PHP 回显空白日期时间字段

Question

这是我的困境。我尝试引用 http://php.net/manual/en/datetime.format.php

但这对我没有帮助，或者我没有正确设置我的代码。

这是我的代码。

$query =  "SELECT id,f_name, l_name, phone_1, phone_2, car_name, email, zipcode, ref_name, dateadded ";
$query .= "FROM customers JOIN carriers ON customers.carrier = carriers.car_id ";
$query .= "JOIN referrals ON customers.referral = referrals.ref_id";        

$select_customers = mysqli_query($connection, $query);

while($row = mysqli_fetch_assoc($select_customers)) {
$customer_id = $row['id'];    
$c_fname     = $row['f_name'];
$c_lname     = $row['l_name'];
$c_email     = $row['email'];
$c_phone1    = $row['phone_1'];
$c_phone2    = $row['phone_2'];
$car_id      = $row['car_id'];    
$c_carrier   = $row['car_name'];
$c_zip       = $row['zipcode'];
$ref_id      = $row['ref_id'];
$c_ref       = $row['ref_name'];
$c_added     = $row['dateadded'];

echo "<tr>";
echo "<td>{$c_fname}</td>";
echo "<td>{$c_lname}</td>";
echo "<td>{$c_email}</td>";
echo "<td>{$c_phone1}</td>";
echo "<td>{$c_phone2}</td>";
echo "<td>{$c_carrier}</td>";
echo "<td>{$c_zip}</td>";
echo "<td>{$c_ref}</td>";   
echo "<td>". date_format($c_added, 'd/m/y') ."</td>";
echo "<td><a class='btn btn-danger' href='customers.php?source=edit_customer&id={$customer_id}'>Edit</a></td>";
}    

它将回显一个空白字段。 “dateadded”的 col 设置为“2016-09-28 10:40:44”的日期时间和值。感谢对我的代码的更正。谢谢。

Answer 1

有很多方法可以做到这一点，这是之前答案的另一种方法:

echo "<td>". date_format(new DateTime($c_added), 'd/m/y') ."</td>";