyyy', $null).ToString('dd-MMMM-yyyy')
Output:
I had a different but related need to convert a number (seconds) into days/hours/seconds etc.
$seconds = 41414141
New-Timespan -seconds $seconds
This is a piece of lua script that displays the time. I cannot separate the numbers ie: time.hour, ":",
to basically show hh:mm:ss
time = os.date("*t")
print(time.hour .. time.min .. time.sec)
There are several ways to do this:
Use string concatenation: print(time.hour .. ":" .. time.min .. ":" .. time.sec)
Use formatting: print(("%02d:%02d:%02d"):format(time.hour, time.min, time.sec))
Use table concatenation: print(table.concat({time.hour, time.min, time.sec}, ":"))
When you really need to format your string, my preference would be for #2. For time = {hour = 1, min = 20, sec = 5} this prints:
1:20:5
01:20:05
1:20:5
For simply printing the time - extract what you want (the time) from the full date stamp string:
> os.date():sub(9)
12:30:39
This works on my PC ;). There may be a different date stamp string in your OS.
G
local date = os.date('*t')
local time = os.date("*t")
print(os.date("%A, %m %B %Y | "), ("%02d:%02d:%02d"):format(time.hour, time.min, time.sec))`
I have a timestamp that looks like 25-OCT-10 04.11.00.000000 AM. I'm trying to convert this to a time format with
Time::Piece->strptime("25-OCT-10 04.11.00.000000 AM","%d-%b-%y %I.%M.%S.%6N %p")
but it keeps throwing errors. I've tried %OS, %SZ. They dont seem to work. Can someone tell me what I'm doing wrong?
Time::Piece doesn't support sub-second times. Try DateTime instead, for example, DateTime::Format::Strptime:
use DateTime::Format::Strptime;
my $parser = DateTime::Format::Strptime->new(
pattern => '%d-%b-%y %I.%M.%S.%6N %p',
);
my $dt = $parser->parse_datetime("25-OCT-10 04.11.00.000000 AM");
strptime won't read that. strptime works with a structure that only goes down to integer seconds, and it doesn't have any formats for recognizing non-integer numerics -- and there's no such format as N in Time::Piece's strptime. If you know that you're always expecting .000000 for a number of microseconds then you could try using ..."%I.%M.%S.000000 %p", otherwise strptime just isn't for you.
How about DateTime::Format::CLDR? A format of "dd-MMM-yy hh.mm.ss.SSSSSS a" seems to work perfectly well with that format.
use DateTime::Format::CLDR;
my $parser = DateTime::Format::CLDR->new(
pattern => "dd-MMM-yy hh.mm.ss.SSSSSS a",
locale => "en_US",
);
my $dt = $parser->parse_datetime("25-OCT-10 04.11.00.000100 AM");
say $dt->iso8601; # 2010-01-25T04:11:00
Edit: just noticed that this doesn't recognize months properly if they're all uppercase -- it recognizes "Oct" but not "OCT". A fixed version is available here and has been sent upstream for merge :)
Update: DateTime::Format::CLDR 1.11 is properly case-insensitive.
In Perl, why I get different results from parsedate(2010-7-2 13:0:0) and parsedate(2010-7-2 13:00:0) ?
The 2010-7-2 13:0:0 string is not in a valid format, and is actually not being parsed at all (it appears) as evidenced by the fact that parsedate("2010-7-2") returns the same value as parsedate("2010-7-2 13:0:0") for me.
Based on the docs, it's simply parsing the YYYY-MM-DD, but not parsing the 13:0:0 at all because it is expecting it to be in HH:MM format and not HH:M format. Basically, you have to use two digits for the minutes in order for it to be valid input.
To handle your date format with more flexibility, try using DateTime::Format::Strptime
my $strp = DateTime::Format::Strptime->new(
pattern => '%Y-%m-%d %T',
locale => 'en_AU',
time_zone => 'Australia/Melbourne',
);
my $dt1 = $strp->parse_datetime('2010-7-2 13:0:0');
my $date_1 = $strp->format_datetime($dt1);
$date_1 is now converted into a well-formatted date format "2010-07-02 13:00:00". Then you can call parsedate($date_1) & get epoch.
I need to print the date in the format of mm/dd/yyyy.
if the date is 4/24/2009 it should print the date as 04/24/2009.
that is zero padding is also needed..
I used date function to get the current date...but the date is getting in the format of m/dd/yyyy...
Tested in the immediate window and is working for me (output as a comment)
Format(Now, "MM/dd/yyyy") '04/29/2009
Format(Date, "MM/dd/yyyy") '04/29/2009
Format(CStr(Now), "MM/dd/yyyy") '04/29/2009
Format(Date$, "MM/dd/yyyy") '04/29/2009
Format(CDate(Date), "MM/dd/yyyy")'04/29/2009
So whether it is string or datetime should not matter.
Edit: Saw your comment to Fredrik. It doesn't matter how it looks like when you save it to the db table (column date format would be a property of the db and not your program's (or vb's) responsibility). Just format the value as and when you retrieve it from the db.
Note that the "/" character in date formatting functions has a special meaning, as "date separator". This means that i may be replaced with the date separator for the current locale that the program is executed in (here in Sweden it would be replaced with "-" for instance). In order to ensure that you do indeed get the "/" character in the output, I think this would work (I don't have a VB installation to verify with):
Format(date, "MM'/'dd'/'yyyy")
just for the record, escaping the slash will work
Format(dt,"MM\/dd\/yyyy")
Try the next code:
Format(dt,"MM/dd/yyyy")
When you enter date in whatever format it will convert default value so do one thing that in access change your data type date/time to text then it can't affect to your work sure.
I also use VB6 and need to format date in my txt report
this works for me
Format$(Now, "yyyy-mm-dd-00.00.00")
but only if I declare date as string
You can try like this also depending upon your requirement
Dim strDate As String
Dim strDate1() As String
strDate = FormatDateTime(Now, vbGeneralDate)
If InStr(strDate, " ") > 0 Then
strDate1 = Split(strDate, " ")
Dim datDate1 As Date
If Month(strDate1(0)) < 10 Then
txtDate.Text = "0" + strDate1(0)
Else
txtDate.Text = strDate1(0)
End If
Else
End If
Formatting DateTime as a string is straightforward. Often we use format patterns like "HH." But methods like ToShortDateString are also useful.
Example. First we get the current time through DateTime.Now. When you execute these code examples, the current DateTime will be different on your computer.
Here: A format string beginning with MMM (for the month) is used. Look at how the pattern matches up to the output of the program.
VB.NET program that uses format string with DateTime
Module Module1
Sub Main()
' Use current time.
' ... Use a format.
' ... Write to console.
Dim time As DateTime = DateTime.Now
Dim format As String = "MMM ddd d HH:mm yyyy"
Console.WriteLine(time.ToString(format))
End Sub
End Module
Output
Feb Tue 21 13:26 2017
strDate = Format(strDate, "yyyy-mm-dd")
BillTime = Format(BillTime, "hh:mm:ss")