Table of Contents
Use NamedParameterJdbcTemplate for parameter replacement
Example implementation
Home Database Mysql Tutorial How Can I Efficiently Use IN() Queries with Spring JDBCTemplate?

How Can I Efficiently Use IN() Queries with Spring JDBCTemplate?

Jan 17, 2025 am 02:23 AM

How Can I Efficiently Use IN() Queries with Spring JDBCTemplate?

Improve IN() query efficiency in Spring JDBCTemplate

When using Spring’s JDBCTemplate for IN() queries, the process of manually building the IN clause can become tedious. To solve this problem, let's explore a more elegant and efficient approach.

Use NamedParameterJdbcTemplate for parameter replacement

Spring provides the NamedParameterJdbcTemplate class, which simplifies IN() queries by allowing parameter substitution. Instead of manually building the IN clause, you can append the parameters to the MapSqlParameterSource and pass them to jdbcTemplate as named parameters.

For example:

Set<Integer> ids = ...;

MapSqlParameterSource parameters = new MapSqlParameterSource();
parameters.addValue("ids", ids);

List<Foo> foo = getJdbcTemplate().query("SELECT * FROM foo WHERE a IN (:ids)",
     parameters, getRowMapper());
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Note: This method requires getJdbcTemplate() to return an instance of NamedParameterJdbcTemplate.

Example implementation

Using NamedParameterJdbcTemplate, you can rewrite the code as follows:

Set<Type> jobTypes = ...;

MapSqlParameterSource parameters = new MapSqlParameterSource();
parameters.addValue("jobTypes", jobTypes.stream().map(Type::convert).collect(Collectors.toList()));

List<Job> jobs = getJdbcTemplate().query("SELECT * FROM job WHERE job_type IN (:jobTypes)",
     parameters, getRowMapper());
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This code eliminates the need to manually construct clauses, making Spring JDBCTemplate's IN() query more efficient and elegant.

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