Sphinx - 将 OR 运算符与字段开始/结束修饰符组合

Question

这个表达式写对了吗？因为我没有收到预期的所有结果：

(@nationality_code ("^MD$" | "^GB$" | "^IT$"))

我应该让所有国籍代码等于“MD”或“GB”或“IT”的用户

在mysql中应该是这样的查询：

SELECT * FROM users WHERE nationality_code IN ('MD', 'GB', 'IT')

更新 如果我只搜索 MD 和 GB，我会收到正确的结果：

我执行的查询是这样的：

SELECT id, nationality_code 
FROM jobs_applied_main 
WHERE MATCH('(@nationality_code ("^MD$" | "^GB$"))') AND job_id = '6257' 
LIMIT 0, 999;

但如果我还添加“^IT$”，我将不再获得国籍为“MD”的用户。

如果我删除开始/结束修饰符，我会收到预期的结果，为什么会出现这种行为？

更新 所以，这里是我的索引转储和 sphinx.conf 文件：sphinx-dump

Answer 1

是的。表达是对的。它对我来说很好用：

[snikolaev@dev01 ~]$ mysql -P9314 -h0
Welcome to the MySQL monitor.  Commands end with ; or \g.
Your MySQL connection id is 1
Server version: 2.7.1 668faae@180620 dev

Copyright (c) 2009-2017 Percona LLC and/or its affiliates
Copyright (c) 2000, 2017, Oracle and/or its affiliates. All rights reserved.

Oracle is a registered trademark of Oracle Corporation and/or its
affiliates. Other names may be trademarks of their respective
owners.

Type 'help;' or '\h' for help. Type '\c' to clear the current input statement.

mysql> select * from idx_min where match('(@nationality_code ("^MD$" | "^GB$" | "^IT$"))');
+------+------------------+------+
| id   | nationality_code | a    |
+------+------------------+------+
|    1 | MD               |  123 |
|    2 | GB               |  123 |
+------+------------------+------+
2 rows in set (0.00 sec)

确保您在索引中确实只有 MD/GB/IT，之前/之后没有任何内容，并且您没有 min_word_len

Answer 2

Sphinx 3.0.3 版本存在问题，因为 Sphinx 2.2.11 可以正常工作