Question

我打算通过querydsl生成SQL查询以进行动态字符串构建。我当前未使用spring数据jpa，我生成了一个Q表，但是我不想通过实例化Q对象来构建字符串，我的要求是通过采用表名/ q bean来动态构建查询。用户在运行时选择的Q bean属性中定义的变量名称。

Oracle表架构：

Create table VW_OUT.AWARDS_VW
(
   GIVER_LAST_NAME varchar2(50),
   GIVER_FIRST_NAME varchar2(50)
);

Maven部门

<dependency>
            <groupId>com.querydsl</groupId>
            <artifactId>querydsl-sql</artifactId>
            <version>4.1.3</version>
        </dependency>

        <dependency>
            <groupId>com.querydsl</groupId>
            <artifactId>querydsl-sql-spring</artifactId>
            <version>4.1.3</version>
        </dependency>

给出Spring Query DSL配置

@Configuration
public class QueryDSLConfig {

    @Autowired
    public DataSource dataSource;

    @Bean
    public PlatformTransactionManager transactionManager() {
        return new DataSourceTransactionManager(dataSource);
    }

    @Bean
    public com.querydsl.sql.Configuration querydslConfiguration() {
        SQLTemplates templates = OracleTemplates.builder()
                .printSchema()
                .quote()
                .build();
        com.querydsl.sql.Configuration configuration = new com.querydsl.sql.Configuration(templates);
        configuration.setUseLiterals(true);
        configuration.setExceptionTranslator(new SpringExceptionTranslator());

        return configuration;
    }

    @Bean
    public SQLQueryFactory queryFactory() {
        Provider<Connection> provider = new SpringConnectionProvider(dataSource);
        return new SQLQueryFactory(querydslConfiguration(), provider);
    }
}

和Q类型之一：

@Generated("com.querydsl.sql.codegen.MetaDataSerializer")
public class QAwardsVw extends com.querydsl.sql.RelationalPathBase<QAwardsVw> {

    private static final long serialVersionUID = 519295604;

    public static final QAwardsVw awardsVw = new QAwardsVw("AWARDS_VW");

    public final StringPath giverFirstName = createString("giverFirstName");

    public final StringPath giverLastName = createString("giverLastName");


    public QAwardsVw(String variable) {
        super(QAwardsVw.class, forVariable(variable), "VW_OUT", "AWARDS_VW");
        addMetadata();
    }

    public QAwardsVw(String variable, String schema, String table) {
        super(QAwardsVw.class, forVariable(variable), schema, table);
        addMetadata();
    }

    public QAwardsVw(Path<? extends QAwardsVw> path) {
        super(path.getType(), path.getMetadata(), "VW_OUT", "AWARDS_VW");
        addMetadata();
    }

    public QAwardsVw(PathMetadata metadata) {
        super(QAwardsVw.class, metadata, "VW_OUT", "AWARDS_VW");
        addMetadata();
    }

        addMetadata(giverFirstName, ColumnMetadata.named("GIVER_FIRST_NAME").withIndex(8).ofType(Types.VARCHAR).withSize(50));
        addMetadata(giverLastName, ColumnMetadata.named("GIVER_LAST_NAME").withIndex(7).ofType(Types.VARCHAR).withSize(50));   
    }
}

我的测试，尝试以通用方式而不是直接从Q bean构建from路径

@RunWith(SpringRunner.class)
@SpringBootTest
public class QuerydslApplicationTests {

    @Autowired
    SQLQueryFactory queryFactory;


    @Test
    public void contextLoads() {
        PathBuilder<Object> fromPath = new PathBuilder<Object>(Object.class, "AWARDS_VW");
        StringPath lnamePath = Expressions.stringPath(fromPath, "giverLastName");
        StringPath fnamePath = Expressions.stringPath(fromPath, "giverFirstName");
        String s = queryFactory.select(fnamePath ).from(fromPath).where(lnamePath .eq("MC")).getSQL().getSQL();


        System.out.println(s);
    }
}

这让我明白了，列名不是数据库列名，似乎是bean属性变量名，并且架构名称也没有打印为VW_OUT.Awards_Vw

select Awards_Vw.giverFirstName
from Awards_Vw
where Awards_Vw.giverLastName= 'MC'

我想要的是

select AWARDS_VW.giver_First_Name
    from AWARDS_VW
    where AWARDS_VW.giver_Last_Name= 'MC'

但是，如果我直接使用Q bean，它将满足我想要的结果，但是我不想像这样通过实例化Q bean构建路径

QWsPersonAwardsVw p = new QWsPersonAwardsVw("p");
String s = queryFactory.select(p.giverFirstName).from(p).where(p.giverLastName.eq("MC")).getSQL().getSQL();

产量：

select "p"."GIVER_FIRST_NAME"
from "ODS_OUT"."AWARDS_VW" "p"
where "p"."GIVER_LAST_NAME" = 'MC'

任何人都有见解或如何正确获取元数据？还是使用spring jpa使这更容易？我们的目标是在运行时生成字符串，以便事先不知道用户提供的表名/ Q变量名。

谢谢。

QueryDSL JDBC-不使用Q类型的动态路径构建

0 个答案: