我使用PHP變量在JScript上輸出$ date =('2016-12-07'),它必須輸出12-07-16,而不是我在JScript alert上收到1997年,而且我也嘗試過現在的日期是('Ym-d'),但仍然收到了1997年。我該如何解決這些問題,因爲我無法在桌子上查詢現在的日期。date返回錯誤的結果在php
<?php
$mysqli = new mysqli('10.237.2.152','root','c0k3float','test',3306);
//Output any connection error
if ($mysqli->connect_error) {
die('Error : ('. $mysqli->connect_errno .') '. $mysqli->connect_error);
}
$date = date('2016-12-07');
echo "<script> alert(".$date."); </script>";
$results = $mysqli->query("SELECT * FROM test.remaining_ham where ddate=".$date." ") or mysqli0;
$count = $results->num_rows;
echo "<table class='datatable table' style='border-spacing: 0 5px;'>
<thead>
<tr>
<th>Emp_Name</th>
<th>Stub</th>
<th>Brickham</th>
<th>Jamon</th>
<th>Fiesta</th>
<th>Total</th>
<th>Status</th>
<th>ddate</th>
</tr>
</thead>";
echo "<tbody>";
while($row = $results->fetch_array()) {
echo "<tr>";
echo "<td>" . $row['Emp_Name'] . "</td>";
echo "<td>" . $row['Stub'] . "</td>";
echo "<td>" . $row['Brickham'] ."</td>";
echo "<td>" . $row['Jamon'] . "</td>";
echo "<td>" . $row['Fiesta'] . "</td>";
echo "<td>" . $row['Total'] . "</td>";
echo "<td>" . $row['status'] ."</td>";
echo "<td>" . $row['ddate'] . "</td>";
echo "</tr>";
}
echo "</tbody>
</table>";
mysqli_close($con);
?>
它看起來像你使用[日期](http://php.net/manual/en/function.date.php)錯誤。第一個參數是你想要的字符串返回的格式。 – Terminus
但我看到它在PHP手冊只是設置格式將輸出當前日期和時間 –
它會!但是,從您發佈的代碼('$ date = date('2016-12-07');')不設置格式。請參閱文檔中已識別字符的表格和示例。 – Terminus