使用jOOQ构建SQL时如何将变量绑定到条件语句？

Question

I'm using jOOQ-3.11.9 to build SQL. The following is my code:

String sql = DSL.using(SQLDialect.MYSQL)
        .select(DSL.asterisk())
        .from(table("service"))
        .where("name = ?", "service1")
        .getSQL();

What I expect is

select * from service where (name = "service1")

But the result is

select * from service where (name = ?)

Is there anything wrong with my code?

Answer 1

You should use the field name from jOOQ-generated classes:

String sql = DSL.using(SQLDialect.MYSQL)
    .select(DSL.asterisk())
    .from(YOURENTITY)
    .where(YOURENTITY.NAME.eq(nameParam))
    .getSQL();

YOURENTITY should be the jOOQ-generated class in your project. nameParam would then be an argument passed to the method wrapping the above query.

jOOQ has pretty good docs with lots of examples: https://www.jooq.org/doc/3.11/manual/sql-building/sql-statements/select-statement/where-clause/

Answer 2

This works as expected. The default Settings.statementType value is StatementType.PREPARED_STATEMENT , so jOOQ by default generates bind value placeholders in your SQL string, which can be used for execution in other tools, such as JDBC, Spring, etc.

You can pass the ParamType.INLINE value to the getSQL() method, or specify Settings.withStatementType(StatementType.STATIC_STATEMENT)

Please consider the explanation in the Javadoc for more details: https://www.jooq.org/javadoc/latest/org/jooq/Query.html#getSQL--