我正在尝试在 Sphinx 中实现过滤器，以便我可以从同一个连接表中过滤掉两个 MVA 属性的结果（我需要的是双重条件，但因为 MVA 只有 2 个字段..），但无论我做什么，似乎我一直得到相同的结果，因为在多属性的第一个字段上总是有一个匹配项？
也许还有另一种解决方案，但我似乎无法找到我想要的结果，我必须简化 4 个表：产品、类别、价格和位置。
数据设置
下面是数据结构的示例：
PRODUCT
+------+-------------+-----------------+---------------+
| ID | CATEGORY_ID | MANUFACTURER_ID | TITLE |
+------+-------------+-----------------+---------------+
| 1000 | 1000 | 1000 | Apple iPhone |
| 1001 | 1000 | 1000 | Apple iPad |
| 1002 | 1000 | 1000 | Apple iPod |
| 1003 | 1001 | 1001 | Do not show |
+------+-------------+-----------------+---------------+
CATEGORY
+------+-------+
| ID | TITLE |
+------+-------+
| 1000 | Apple |
| 1001 | Other |
+------+-------+
PRICE
+------+--------+---------+
| ID | USERID | PRICE |
+------+--------+---------+
| 1000 | 1000 | 359.00 |
| 1001 | 1001 | 1058.30 |
| 1002 | 1002 | 1078.00 |
| 1003 | 1003 | 1160.45 |
| 1004 | 1004 | 1180.00 |
| 1005 | 1000 | 1190.00 |
| 1006 | 1000 | 228.76 |
+------+--------+---------+
POSITION
+------+------------+--------+------+
| ID | PRODUCT_ID | USERID | RANK |
+------+------------+--------+------+
| 1000 | 1000 | 1000 | 1 |
| 1001 | 1001 | 1001 | 1 |
| 1002 | 1001 | 1002 | 2 |
| 1003 | 1001 | 1003 | 3 |
| 1004 | 1001 | 1004 | 4 |
| 1005 | 1001 | 1000 | 5 |
| 1006 | 1002 | 1000 | 1 |
+------+------------+--------+------+
狮身人面像设置：
source product
{
type = mysql
sql_host = localhost
sql_user = ...
sql_pass = ...
sql_db = ...
sql_query_pre = SET NAMES utf8
sql_query = SELECT P.ID AS ID, C.ID AS SEARCH_CAT_ID, P.CATEGORY_ID, P.MANUFACTURER_ID, P.TITLE AS TITLE_SORT FROM CATEGORY C, PRODUCT P WHERE P.CATEGORY_ID=C.ID
sql_attr_uint = CATEGORY_ID
sql_attr_uint = MANUFACTURER_ID
sql_attr_string = TITLE_SORT
sql_attr_multi = uint POS_RANK from query; SELECT PRODUCT_ID, RANK FROM POSITION
sql_attr_multi = uint POS_USERID from query; SELECT PRODUCT_ID, USERID FROM POSITION
}
index product
{
source = product
path = ...
docinfo = extern
min_word_len = 1
}
MySQL转储：
CREATE TABLE IF NOT EXISTS `CATEGORY` (
`ID` int(11) NOT NULL,
`TITLE` varchar(255) NOT NULL,
PRIMARY KEY (`ID`)
) ENGINE=InnoDB DEFAULT CHARSET=latin1;
INSERT INTO `CATEGORY` (`ID`, `TITLE`) VALUES
(1000, 'Apple'),
(1001, 'Other');
CREATE TABLE IF NOT EXISTS `POSITION` (
`ID` int(11) NOT NULL,
`PRODUCT_ID` int(11) NOT NULL,
`USERID` int(11) NOT NULL,
`RANK` int(11) NOT NULL,
PRIMARY KEY (`ID`),
KEY `USERID` (`USERID`),
KEY `PRODUCT_ID` (`PRODUCT_ID`)
) ENGINE=InnoDB DEFAULT CHARSET=latin1;
INSERT INTO `POSITION` (`ID`, `PRODUCT_ID`, `USERID`, `RANK`) VALUES
(1000, 1000, 1000, 1),
(1001, 1001, 1001, 1),
(1002, 1001, 1002, 2),
(1003, 1001, 1003, 3),
(1004, 1001, 1004, 4),
