我想显示一个表与PHP MYSQL的FIELD是日期和行是TIME从MySQL数据库与未知/无限记录,一个与不同的时间为同一日期,通过查询它的约会时间。多个日期日期时间
我的mysql日期将dateTime存储在同一列中,但我正在分割它并尝试单独显示它们。但我似乎无法显示日期只有一次和多次,这只是两个。
$sql_result = mysqli_query($connection, "SELECT DATE(date_time) AS date_part, TIME(date_time) AS time_part FROM $table WHERE date_time LIKE '$date_input%'");
if (mysqli_num_rows($sql_result) == 0)
{
echo "<p>No bookings exist.</p>";
}
else {
echo "<h3>Results for booked " . $table . " Appointments:</h3>";
echo "<h3>" . $formattedDate ."</h3>";
while ($row = mysqli_fetch_array($sql_result))
{
echo $row['date_part'];
$array_time = array($row['time_part']);
foreach ($array_time as $time_output)
{
echo $time_output;
}
}
}
我的输出是这样的:
2013-12-0809:00:002013-12-0810:00:002013-12-0811:00:002013-12-0812:00:002013-12-0814:00:002013-12-0815:00:002013-12-0816:00:002013-12-0817:00:002013-12-0909:00:002013-12-0809:00:00
不过,我想这样的:
2013-12-08 09:00:0010:00:0011:00:0012:00:0014:00:0015:00:0016:00:0017:00:0009:00:000
2013-12-09 9:00:00
你的逻辑非常棒。这工作完美。尽管第一部分我不得不将“time_part”输入$ array [$ row ['date_part']] [$ row ['TIME_PART']] = 1;为它执行(否则错误消息:无效索引)。非常感谢您的帮助! – ESavage
噢对不起,很高兴你知道它,我忘了写在单引号:) –