“对非对象”上的成员函数格式()的调用“在PHP转换日期时 [英] "Call to a member function format() on a non-object" when converting date in PHP
问题描述
首先,我是PHP的初学者,请耐心等待。
我认为转换日期是一个非常基本的操作,不应该那么难。但是无论如何,我仍然无法摆脱这个错误信息:
调用非对象上的成员函数格式() c $ c>
所以,我去谷歌搜索,并得到一些好的来源,如这个问题。
我试图做类似这样的事情,但是我失败了。这是我的代码:
$ temp = new DateTime();
/ * ERROR HERE * / $ data_umat ['tanggal_lahir'] = $ data_umat ['tanggal_lahir'] - >格式('Y-m-d');
$ data_umat ['tanggal_lahir'] = $ temp;
所以,我做了试用&错误,我发现如果我这样做:
$ data_umat ['tanggal_lahir'] = date(Ymd H:i :s);
日期将成功转换,但它总是返回今天的日期(我不想要) p>
我只想转换日期,所以2010年10月22日将是2013-10-22。
你正在调用方法 format() / code>非对象。尝试这样:
$ data_umat ['tanggal_lahir'] = new DateTime('10 / 22/2013');
$ data_umat ['tanggal_lahir'] = $ data_umat ['tanggal_lahir'] - >格式('Y-m-d');
或单班:
$ data_umat ['tanggal_lahir'] = date_create('10 / 22/2013') - >格式('Ym-d');
First, im a beginner in PHP so please bear with me.
I think converting date is a very basic operation, and shouldnt be that hard. But anyway, i still cant get rid of this error message :
Call to a member function format() on a non-object
So, i go on googling and get some good source like this question.
I tried to do something similiar like that question, but i failed. This is my code :
$temp = new DateTime();
/*ERROR HERE*/ $data_umat['tanggal_lahir'] = $data_umat['tanggal_lahir']->format('Y-m-d');
$data_umat['tanggal_lahir'] = $temp;
So, i did trial & errors, and i found out if i do this :
$data_umat['tanggal_lahir'] = date("Y-m-d H:i:s");
The date will successfully converted, BUT it always return today's date (which i dont want).
All i want is just to convert the date so 10/22/2013 will be 2013-10-22.
Any help is appreciated, Thanks for your time.
You are calling method format()
on non-object. Try this:
$data_umat['tanggal_lahir'] = new DateTime('10/22/2013');
$data_umat['tanggal_lahir'] = $data_umat['tanggal_lahir']->format('Y-m-d');
or one-liner:
$data_umat['tanggal_lahir'] = date_create('10/22/2013')->format('Y-m-d');
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