Question

首先，感谢您的家伙花时间阅读本文。

现在我有两个表，数据如下：

表名：table1

+-----------+----------+---------------------------+
| tid       | area     | Subject                   |
+-----------+----------+---------------------------+
| 1         | US       | The one restaurant        |
| 2         | US       | Landmark hotel            |
| 3         | US       | Tholo restaurant          |
| 4         | CA       | GE bar                    |
+-----------+----------+---------------------------+

表名：table2

+--------+---------+---------------------------+---------------+
| tid    | area    | Value                     | optionid      |
+--------+---------+---------------------------+---------------+
| 1      | US      | the one restaurant desc   | restaurant    |
| 1      | US      | the one rest. contact     | recontact     |
| 1      | US      | the one rest. address     | readdress     |
| 2      | US      | landmark hotel desc       | hotel         |
| 2      | US      | landmark hotel.contact    | hocontact     |
| 2      | US      | landmark hotel.address    | hoaddress     |
| 3      | US      | Tholo restaurant.desc     | restaurant    |
| 3      | US      | Tholo restaurant.contact  | recontact     |
| 3      | US      | Tholo restaurant.address  | readdress     |
| 4      | CA      | GE bar.desc               | bar           |
| 4      | CA      | GE bar.contact            | bacontact     |
| 4      | CA      | GE bar.address            | baaddress     |
+--------+---------+---------------------------+---------------+

我想在用户查询区域= US

时显示如下数据

|tid  | Subject              | description             | contact                     |    Address               |area|
+-----+----------------------+-------------------------+-----------------------------+--------------------------+----+
| 1   | The one restaurant   | the one restaurant desc | the one rest.contact        | the one rest.address     | US |
| 2   | Landmark hotel       | landmark hotel desc     | landmark hotel contact      | landmark.address         | US |
| 3   | Tholo restaurant     | Tholo restaurant.desc   | Tholo restaurant.contact    | Thoro restaurant.address | US |

现在我只能成功别名和联盟，不能加入主题，我的代码如下：

$query = mysql_query("SELECT value AS restaurant_description FROM table2 WHERE fid = US AND optionid = restaurant UNION SELECT value AS restaurant_contact FROM table2 WHERE fid = US AND optionid = recontact UNION SELECT value AS restaurant_address FROM table2 WHERE fid = US AND optionid = readdress UNION SELECT value AS hotel FROM table2 WHERE fid = US AND optionid = hotel UNION SELECT value AS hotel_description FROM table2 WHERE fid = US AND optionid = hotel UNION SELECT value AS hotel_contact FROM table2 WHERE fid = US AND optionid = hocontact UNION SELECT value AS hotel_address FROM table2 WHERE fid = US AND optionid = hoaddress UNION SELECT value AS bar_description FROM table2 WHERE fid = US AND optionid = bar UNION SELECT value AS bar_contact FROM table2 WHERE fid = US AND optionid = bacontact UNION SELECT value AS bar_address FROM table2 WHERE fid = US AND optionid = baaddress"); 

    while($row = mysql_fetch_array($query)) {
        echo $row['hotel_description'];
        echo $row['hotel_lat'];
        echo $row['subject'];
}

我已尝试下面的代码，但它无法显示我需要的数据。

$query = mysql_query("SELECT value AS restaurant_description FROM table2 WHERE fid = US AND optionid = restaurant UNION SELECT value AS restaurant_contact FROM table2 WHERE fid = US AND optionid = recontact UNION SELECT value AS restaurant_address FROM table2 WHERE fid = US AND optionid = readdress UNION SELECT value AS hotel FROM table2 WHERE fid = US AND optionid = hotel UNION SELECT value AS hotel_description FROM table2 WHERE fid = US AND optionid = hotel UNION SELECT value AS hotel_contact FROM table2 WHERE fid = US AND optionid = hocontact UNION SELECT value AS hotel_address FROM table2 WHERE fid = US AND optionid = hoaddress UNION SELECT value AS bar_description FROM table2 WHERE fid = US AND optionid = bar UNION SELECT value AS bar_contact FROM table2 WHERE fid = US AND optionid = bacontact UNION SELECT value AS bar_address FROM table2 WHERE fid = US AND optionid = baaddress UNION SELECT subject FROM table1 WHERE fid = US AND table1.tid = table2.tid");

Answer 1

这更像是创建pivot表。如果您只需要recontact，restaurant和readdress值，则可以使用以下技术。

select
t1.tid,
t1.subject,
t2.area,
max(case when t2.optionid = 'restaurant' then t2.Value end) as `description`,
max(case when t2.optionid = 'recontact' then t2.Value end ) as `contact`,
max(case when t2.optionid = 'readdress' then t2.Value end ) as `Address`
from table1 t1
join table2 t2 on t1.tid = t2.tid
group by t1.tid;

Answer 2

$query = mysql_query("SELECT value AS restaurant_description FROM table2 WHERE `fid` ="US" AND `optionid` = "restaurant" UNION SELECT value AS restaurant_contact FROM table2 WHERE `fid` = "US" AND `optionid` = "recontact" UNION SELECT value AS restaurant_address FROM `table2` WHERE `fid` ="US" AND `optionid` = "readdress" UNION SELECT value AS hotel FROM `table2` WHERE `fid` = "US" AND `optionid` = "hotel" UNION SELECT value AS hotel_description FROM `table2` WHERE `fid` = "US" AND `optionid` = "hotel" UNION SELECT value AS hotel_contact FROM `table2` WHERE `fid` = "US" AND `optionid` = "hocontact" UNION SELECT value AS hotel_address FROM `table2` WHERE `fid` = "US" AND `optionid` = "hoaddress" UNION SELECT value AS bar_description FROM `table2` WHERE `fid` = "US" AND `optionid` = "bar" UNION SELECT value AS bar_contact FROM table2 WHERE `fid` = "US" AND `optionid` = bacontact UNION SELECT value AS bar_address FROM `table2` WHERE `fid` ="US" AND `optionid` = "baaddress" UNION SELECT subject FROM `table1` WHERE `fid` = "US" AND `table1`.tid = `table2`.tid");

首先像这样更改您的查询并在显示任何错误时回复它，而不是让我知道

php如何使用Condition获取多个表列值

2 个答案: