I am currently using Zend framework 3 with zend log module. I noticed the Timestamp column of zend log table is not unix timestamp.
I have tried to search solution but all I could see the way to change format. It there a way to change it and store to unix timestamp?
You can set your prefered time using setDateTimeFormat() to Formatter, as in below-
$logger = new \Zend\Log\Logger();
$formatter = new \Zend\Log\Formatter\Simple();
$formatter->setDateTimeFormat('Y-m-d'); // as per your choice
$writer = new \Zend\Log\Writer\Stream('php://output');
$writer->setFormatter($formatter);
$logger->addWriter($writer);
Related
I'm migrating a whole bunch of web pages that were written in classic asp over to a new server, and have discovered many references to the simple date() function, like:
if cint(left(date,instr(date,"/")-1)) < 9 then blah blah
I'm getting errors because the new server's default date format is returning yyyy-mm-dd, and the code above is expecting it to be in dd/mm/yyyy format.
Rather than manually fixing every occurrence, of which there could be hundreds, I'm looking to see if I can change the default date format for asp so that date() returns dd/mm/yyyy. I thought by simply changing the system's short date format would do the trick, but even after restarting the server it's still showing yyyy-mm-dd.
Is there a setting somewhere where you can specify the default date format when using the date() function?
This worked for me:
change global.asa, in the Sub Session_OnStart, add a line
Session.LCID=1033
I was wondering if Pandas can somehow figure out what today's date is allowing me to automate the naming of the html file I create when I use the "df.to_html" method.
Basically I"m trying to read a website using method "pd.read_html", and then save the dataframe as an html file, daily. The name of the html file will be the day's date. (So today is 9/28/2016 and tomorrow will be 10/01/16 and so on ) I'm not particular about the format of the date, so Sept or 09, whichever is okay.
I'm trying to automate this as much as possible, and so far the best I've gotten is, using ".format" which allows me some flexiblity. But I don't know how I can further automate the process.
import pandas as pd
df = pd.read_html('random site')
today_date = 'saved data/{}.html'.format('Sept 28') # I'm saving it in the folder "saved data" with the name as today's date.html.
df.to_html(today_date)
Thanks.
See the datetime module.
specifically: datetime.date.today().isoformat() gives you a string with the current date in ISO 8601 format (‘YYYY-MM-DD’)
I need to set end_at attribute 30 days from current date. how can i do that in laravel 4.
When I have used bellow code I am getting error saying "Class 'Date' not found"
Please help me to fix this.
$sub->end_at = new Date('+30 days');
There is no Date class in PHP, there is only a DateTime class which you could use.
But since you're using Laravel, which uses the Carbon library by default, you can use that to handle dates because it has a better API. In your case you can do this:
use Carbon\Carbon;
...
$sub->end_at = Carbon::now()->addDays(30)->toIso8601String();
If you're trying to update a Eloquent model, then you can take advantage of Eloquent's integrated date/time column handling. In your model you can add the dates property with this value:
protected $dates = ['end_at'];
and now when assigning a timestamp to the end_at column, Laravel will automatically transform and save it to the correct format in your database. So you'll only need to use this:
$sub->end_at = Carbon::now()->addDays(30);
This will return that in the format appropriate for MySQL
$sub->end_at = date('Y-m-d H:i:s', strtotime('+30 days'));
I'm working on an app that allows the user to edit several dates in a form. The dates are rendered in the European format (DD-MM-YYYY) while the databases uses the default YYYY-MM-DD format.
There are several ways to encode/decode this data back and forth from the database to the user, but they all require a lot of code:
Use a helper function to convert the date before saving and after retrieving (very cumbersome, requires much code)
Create a separate attribute for each date attribute, and use the setNameAttribute and getNameAttribute methods to decode/encode (also cumbersome and ugly, requires extra translations/rules for each attribute)
Use JavaScript to convert the dates when loading and submitting the form (not very reliable)
So what's the most efficient way to store, retrieve and validate dates and times from the user?
At some point, you have to convert the date from the view format to the database format. As you mentioned, there are a number of places to do this, basically choosing between the back-end or the front-end.
I do the conversion at the client side (front-end) using javascript (you can use http://momentjs.com to help with this). The reason is that you may need different formats depending on the locale the client is using (set in the browser or in his profile preferences for example). Doing the format conversion in the front-end allows you to convert to these different date formats easily.
Another advantage is that you can then use the protected $dates property in your model to have Laravel handle (get and set) these dates automatically as a Carbon object, without the need for you to do this (see https://github.com/laravel/framework/blob/master/src/Illuminate/Database/Eloquent/Model.php#L126).
As for validation, you need can then use Laravel's built-in validation rules for dates, like this:
'date' => 'required|date|date_format:Y-n-j'
While client-side is good for UX, it doesn't let you be sure, all will be good.
At some point you will need server-side validation/convertion anyway.
But here's the thing, it's as easy as this:
// after making sure it's valid date in your format
// $dateInput = '21-02-2014'
$dateLocale = DateTime::createFromFormat('d-m-Y', $dateInput);
// or providing users timezone
$dateLocale =
DateTime::createFromFormat('d-m-Y', $dateInput, new DateTime('Europe/London'));
$dateToSave = $dateLocale
// ->setTimeZone(new TimeZone('UTC')) if necessary
->format('Y-m-d');
et voila!
Obviously, you can use brilliant Carbon to make it even easier:
$dateToSave = Carbon::createFromFormat('d-m-Y', $dateInput, 'Europe/London')
->tz('UTC')
->toDateString(); // '2014-02-21'
Validation
You say that Carbon throws exception if provided with wrong input. Of course, but here's what you need to validate the date:
'regex:/\d{1,2}-\d{1,2}-\d{4}/|date_format:d-m-Y'
// accepts 1-2-2014, 01-02-2014
// doesn't accept 01-02-14
This regex part is necessary, if you wish to make sure year part is 4digit, since PHP would consider date 01-02-14 valid, despite using Y format character (making year = 0014).
The best way I found is overriding the fromDateTime from Eloquent.
class ExtendedEloquent extends Eloquent {
public function fromDateTime($value)
{
// If the value is in simple day, month, year format, we will format it using that setup.
// To keep using Eloquent's original fromDateTime method, we'll convert the date to timestamp,
// because Eloquent already handle timestamp.
if (preg_match('/^(\d{2})\/(\d{2})\/(\d{4})$/', $value)) {
$value = Carbon\Carbon::createFromFormat('d/m/Y', $value)
->startOfDay()
->getTimestamp();
}
return parent::fromDateTime($value);
}
}
I'm new in PHP, so I don't know if it's the best approach.
Hope it helps.
Edit:
Of course, remember to set all your dates properties in dates inside your model. eg:
protected $dates = array('IssueDate', 'SomeDate');
I have a form, that saves date of birth in the database. This form passes three variables to php: $ayear $amonth $adate
Now I need to save these in the database in a format: 00/00/0000 (or 00/00/00, if 00/00/0000 is not supported).
I need to create a date() variable, but I dont know how. All the examles on the internet use time() or now() which are not usefull in my case. Can someone give me the right function?
Thank you for your time!
If you're using MySQL you'll want to save as YYYY-MM-DD with the field type set to date. I'd save this variables value to the database.
$birthday = implode('-', array($year, $month, $day));