Question

我需要排除第二个查询（来自第二个表的数据），类似于MINUS。但由于MySQL不支持MINUS，我决定选择LEFT JOIN。但它给我一个数据库错误。

enter image description here

模型

function searchRooms($room_type, $start_date, $end_date) {
// Line 6 is the below one:
    $query = $this->db->query(
        "SELECT     room_number
        FROM        rooms
        WHERE       room_type = '$room_type'
        ORDER BY    room_number ASC

        LEFT JOIN

        SELECT      room_number
        FROM        new_reservation
        WHERE       start_date >= '$start_date'
        AND         end_date <= '$end_date'
        ORDER BY    room_number ASC
        ");
    return $query->result();

}

控制器

function search() {
        $room_type = $this->input->post('room_type');
        $start_date = $this->input->post('start_date');
        $end_date = $this->input->post('end_date');
        $this->reservations_model->searchRooms($room_type, $start_date, $end_date);

        $this->index();
}

Answer 1

LEFT JOIN的语法错误，应该是以下形式：

SELECT     room_number
FROM        rooms
LEFT JOIN .... ON ...
WHERE  ...
ORDER BY ...

或者您在寻找UNION而不是LEFT JOIN？

虽然在你的情况下你可能只是反转第二个查询的条件，并在第一个查询中将其用作WHERE条件（如果我理解正确的话......）。

Answer 2

如果要从第二个查询中排除结果，那么

  SELECT room_number
    FROM rooms
   WHERE room_type = '$room_type'
     AND room_number not in (SELECT room_number
                          FROM new_reservation
                         WHERE start_date >= '$start_date'
                           AND end_date <= '$end_date')
ORDER BY room_number ASC

使用join而不是子查询

  SELECT a.room_number
    FROM rooms a left outer join (SELECT room_number
                          FROM new_reservation
                         WHERE start_date >= '$start_date'
                           AND end_date <= '$end_date') b
      ON a.room_number = b.room_number
   WHERE a.room_type = '$room_type'
     AND b.room_number is null
ORDER BY a.room_number ASC

如果您不想使用内嵌视图，请再次

  SELECT a.room_number
    FROM rooms a left outer join new_reservation b
      ON a.room_number = b.room_number
   WHERE a.room_type = '$room_type'
     AND b.start_date >= '$start_date'
     AND b.end_date <= '$end_date'
     AND b.room_number is null
ORDER BY a.room_number ASC

Answer 3

在MySQL中，您是JOIN表，而不是SELECT语句。当JOINed表中可能没有匹配的行时，使用LEFT JOIN。

您想要加入并仅选择所需的行。

试试这个：

SELECT rooms.room_number
FROM rooms
JOIN new_reservation ON rooms.room_number = new_reservation.room_number
WHERE rooms.room_type = '$room_type'
AND NOT (start_date >= '$start_date' AND end_date <= '$end_date')
ORDER BY room_number ASC

Answer 4

在codeigniter中你可以像这样使用

$this->db->query("select tb.required_field,tb.another_require_field from table_name as tb
    left join another_table_name as atb on atb.common_column = tb.common_column 
      where tb.conditional_colun= condition");

这是我在我的应用程序中使用的语法

CodeIgniter：$ this-＆gt; db-＆gt;查询中的LEFT JOIN返回数据库错误

4 个答案: