douwen1313 2019-04-05 09:14
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如何使用php从数据库中提取日期

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.

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1条回答 默认 最新

  • dousa1630 2019-04-05 09:28
    关注

    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'));
    
    本回答被题主选为最佳回答 , 对您是否有帮助呢?
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