Displaying date from mysql in php - php

Question

So i want to get one and only the first value from a date column in my database, and put it in my PHP file, this is the snippet of the code to get the date:
$one = 1;
$Lihat="SELECT * FROM event where status = '$one'";
$Tampil = mysqli_query($db, $Lihat);
while ( $hasil = mysqli_fetch_array ($Tampil)) {
$query_start_date = "SELECT tgl_event
FROM jadwal_acara
WHERE id_event = '$id_event' AND status = $one
ORDER BY tgl_event ASC
LIMIT 1";
$result = mysqli_query($db, $query_start_date);
$start_date = mysqli_fetch_field($result);
<td class="tgl_mulai"><?php echo date('d-m-Y', strtotime($start_date));?></td>
The problem is that when i tried to echo it in my table row:
It returns an error that said:
strtotime() expects parameter 1 to be string, object given.
Does anyone know how to change it into a string or another method to get the desired display result? Thank you.
EDIT:
$start_date = var_dump($start_date); returns a NULL value when i tested it.
Notice: Undefined property: stdClass::$tgl_event in
NULL
But i got other rows that displayed the correct name and event ID, it's just the date that won't display correctly.
it now displays :01-01-1970

Answer 1

mysqli_fetch_field() returns an object. Instead use mysqli_fetch_assoc
Try this:
if (mysqli_connect_errno()) {
printf("Connect failed: %s\n", mysqli_connect_error());
exit();
}
$query = "SELECT tgl_event FROM jadwal_acara WHERE id_event = ".$id_event." AND status = 1 ORDER BY tgl_event ASC LIMIT 1";
if ($result = mysqli_query($db, $query)) {
while ($finfo = mysqli_fetch_assoc($result)) {
$date = date('m-d-Y', strtotime($finfo['tgl_event']));
}
mysqli_free_result($result);
}
mysqli_close($db);
And then in your HTML:
<td class="tgl_mulai"><?php echo $date; ?></td>