Question

您好我试图将sql的日期输出从yyyy / mm / dd更改为dd / mm / yyyy

当我通过表单上的日期选择器输入日期时，日期不会保存在mySQL中。（wil输入only 0000-00-00）

如果我手动将日期输入MySQL，那么它只会显示今天的日期。

我正在使用以下代码：

$result = mysql_query("SELECT * FROM overboekingen") 
while($row = mysql_fetch_array( $result )) {
echo '<td>' . $row['datum_overboeking'] = date('d-m-Y') . '</td>';

Answer 1

试试这种类型的解决方案

$TodayDate = "2014-03-07"; //OR the date from database. It means the date you will get from database
$FormatedDate = date("d-m-Y", strtotime($TodayDate));

Answer 2

这样你可以把条件

$result = mysql_query("SELECT * FROM overboekingen") 
while($row = mysql_fetch_array( $result ))
{
   $databaseDate = $row['YOUR DATABASE DATE FIELD NAME']; 
   $FormatedDate = date("d-m-Y", strtotime($databaseDate));
   echo '<td>' . $row['datum_overboeking'] = $FormatedDate . '</td>';
}

Answer 3

我更进了一步，我现在可以在数据库中正确地插入数据了，只有当我在我的web应用程序中回显它再次将它拉出SQL时，它才能正确显示mySQL的日期。

这段代码似乎不正确：

echo '<td>' . $row['datum_overboeking'] = strtotime($row), date('m-d-Y') . '</td>'

Answer 4

尝试使用此过程以数据库显示具有所需日期格式的日期

$result = mysql_query("SELECT * FROM overboekingen") 
while($row = mysql_fetch_array( $result ))
{
   $databaseDate = $row['YOUR DATABASE DATE FIELD NAME']; 
   $FormatedDate = date("d-m-Y", strtotime($databaseDate));
   echo '<td>' . $row['datum_overboeking'] = $FormatedDate . '</td>';
}

Answer 5

愿这会帮到你

$TodayDate = "2014-03-07"; //OR the date from database. It means the date you will get from database
$FormatedDate = date("d-m-Y", strtotime($TodayDate));

将日期输出从sql从yyyy / mm / dd更改为dd / mm / yyyy

5 个答案: