Question

嗨，我有以下表格 - Location Schedule

我希望通过location_id根据时间表从此表中获取campaign_id，如果根据当前时间表找不到campaign_id，则我需要默认的campaign_id（默认值> 0）。我需要你的帮助。如果你给我建议任何优化表，我会改变我的表。

抱歉我的英语不好。

Answer 1

未经测试。我认为它的帮助完全适合你

select * from (
               select *,count(*) as cnt,1 as ord from table1 t 
               inner join schedule s on s.location_id=t.location_id 
               and now() between start_date and end_date
               union all
               select *,count(*) as cnt,1 as ord from table1 t 
               inner join schedule s on s.location_id=t.location_id 
               and default_id > 0
             ) as tt where ord= case when (ord=1 and cnt > 0) then 1 else 2 end

Answer 2

DB：

mysql> select * from schedule_table;
+----+-------------+-------------+
| id | campaign_id | location_id |
+----+-------------+-------------+
|  1 |           5 |          10 |
|  2 |          10 |           5 |
|  3 |           7 |          11 |
|  4 |           6 |          12 |
|  5 |           8 |          13 |
+----+-------------+-------------+
5 rows in set (0.00 sec)

x.php

<?php

$id = $argv[1];
$dbh = new PDO('mysql:dbname=test;host=127.0.0.1', 'root');
$sth = $dbh->prepare(
    '
    SELECT
        d.campaign_id AS default_id,
        a.campaign_id AS actual_id
    FROM schedule_table AS d
    LEFT JOIN schedule_table As a ON a.location_id = :id
    WHERE
        d.campaign_id > 0
    ORDER BY d.campaign_id
    LIMIT 1
    '
);
$sth->bindParam(':id', $id, PDO::PARAM_INT);
$sth->execute();
$result = $sth->fetch(PDO::FETCH_ASSOC);
var_export($result);

结果：

php x.php 12
    array (
  'default_id' => '5',
  'actual_id' => '6',
)
php x.php 125
    array (
  'default_id' => '5',
  'actual_id' => NULL,
)

根据计划获取数据

2 个答案: