php - 使用第一个查询中的变量从第二个查询中获取结果时出错

Question

我正确地从第一个查询中获得结果但是，当我想在 WHERE 条件下的第二个查询中使用它们时，我收到查询错误 1064。如果我删除 WHERE 它将正常工作。另外，当我尝试在第二个查询代码中回显内部变量时，它将打印。这些变量只能在第二个查询的 WHERE 中工作

	$queryDate = "SELECT date , time from DATES where ID = $ID";
	$result = mysqli_query($connection, $queryDate);
	
	if ($result->num_rows > 0) {
	     while($row = $result->fetch_assoc()) {
		
			$date = $row['date']; 
			$time = $row['time'];

		 } 
	}
	
	$queryCom = "SELECT * from DATES, BOOKING where dates.time = $time and booking.IDofFullDate= $date";
	$result1 = mysqli_query($connection, $queryCom);
	if (!$result1)
				{
	
					die("Query Faile".  mysqli_errno($connection));   
				}
	
	if ($result1->num_rows >0) {
			  echo $date;
			  echo $time;
	}

Answer 1

在您的第二个查询中尝试以下操作:

$queryCom = "SELECT * from DATES as dates, BOOKING as booking where dates.time = $time and booking.IDofFullDate= $date";