(1005, 1001, 1000, 5),
(1006, 1002, 1000, 1);
CREATE TABLE IF NOT EXISTS `PRICE` (
`ID` int(11) NOT NULL,
`USERID` int(11) NOT NULL,
`PRICE` decimal(9,2) NOT NULL,
PRIMARY KEY (`ID`),
KEY `USERID` (`USERID`)
) ENGINE=InnoDB DEFAULT CHARSET=latin1;
INSERT INTO `PRICE` (`ID`, `USERID`, `PRICE`) VALUES
(1000, 1000, '359.00'),
(1001, 1001, '1058.30'),
(1002, 1002, '1078.00'),
(1003, 1003, '1160.45'),
(1004, 1004, '1180.00'),
(1005, 1000, '1190.00'),
(1006, 1000, '228.76');
CREATE TABLE IF NOT EXISTS `PRODUCT` (
`ID` int(11) NOT NULL,
`CATEGORY_ID` int(11) NOT NULL,
`MANUFACTURER_ID` int(11) NOT NULL,
`TITLE` varchar(255) NOT NULL,
PRIMARY KEY (`ID`),
KEY `CATEGORY_ID` (`CATEGORY_ID`),
KEY `MANUFACTURER_ID` (`MANUFACTURER_ID`)
) ENGINE=InnoDB DEFAULT CHARSET=latin1;
INSERT INTO `PRODUCT` (`ID`, `CATEGORY_ID`, `MANUFACTURER_ID`, `TITLE`) VALUES
(1000, 1000, 1000, 'Apple iPhone'),
(1001, 1000, 1000, 'Apple iPad'),
(1002, 1000, 1000, 'Apple iPod'),
(1003, 1001, 1001, 'Do not show');
ALTER TABLE `POSITION`
ADD CONSTRAINT `POSITION_ibfk_1` FOREIGN KEY (`PRODUCT_ID`) REFERENCES `POSITION` (`ID`) ON DELETE CASCADE;
ALTER TABLE `PRODUCT`
ADD CONSTRAINT `PRODUCT_ibfk_1` FOREIGN KEY (`CATEGORY_ID`) REFERENCES `CATEGORY` (`ID`) ON DELETE CASCADE;
测试 1
过滤器 USERID 的 SphinxQL 查询：
SELECT ID FROM product WHERE MATCH('1000') AND POS_USERID IN (1000) ORDER BY WEIGHT() DESC LIMIT 0,20 FACET POS_RANK LIMIT 5;
USERID 的结果（这次第一组似乎没问题，第二组返回所有内容）：
+------+
| id |
+------+
| 1000 |
| 1001 |
| 1002 |
+------+
+----------+----------+
| pos_rank | count(*) |
+----------+----------+
| 5 | 1 |
| 4 | 1 |
| 3 | 1 |
| 2 | 1 |
| 1 | 3 |
+----------+----------+
我所期望的：
- 产品编号 (1000,1001,1002)
- 位置排名（1 与 3 计数，5 与 1 计数）
测试 2
过滤器 USERID 1000 和 RANK 5 的 SphinxQL 查询：
SELECT ID FROM product WHERE MATCH('1000') AND POS_USERID IN (1000) AND POS_RANK IN (5) ORDER BY WEIGHT() DESC LIMIT 0,20 FACET POS_RANK LIMIT 5;
USERID 和 RANK 的结果（这次第一组似乎没问题，第二组返回所有内容）：
+------+
| id |
+------+
| 1001 |
+------+
+----------+----------+
| pos_rank | count(*) |
+----------+----------+
| 5 | 1 |
| 4 | 1 |
| 3 | 1 |
| 2 | 1 |
| 1 | 1 |
+----------+----------+
我所期望的：
- 产品编号 (1001)
- 职位排名（5，计数为 1）
测试 3
过滤器 USERID 1000 和 RANK 4 的 SphinxQL 查询：
SELECT ID FROM product WHERE MATCH('1000') AND POS_USERID IN (1000) AND POS_RANK IN (4) ORDER BY WEIGHT() DESC LIMIT 0,20 FACET POS_RANK LIMIT 5;
USERID 和 RANK 的结果（第一个返回产品，第二个返回所有内容）：
+------+
| id |
+------+
| 1001 |
+------+
+----------+----------+
| pos_rank | count(*) |
+----------+----------+
| 5 | 1 |
| 4 | 1 |
| 3 | 1 |
| 2 | 1 |
| 1 | 1 |
+----------+----------+
我所期望的：
- 产品 ID（空集）
- 职位排名（空集）
我希望你们能理解我想要达到的目标并能帮助我吗？