blocks|key|455082|text|用-替换/|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|455083|$date1+=+strtr($_REQUEST['date'],+'/',+'-');
echo+date('Y-m-d',+strtotime($date1));|code-block|syntax|javascript|455084|entityMap^0|1|1|4|1|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]|$9|P|A|Q|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|R|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|S|8|@]|D|@]|E|$]]]|L|$]]

Replace <code>/</code> with <code>-</code>:

<pre><code>$date1 = strtr($_REQUEST['date'], '/', '-');
echo date('Y-m-d', strtotime($date1));
</code></pre>

blocks|key|669407|text|1970年1月1日是所谓的Unix时代。这是他们开始计算Unix+time的日期。如果将此日期作为返回值获取，则通常意味着从日期到Unix时间戳的转换返回了(接近)零的结果。所以日期转换不会成功。很可能是因为它接收了错误的输入。|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|669408|换句话说，您的strtotime($date1)返回0，这意味着$date1是以strtotime函数不支持的格式传递的。|style|CODE|669409|entityMap|0|LINK|mutability|MUTABLE|url|http://en.wikipedia.org/wiki/Unix_time^0|S|9|0|0|7|H|W|6|0^^$0|@$1|2|3|4|5|6|7|P|8|@]|9|@$A|Q|B|R|1|S]]|C|$]]|$1|D|3|E|5|6|7|T|8|@$A|U|B|V|F|G]|$A|W|B|X|F|G]]|9|@]|C|$]]|$1|H|3|-4|5|6|7|Y|8|@]|9|@]|C|$]]]|I|$J|$5|K|L|M|C|$N|O]]]]

January 1, 1970 is the so called Unix epoch. It's the date where they started counting the <a href="http://en.wikipedia.org/wiki/Unix_time">Unix time</a>. 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.

In other words, your <code>strtotime($date1)</code> returns 0, meaning that <code>$date1</code> is passed in an unsupported format for the strtotime function.

blocks|key|669418|text|最后，我找到了解决这个问题的一行代码。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|669419|date('d/m/Y',+strtotime(str_replace('.',+'-',+$row['DMT_DATE_DOCUMENT'])));|code-block|syntax|javascript|669420|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

finally i have found a one line code to solve this problem

<pre><code>date('d/m/Y', strtotime(str_replace('.', '-', $row['DMT_DATE_DOCUMENT'])));
</code></pre>

blocks|key|669450|text|$date1+=+$_REQUEST['date'];

if($date1)+{
++++$date1+=+date(+'Y-m-d',+strtotime($date1));
}+else+{
++++$date1+=+'';
}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|669451|当$date中存在有效的date()时，它将正确显示，如果没有，则不显示任何内容。|unstyled|offset|length|style|CODE|669452|为我解决了这个问题。|669453|entityMap^0|0|1|5|C|6|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|P|8|@$G|Q|H|R|I|J]|$G|S|H|T|I|J]]|9|@]|A|$]]|$1|K|3|L|5|F|7|U|8|@]|9|@]|A|$]]|$1|M|3|-4|5|F|7|V|8|@]|9|@]|A|$]]]|N|$]]

<pre><code>$date1 = $_REQUEST['date'];

if($date1) {
 $date1 = date( 'Y-m-d', strtotime($date1));
} else {
 $date1 = '';
}
</code></pre>

This will display properly when there is a valid <code>date()</code> in <code>$date</code> and display nothing if not. 
Solved the issue for me.

blocks|key|669465|text|$inputDate+=+'07/05/-0001';
$dateStrVal+=+strtotime($inputDate);
if(empty($dateStrVal))
{
++echo+'Given+date+is+wrong';+
}
else{
+echo+'Date+is+correct';
}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|669466|O/P+:给定日期错误|unstyled|669467|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|J|8|@]|9|@]|A|$]]|$1|G|3|-4|5|F|7|K|8|@]|9|@]|A|$]]]|H|$]]

<pre><code>$inputDate = '07/05/-0001';
$dateStrVal = strtotime($inputDate);
if(empty($dateStrVal))
{
 echo 'Given date is wrong'; 
}
else{
 echo 'Date is correct';
}
</code></pre>

O/P : Given date is wrong

blocks|key|455221|text|另一种解决方法：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|455222|将日期选择器dd/mm/yyyy转换为yyyy-mm-dd|offset|length|style|CODE|455223|$startDate+=+trim($_POST['startDate']);
$startDateArray+=+explode('/',$startDate);
$mysqlStartDate+=+$startDateArray[2]."-".$startDateArray[1]."-".$startDateArray[0];
$startDate+=+$mysqlStartDate;|code-block|syntax|javascript|455224|entityMap^0|0|6|A|J|A|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|P|8|@$D|Q|E|R|F|G]|$D|S|E|T|F|G]]|9|@]|A|$]]|$1|H|3|I|5|J|7|U|8|@]|9|@]|A|$K|L]]|$1|M|3|-4|5|6|7|V|8|@]|9|@]|A|$]]]|N|$]]

Another workaround:

Convert datepicker <code>dd/mm/yyyy</code> to <code>yyyy-mm-dd</code>

<pre><code>$startDate = trim($_POST['startDate']);
$startDateArray = explode('/',$startDate);
$mysqlStartDate = $startDateArray[2]."-".$startDateArray[1]."-".$startDateArray[0];
$startDate = $mysqlStartDate;
</code></pre>

blocks|key|455230|text|问题是当你的数据设置为000-00-00或空的时候，你必须仔细检查并给出正确的信息，这个问题就会消失。我希望这能帮到你。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|455231|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

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.

I have a form in which date format is <code>dd/mm/yyyy</code> . For searching database , I hanverted the date format to <code>yyyy-mm-dd</code> . But when I <code>echo</code> it, it showing <code>1970-01-01</code> . The PHP code is below:

<pre><code>$date1 = $_REQUEST['date']; 
echo date('Y-m-d', strtotime($date1));
</code></pre>

Why is it happening? How can I format it to <code>yyyy-mm-dd</code>?

PHP date showing '1970-01-01 ' after conversion

数据库

Unix

我有一个日期格式为dd/mm/yyyy的表单。为了搜索数据库，我把日期格式交给了yyyy-mm-dd。但当我echo它时，它显示的是1970-01-01。PHP代码如下：$date1 = $_REQUEST['date'];     echo date('Y-m-d', strtotime($date1));为什么会发...

问转换后显示'1970-01-01‘的PHP日期
EN

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