On my db migration i have a "dob" field:
$table->date('DOB')->nullable();
I have a string representing a date: "12/12/1960" in the input. I tested "dob" and the content is there. But when I try to set it to the "dob" field in the database..
$member->DOB = date('m/d/Y',Request::input('dob'));
The datebase field becomes 0000-00-00
What am I doing wrong?
The database most likely wants it in YYYY-MM-DD format. So just changing the format should work. But to get from 12/12/1960 to there, you need to parse it.
$oldDateString = '12/12/1960';
$newDateString = DateTime::createFromFormat('m/d/Y', $oldDateString)->format('Y-m-d');
Look at definition
string date ( string format [, int timestamp] )
Second parameter is timestamp, not string
Use DateTime PHP5 functions or Laravel Carbon
E.g. date_create_from_format('m/d/Y', Request::input('dob'))
->format('Y-m-d H:i:s');
Related
I am using ASP.NET Core 3.1. I am trying to bind dates in a query string to a PageModel using GET. Here is my query string:
?id=15+startDate=1900-01-01+endDate=1900-01-01
Here is the signature of my OnGet:
public void OnGet(int id, DateTime startDate, DateTime endDate)
The values are always 0, 1/1/0001 00:00:00, 1/1/0001 00:00:00 respectively.
If I remove the dates and just pass the id, it works fine. Why do the dates in the query string break everything including the id? I have also tried changing the data types from DateTime to string and it still breaks with the string parameters always being null.
The separator for parameters in a query string is &, not + as your example has. Corrected:
?id=15&startDate=1900-01-01&endDate=1900-01-01
An arguably better approach with datetime parameters is to accept them as strings, then use DateTime.TryParse to parse with control over expected format. That way you'll always have the opportunity to inspect the string value provided and decide if you want to convert it to a DateTime, and how.
I want to extract the filename string as a DateTime column.
The code for it as follows:
#data =
EXTRACT
...
filename_date DateTime
FROM "/input/vga_{filename_date}.txt"
USING Extractors.Tsv(skipFirstNRows:1);
filename = vga_20171201.txt
whenever i have used datatype as string or int, it's work for me.
You have to specify .net date format strings along with the virtual column name to get that behaviour, like this:
#data =
EXTRACT someData string,
filename_date DateTime
FROM "/input/vga_{filename_date:yyyy}{filename_date:MM}{filename_date:dd}.txt"
USING Extractors.Tsv(skipFirstNRows : 1);
I have a series of files that are named like 1601.gz to represent January of 2016. {date:yyMM}.gz or {date:yy}{date:MM}.gz don't seem to
How to get only date on database column, I have difficulties using Carbon on controller:
$data['nowUser'] = User::where('date', Carbon::today()->toDateString())->get();
Date column looks like this in the database:
That is a DATETIME column, so there's no need for additional formatting the Carbon instance. However you need to use whereDate if you want the fetch all users for which the date column contains today's date:
$data['nowUser'] = User::whereDate('date', '=', Carbon::today())->get();
Because when you pass Carbon::today() to the Query Builder method, the __toString method will be automatically called and return a DATETIME string with the format from Carbon::DEFAULT_TO_STRING_FORMAT, which is exactly that MySQL format Y-m-d H:i:s.
I am using laravel 5.0. I am getting data from controller as json. I am having value like this.
{"timstamp_date":"1434360957"},
I need to convert this unix timestamp value as Normal Date Like (15-06-2015) or (15-March-2015).
I have used Date(timstamp_date) but it is showing current time only. Not my timstamp date
You could use:
date("d-m-Y H:i:s", 1434360957);
EDIT
You could try;
var dateTime = new Date(1434360957*1000);
var formatted = dateTime.toGMTString();
https://jsfiddle.net/sp57Lnpf/
Use the date function. You need to specify the format as the first parameter:
date("d-m-Y", $timestamp_date)
http://php.net/manual/en/function.date.php
Laravel also comes with Carbon you could use that if you wanted to for further manipulation of the data if you so required it.
http://carbon.nesbot.com/docs/
We have some data in a DataTable and we are using the query like this to get what we need.
IEnumerable<Results> subResult = from query in datatable.AsEnumerable()
select new Results
{
Name = query.Field<string>("Name"),
Date = query.Field<DateTime?>("Date")
}
This above query returns what i need but date in full format (Ex: m/dd/yyyy hh:min:sec am/pm) but we need only date part of it(only mm/dd/yyyy need to be pulled in). When looked in the properties of this, couldn't find an implicit way to get it, please help me in getting this result. Thanks.
The DateTime class has all the properties you just mentioned. You should try to fix your display with the ToString("anyformatyouwant") formats.
What about this?
Date = query.Field<DateTime?>("Date").Date
This will give you just the date part:
IEnumerable<Results> sub-result= from query in datatable.AsEnumerable()
where new Results
{
Name = query.Field<string>("Name"),
Date = query.Field<DateTime?>("Date").Date
}
However if you just want to display the date somewhere (meaning you're not using this to do date grouping, etc. then I would just specify the display format in your UI layer.
AFAIK there is no Date Class in .net, if you want pure date you should write a Class for that, otherwise you can pull out the part as a string
DateTime.Now.Date.ToString("yyyy/MM/dd")
gives you the current date in string type, but if you get the .Date part it gives you a DateTime object which still has a time part showing AM 12:00:00:0 blah blah