从 Java/Spring 项目执行 SQL 语句时出错答案

【问题标题】：SQL sentence causes error when executed from Java/Spring Project从 Java/Spring 项目执行 SQL 语句时出错
【发布时间】：2018-07-05 20:43:07
【问题描述】：

在我当前的 spring 项目中，我已经为我的 UserDetailsService bean 配置了带有这个 SQL 语句的 spring-security xml：

<bean id="userDetailsService" class="org.springframework.security.core.userdetails.jdbc.JdbcDaoImpl">
    <property name="dataSource" ref="dataSource"></property>
    <property name="usersByUsernameQuery" value="select login, senha, enabled from usuario where login = ?"></property>
    <property name="authoritiesByUsernameQuery" value="SELECT t1.username, t2.authority FROM (SELECT u.login as username, c.nome as credencial FROM usuario u, usuario_credencial uc, credencial c WHERE u.id = uc.usuario_id and c.id = uc.credenciais_id) as t1 INNER JOIN (SELECT c.nome as credencial, a.nome as authority FROM credencial c, credencial_autorizacao ca, autorizacao a WHERE c.id = ca.credencial_id and a.id = ca.autorizacoes_id) as t2 ON t1.credencial = t2.credencial WHERE t1.username = ?;"></property>
</bean>

SQL语句：

SELECT t1.username, t2.authority
FROM (
    SELECT u.login as username, c.nome as credencial
    FROM usuario u, usuario_credencial uc, credencial c
    WHERE u.id = uc.usuario_id and c.id = uc.credenciais_id
    ) as t1
        INNER JOIN
    (
    SELECT c.nome as credencial, a.nome as authority
    FROM credencial c, credencial_autorizacao ca, autorizacao a
    WHERE c.id = ca.credencial_id and a.id = ca.autorizacoes_id
    ) as t2
        ON t1.credencial = t2.credencial
WHERE t1.username = ?;

当我在 tomcat 上部署和运行应用程序时导致此错误（以下文本来自 postgresql 日志文件）：

2018-01-26 07:53:41.644 -03 [20491] postgres@mydata ERROR:  syntax error at or near "INNER" at character 160
2018-01-26 07:53:41.644 -03 [20491] postgres@mydata STATEMENT:  (SELECT u.login as username, c.nome as credencial FROM usuario u, usuario_credencial uc, credencial c WHERE u.id = uc.usuario_id and c.id = uc.credenciais_id) INNER JOIN (SELECT c.nome as credencial, a.nome as authority FROM credencial c, credencial_autorizacao ca, autorizacao a WHERE c.id = ca.credencial_id and a.id = ca.autorizacoes_id)
2018-01-26 07:54:52.882 -03 [20491] postgres@mydata ERROR:  syntax error at or near "as" at character 160
2018-01-26 07:54:52.882 -03 [20491] postgres@mydata STATEMENT:  (SELECT u.login as username, c.nome as credencial FROM usuario u, usuario_credencial uc, credencial c WHERE u.id = uc.usuario_id and c.id = uc.credenciais_id) as t1 INNER JOIN (SELECT c.nome as credencial, a.nome as authority FROM credencial c, credencial_autorizacao ca, autorizacao a WHERE c.id = ca.credencial_id and a.id = ca.autorizacoes_id) as t2
2018-01-26 07:55:14.275 -03 [20491] postgres@mydata ERROR:  syntax error at or near "INNER" at character 158
2018-01-26 07:55:14.275 -03 [20491] postgres@mydata STATEMENT:  SELECT u.login as username, c.nome as credencial FROM usuario u, usuario_credencial uc, credencial c WHERE u.id = uc.usuario_id and c.id = uc.credenciais_id INNER JOIN SELECT c.nome as credencial, a.nome as authority FROM credencial c, credencial_autorizacao ca, autorizacao a WHERE c.id = ca.credencial_id and a.id = ca.autorizacoes_id
2018-01-26 07:56:55.362 -03 [20520] postgres@mydata ERROR:  syntax error at or near "as" at character 160
2018-01-26 07:56:55.362 -03 [20520] postgres@mydata STATEMENT:  (SELECT u.login as username, c.nome as credencial FROM usuario u, usuario_credencial uc, credencial c WHERE u.id = uc.usuario_id and c.id = uc.credenciais_id) as t1 INNER JOIN (SELECT c.nome as credencial, a.nome as authority FROM credencial c, credencial_autorizacao ca, autorizacao a WHERE c.id = ca.credencial_id and a.id = ca.autorizacoes_id) as t2 ON t1.credencial = t2.credencial

谁能给我提示一下是什么导致了这个错误？

【问题讨论】：

没有具体导致错误，但混合逗号连接和JOIN 语法使其难以解析。至少对我来说。
不相关，但是：为什么在派生表中使用古老而过时的隐式连接，而在它们之间使用现代显式连接运算符？
查询看起来很好，如果你在SQL客户端手动运行它是否有效？
@a_horse_with_no_name 是的，我在 pgAdmin3 中运行它。另外，句子执行没有错误。当我尝试完整的句子时（只需复制并粘贴上面的代码，就会出错）。
credential 是 SELECTs 的一部分，会考虑只使用一个涉及所有 5 个表的大型联接，而不是 2x3。使整个查询更易于阅读和理解。

标签： java sql spring postgresql inner-join

【解决方案1】：

提出一个更简单的查询：

SELECT u.login as username, a.nome as authority
FROM usuario u
JOIN usuario_credencial uc     ON u.id = uc.usuario_id
JOIN credencial c              ON c.id = uc.credenciais_id
JOIN credencial_autorizacao ca ON c.id = ca.credencial_id
JOIN autorizacao a             ON a.id = ca.autorizacoes_id
WHERE u.login = ?;

【讨论】：