因为您没有提供表架构，所以我必须使用占位符列名等。
如前面的注解所述，您可以使用mysql函数 BETWEEN 看看是否有东西在给定的时间之间（也适用于日期）。它的作用是返回一个布尔值。如果为真，则为1；如果为假，则为0。
假设您的表有一个日期行，以便查看特定日期的预订，您可以编写以下示例：

$sql="SELECT
        COUNT(column_id)
      FROM
        table_name
      WHERE 
        TIME($orario_inizio) BETWEEN '10:30' AND '12:30'
        AND date_column=CURDATE()";
$result_set=mysqli_query($conn, $sql);
$checkIfValid=mysqli_fetch_array($conn, $result_set)[0];

if($checkIfValid > 0) {
    //throw error
}
else {
    //perform insert query
}

你可以调整逻辑，让它更符合你的喜好。例如， CURDATE() 只需要当前日期，但也许您有一个特定的预订日期，您可以使用？
这个sql基本上是根据时间在上午10:30到12:30之间，并且日期是今天（可以是给定的预订日期）来计算您选择的id（用户id或诸如此类）。
现在，如果这个sql找到任何记录，它将被放入计数中。所以它的值不是0就是更高。如果计数大于0，则表示我们正忙，或者应用了逻辑，这意味着您无法预订该天/时间段，因此我们将抛出一个错误。但是，如果计数不大于0，则继续 INSERT 因为那时一切都会好起来的。

考虑以下几点：

DROP TABLE IF EXISTS my_table;

CREATE TABLE my_table(booking_id SERIAL PRIMARY KEY,booking_start INT NOT NULL);

INSERT INTO my_table (booking_start) VALUES (3),(7),(10);

SET @n = 9;

INSERT INTO my_table (booking_start)
SELECT @n
  FROM (SELECT 1) x
  LEFT
  JOIN my_table y
    ON y.booking_start BETWEEN @n AND @n+2
 WHERE y.booking_start IS NULL
 LIMIT 1;
Query OK, 0 rows affected (0.01 sec)
Records: 0  Duplicates: 0  Warnings: 0

SELECT * FROM my_table;
+------------+---------------+
| booking_id | booking_start |
+------------+---------------+
|          1 |             3 |
|          2 |             7 |
|          3 |            10 |
+------------+---------------+
3 rows in set (0.00 sec)

-- Nothing happens, because 10 is between 9 and 11.

SET @n = 8;

INSERT INTO my_table (booking_start)
SELECT @n
  FROM (SELECT 1) x
  LEFT
  JOIN my_table y
    ON y.booking_start BETWEEN @n AND @n+2
 WHERE y.booking_start IS NULL
 LIMIT 1;

SELECT * FROM my_table;
+------------+---------------+
| booking_id | booking_start |
+------------+---------------+
|          1 |             3 |
|          2 |             7 |
|          3 |            10 |
+------------+---------------+
3 rows in set (0.00 sec)

-- Nothing happens, because 10 is between 8 and 10.

SET @n = 7;

INSERT INTO my_table (booking_start)
SELECT @n
  FROM (SELECT 1) x
  LEFT
  JOIN my_table y
    ON y.booking_start BETWEEN @n AND @n+2
 WHERE y.booking_start IS NULL
 LIMIT 1;

SELECT * FROM my_table;
+------------+---------------+
| booking_id | booking_start |
+------------+---------------+
|          1 |             3 |
|          2 |             7 |
|          3 |            10 |
+------------+---------------+
3 rows in set (0.00 sec)

-- Nothing happens, because 7 is between 7 and 9.

SET @n = 4;

INSERT INTO my_table (booking_start)
SELECT @n
  FROM (SELECT 1) x
  LEFT
  JOIN my_table y
    ON y.booking_start BETWEEN @n AND @n+2
 WHERE y.booking_start IS NULL
 LIMIT 1;
Query OK, 1 row affected (0.00 sec)
Records: 1  Duplicates: 0  Warnings: 0

SELECT * FROM my_table;
+------------+---------------+
| booking_id | booking_start |
+------------+---------------+
|          1 |             3 |
|          2 |             7 |
|          3 |            10 |
|          4 |             4 |
+------------+---------------+
4 rows in set (0.00 sec)

-- Success, because there is no value between 4 and 6.