如何使用php从数据库中提取日期
问题描述：
I'm trying to create a weekly work schedule. When I try to select the shifts from the database my date value is selecting the current time but the date (Y-m-d) from the table.
This is my code:
while ($row = mysqli_fetch_array($res))
{
$dt = new DateTime();
$dt->setISODate($year, $week);
$fetch_mID = $row['ID_EMPLOYEE'];
$fetch_fn = $row['Firstname'];
$fetch_en = $row['Lastname'];
echo "<tr>";
echo "<td>" . $fetch_fn . " " . $fetch_en . "</td>";
do {
$obj = new ReflectionObject($dt);
$pro = $obj->getProperty('date');
$date = $pro->getValue($dt);
echo $date; //Output = 'Y-m-d H:m:s'
$shift = $conn->query("SELECT ShiftDate, ShiftStart, ShiftEnd FROM shifts WHERE ID_EMPLOYEE = '$fetch_mID' AND Date(ShiftDate) = '$date'");
$fetch = mysqli_fetch_array($shift);
$dt->modify('+1 day');
} while ($week == $dt->format('W'));
I would expect the $date to output Y-m-d but it outputs Y-m-d H:m:s. And the time is the current time.
答
The solution is to convert $date to date object and then format the date with the new format Y-m-d.
$time = strtotime($date);
$newformat = date('Y-m-d',$time);
echo $newformat; // output 2019-04-05
You new code seems to b something like :
while ($row = mysqli_fetch_array($res))
{
$dt = new DateTime();
$dt->setISODate($year, $week);
$fetch_mID = $row['ID_EMPLOYEE'];
$fetch_fn = $row['Firstname'];
$fetch_en = $row['Lastname'];
echo "<tr>";
echo "<td>" . $fetch_fn . " " . $fetch_en . "</td>";
do {
$obj = new ReflectionObject($dt);
$pro = $obj->getProperty('date');
$date = $pro->getValue($dt);
//new code
$newformat = date('Y-m-d',strtotime($date));
echo $newformat; // output 2019-04-05
// end new code
$shift = $conn->query("SELECT ShiftDate, ShiftStart, ShiftEnd FROM shifts WHERE ID_EMPLOYEE = '$fetch_mID' AND Date(ShiftDate) = '$date'");
$fetch = mysqli_fetch_array($shift);
$dt->modify('+1 day');
} while ($week == $dt->format('W'));