3
I have the following line
echo date('Y-m-d', strtotime("28/09/2016")); // 1970-01-01
How do I return 2016-09-28? If the day is less than 12, returns...
3
I have the following line
echo date('Y-m-d', strtotime("28/09/2016")); // 1970-01-01
How do I return 2016-09-28? If the day is less than 12, returns...
4
Can do using the object DateTime
of PHP:
$formato = 'd/m/Y'; // define o formato de entrada para dd/mm/yyyy
$data = DateTime::createFromFormat($formato, "28/09/2016"); // define data desejada
echo $data->format('Y-m-d'); // formata a saída
Read more about the supported formats in PHP - DATE
3
You need to pass as parameter to the strtotime
a string with separator using -
(hyphenate) for example;
$date = str_replace("/", "-", "28/09/2016");
echo date("Y-m-d", strtotime($date));
Browser other questions tagged php strtotime
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the string inside the
strtotime
has to have the separator with-
also, try this.– RFL
The funny thing is that if I put a day shorter than 12 months, returns normal
– Sr. André Baill
That was it Rafael, echo date('Y-m-d', strtotime(str_replace("/", "-", "28/09/2016")); thanks buddy. Call me in skype K7, you add!
– Sr. André Baill
Put as an answer, to give the acceptance here.
– Sr. André Baill