SELECT * FROM table WHERE * LIKE (SELECT * FROM table WHERE * = *)

Question

我有一個單獨的城市表，其中有城市代碼，我還有一個主表，其中包含一個“地址”列，它只是一個短地址。 我想要的是 select 代碼類似於“地址”列中數據的城市。

$cityCode=$db->query("SELECT city AS bot FROM city_table WHERE city LIKE (SELECT address FROM people WHERE people_id = $zz)");

$cityCode=$cityCode->num_rows > 0 ? $cityCode->fetch_array()['bot'] : "NOT LIKE";

city_table:

| province                | city                     |
| ----------------------- | ------------------------ |
| ILOCOS NORTE/012800000  | CITY OF BATAC/012805000  |

people:

| people_id               | address                  |
| ----------------------- | ------------------------ |
| 1                       | P-2, Brgy. 20, Batac City|

If the address contains "Batac", I want to echo it as 'CITY OF BATAC/012805000'

如何使這項工作？

Answer 1

您需要在城市周圍添加%通配符，並將其用作LIKE模式以與address匹配。

$cityCode=$db->query("
    SELECT c.city AS bot 
    FROM city_table AS c
    JOIN people AS p ON p.address LIKE CONCAT('%', c.city, '%')
    WHERE p.people_id = $zz");

Answer 2

像這樣在 where 子句中使用Exists ：

SELECT city AS bot FROM city_table c 
WHERE EXISTS  (SELECT 1 FROM people WHERE people_id = $zz AND address LIKE '%'+c.city+'%' )

Answer 3

$cityCode=$db->query("SELECT city AS bot,

 FROM city_table WHERE city LIKE %(SELECT address FROM people WHERE 
 people_id = $zz)%

我已經更新了解決方案。 希望它有效。