转换后 PHP 日期显示“1970-01-01”
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PHP date showing '1970-01-01 ' after conversion
提问by AssamGuy
I have a form in which date format is dd/mm/yyyy. For searching database , I hanverted the date format to yyyy-mm-dd. But when I echoit, it showing 1970-01-01. The PHP code is below:
我有一个日期格式为dd/mm/yyyy. 为了搜索数据库，我将日期格式转换为yyyy-mm-dd. 但是当我echo它时，它显示1970-01-01. PHP代码如下：
$date1 = $_REQUEST['date'];
echo date('Y-m-d', strtotime($date1));
Why is it happening? How can I format it to yyyy-mm-dd?
为什么会发生？我怎样才能将其格式化为yyyy-mm-dd？
回答by Cyclonecode
Replace /with -:
替换/为-：
$date1 = strtr($_REQUEST['date'], '/', '-');
echo date('Y-m-d', strtotime($date1));
回答by Oldskool
January 1, 1970 is the so called Unix epoch. It's the date where they started counting the Unix time. If you get this date as a return value, it usually means that the conversion of your date to the Unix timestamp returned a (near-) zero result. So the date conversion doesn't succeed. Most likely because it receives a wrong input.
1970 年 1 月 1 日是所谓的 Unix 纪元。这是他们开始计算Unix 时间的日期。如果您将此日期作为返回值，通常意味着将您的日期转换为 Unix 时间戳会返回（接近）零结果。所以日期转换不成功。很可能是因为它收到了错误的输入。
In other words, your strtotime($date1)returns 0, meaning that $date1is passed in an unsupported format for the strtotime function.
换句话说，您strtotime($date1)返回 0，这意味着它$date1以不支持的格式传递给 strtotime 函数。
回答by Kaushal Roy
$inputDate = '07/05/-0001';
$dateStrVal = strtotime($inputDate);
if(empty($dateStrVal))
{
echo 'Given date is wrong';
}
else{
echo 'Date is correct';
}
O/P : Given date is wrong
O/P : 给定日期错误
回答by Dnyaneshwar Harer
Another workaround:
另一种解决方法：
Convert datepicker dd/mm/yyyyto yyyy-mm-dd
将日期选择器转换dd/mm/yyyy为yyyy-mm-dd
$startDate = trim($_POST['startDate']);
$startDateArray = explode('/',$startDate);
$mysqlStartDate = $startDateArray[2]."-".$startDateArray[1]."-".$startDateArray[0];
$startDate = $mysqlStartDate;
回答by jerryurenaa
The issue is when your data is set to 000-00-00 or empty you must double-check and give the correct information and this issue will go away. I hope this helps.
问题是当您的数据设置为 000-00-00 或为空时，您必须仔细检查并提供正确的信息，这个问题就会消失。我希望这有帮助。
回答by Dayz
Use below code for php 5.3+:
对 php 5.3+ 使用以下代码：
$date = new DateTime('1900-02-15');
echo $date->format('Y-m-d');
Use below code for php 5.2:
对 php 5.2 使用以下代码：
$date = new DateTime('1900-02-15');
echo $date->format('Y-m-d');
回答by Tye Lucas
$date1 = $_REQUEST['date'];
if($date1) {
$date1 = date( 'Y-m-d', strtotime($date1));
} else {
$date1 = '';
}
This will display properly when there is a valid date()in $dateand display nothing if not.
Solved the issue for me.
当有有效date()输入时，这将正确$date显示，否则不显示任何内容。
为我解决了这个问题。
回答by Bakhtawar GIll
finally i have found a one line code to solve this problem
最后我找到了一行代码来解决这个问题
date('d/m/Y', strtotime(str_replace('.', '-', $row['DMT_DATE_DOCUMENT'])));
