Php 将时间转换为日期
我有剧本Php 将时间转换为日期,php,Php,我有剧本 idea_slug=$_POST['idea_slug']; $data['idea_date']=$this->db->query("SELECT * FROM td_idea WHERE idea_slug='$idea_slug'")->result_array(); $submit_date=$data['idea_date'][0]['idea_submit_date']; $datetime1 = DateTime::create
idea_slug=$_POST['idea_slug'];
$data['idea_date']=$this->db->query("SELECT * FROM td_idea WHERE idea_slug='$idea_slug'")->result_array();
$submit_date=$data['idea_date'][0]['idea_submit_date'];
$datetime1 = DateTime::createFromFormat('d/m/Y', $submit_date);
$submit_date= $datetime1->format('Y-m-d');
$submit_date = strtotime($submit_date);
$end = strtotime('+90 days', $submit_date);
现在我想将$end转换为Y/m/d格式
我怎样才能这样做呢?像这样:
echo date('Y/m/d', $end);
“替代方案”是。您可以像下面这样做
$date_format = date('Y/m/d', $end);
它是simpel=),您应该检查DateTime类
$datetime1 = DateTime::createFromFormat('d/m/Y', $submit_date);
$datetime1->modify('+90 day');
$end = $datetime1->format('d/m/Y');
干杯为什么要创建DateTime对象,然后将其转换为unix时间戳?毫无意义(这是我今天第二次看到它)
$datetime1 = DateTime::createFromFormat('d/m/Y', $submit_date);
$datetime1->modify('+90 day');
$end = $datetime1->format('d/m/Y');