我正在制作用户进行预订的日历系统(在不同的页面上)。在此页面上,用户单击相关框以选择新预订的开始时间。我正在检查数据库中是否有现有的预订,并使那些无法点击以防止双重预订。
我的问题是,为什么当天只有第一张预订出现在桌面上?
变量$booked
是否仍为真,因此不显示其他预订?我无法理解它!
MySQL的:
$query="SELECT * FROM bookings WHERE DateBooked = '{$year}-{$month}-{$selectedday}' AND Approved = 1";
$result = mysql_query($query);
$todayarray = mysql_fetch_assoc($result);
和PHP:
while ($room <= $roomcount) {
echo "\n<div class=\"roomtimes\">";
echo "\n<table border=1>";
echo "\n<tr><th class=\"titlecell\">Room $room</th></tr>";
$cellnum = 10;
while ($cellnum <= 22) {
if (($todayarray['StartTime'] <= $cellnum) && ($todayarray['EndTime'] >= $cellnum) && ($todayarray['Room'] == $room)) {
$booked = true;
} else {
$booked = false;
}
echo "\n<tr>";
if ($booked) {
echo "\n<td class=\"blankcell";
} else {
echo "\n<td class=\"linkcell";
}
if ($selectedtime == $cellnum) {
echo " selectedcell";
}
echo "\">";
if ($booked) {
echo "$cellnum:00 --BOOKED--";
} else {
echo "<a href=\"newbooking.php?m=$selectedmonth&d=$selectedday&t=$cellnum&r=$room\">$cellnum:00</a>";
}
echo "</td>";
echo "\n</tr>";
$cellnum++;
}
$room++;
echo "\n</table>";
echo "\n</div>";
}
echo "\n</div>";
?>
答案 0 :(得分:0)
问题是您只是从表中查看结果集的第一行。您只需调用mysql_fetch_assoc($result)
一次,此函数一次只返回一个结果行。您需要遍历从查询调用中获得的$result
。您可以使用类似于此的while循环:
while ($row = mysql_fetch_assoc($result) {
//this can be your php code from above
}