Question

我正在尝试创建一个可搜索的界面，我的后端包含三个MySQL数据库表：

tbl_country

mysql> select * from tbl_country;
+--------------+---------------+
| country_code | country       |
+--------------+---------------+
| AFG          | AFGHANISTAN   |
| AUS          | AUSTRALIA     |
| CAN          | CANADA        |
| GBR          | GREAT BRITAIN |
| IND          | INDIA         |
| USA          | USA           |
+--------------+---------------+
6 rows in set (0.00 sec)

tbl_state

mysql> select * from tbl_state;
+----------+------------------+--------------+
| state_id | state            | country_code |
+----------+------------------+--------------+
|        1 | Maharashtra      | IND          |
|        2 | Delhi            | IND          |
|        3 | West Bengali     | IND          |
............
|       51 | Queensland       | AUS          |
+----------+------------------+--------------+
33 rows in set (0.00 sec)

tbl_city

mysql> select * from tbl_city;
+---------+----------+--------------------+
| city_id | state_id | city               |
+---------+----------+--------------------+
|       1 |        1 | Mumbai (Bombay)    |
|       2 |        1 | Nagpur             |
.......
|      84 |       37 | Edinburgh          |
|     122 |       44 | Cardiff            |
|     127 |       46 | Melbourne          |
|     130 |       48 | Perth              |
|     131 |       49 | Adelaide           |
|     132 |       50 | Canberra           |
|     133 |       51 | Brisbane           |
|     134 |       51 | Gold Coast         |
+---------+----------+--------------------+
54 rows in set (0.00 sec)

我的MySQL数据库架构和值如SQL Fiddle

所示

我的搜索需要从我可以根据任何给定城市或州详细信息搜索所有国家/地区获取单个界面。

我正在尝试使用这个正确提供我想要获取的输出的查询：

SELECT 
-- tbl_country
c.country_code, c.country,

-- tbl_state
CAST(GROUP_CONCAT(s.state SEPARATOR ',') AS CHAR) as state,
-- tbl_city
CAST(GROUP_CONCAT(ct.city SEPARATOR ',') AS CHAR) as city

-- tbl_papers
FROM tbl_country c
LEFT JOIN tbl_state s ON s.country_code = c.country_code
LEFT JOIN tbl_city ct ON ct.state_id = s.state_id

GROUP BY (c.country_code);

我正在使用Datatable插件，它非常强大且易于使用，但每当我尝试使用搜索过滤器时，它现在将作为数据表的Server Side PHP class尝试创建一个动态查询，其中where条件基于绑定列。

查询绑定数据表，如第一行所示，因为没有搜索：

动态查询是这样的：

SELECT    sql_calc_found_rows `c`.`country_code`,
          `c`.`country`,
          group_concat(`s`.`state` separator "," ) AS state,
          group_concat(`ct`.`city` separator ",")  AS city
FROM      `tbl_country`                            AS `c`
LEFT JOIN `tbl_state`                              AS `s`
ON        (
                    `s`.`country_code` = `c`.`country_code`)
LEFT JOIN `tbl_city` AS`ct`
ON        `ct`.`state_id` = `s`.`state_id`
WHERE     (
                    `c`.`country_code` LIKE :binding_0
          OR        `c`.`country` LIKE :binding_1
          OR        group_concat(`s`.`state` separator ",") LIKE :binding_2
          OR        group_concat(`ct`.`city` separator ",") LIKE :binding_3)
GROUP BY  `c`.`country_code`
ORDER BY  `c`.`country_code` ASC limit 0,
          10

这显然会给出如下所示的错误：

无论如何，我可以使用PHP和datatable插件根据上述场景中给定的城市和州和国家详细信息进行过滤/搜索。可能是我需要更改我的查询结构！

Answer 1

我假设您正在使用ssp.class.php来处理服务器端的数据。

类$table不支持包含JOIN和子查询的复杂查询，但有一种解决方法。诀窍是使用如$table = <<<EOT ( SELECT c.country_code, c.country, CAST(GROUP_CONCAT(s.state SEPARATOR ',') AS CHAR) as state, CAST(GROUP_CONCAT(ct.city SEPARATOR ',') AS CHAR) as city FROM tbl_country c LEFT JOIN tbl_state s ON s.country_code = c.country_code LEFT JOIN tbl_city ct ON ct.state_id = s.state_id GROUP BY (c.country_code) ) temp EOT; $primaryKey = 'id'; $columns = array( array( 'db' => 'country_code', 'dt' => 0 ), array( 'db' => 'country', 'dt' => 1 ), array( 'db' => 'state', 'dt' => 2 ), array( 'db' => 'city', 'dt' => 3 ) ); $sql_details = array( 'user' => '', 'pass' => '', 'db' => '', 'host' => '' ); require( 'ssp.class.php' ); echo json_encode( SSP::simple( $_GET, $sql_details, $table, $primaryKey, $columns ) );定义中所示的子查询。

ssp.class.php

您还需要修改FROM `$table`并将FROM $table的所有实例替换为ssp.class.php以删除反引号。

还有github.com/emran/ssp存储库，其中包含支持JOIN的增强型it('should associate an employee with the company', (done) => { var employee = new Employee(); company.employees.push(employee); request(app) .put(`/api/company/${company._id}`) .send(company) .expect(httpStatus.OK) .then((res) => { expect(res.body.employees).to.be.an('array') expect(res.body.employees).to.contain(employee.id) done(); }) .catch(done); });。

LINKS

有关详细信息，请参阅jQuery DataTables: Using WHERE, JOIN and GROUP BY with ssp.class.php。

如何使用jQuery datatable插件实现这种复杂的搜索？

1 个答案:

LINKS