如何使用Spring的JDBCTemplate有效执行IN()SQL查询? [英] How to execute IN() SQL queries with Spring's JDBCTemplate effectively?

问题描述

我想知道是否有一种更优雅的方法来使用Spring的JDBCTemplate进行IN()查询.目前，我正在执行以下操作:

I was wondering if there is a more elegant way to do IN() queries with Spring's JDBCTemplate. Currently I do something like that:

StringBuilder jobTypeInClauseBuilder = new StringBuilder();
for(int i = 0; i < jobTypes.length; i++) {
    Type jobType = jobTypes[i];

    if(i != 0) {
        jobTypeInClauseBuilder.append(',');
    }

    jobTypeInClauseBuilder.append(jobType.convert());
}

这是很痛苦的，因为如果我有9行仅用于为IN()查询构建子句.我想要类似预准备语句的参数替换

Which is quite painful since if I have nine lines just for building the clause for the IN() query. I would like to have something like the parameter substitution of prepared statements

推荐答案

您想要一个参数源:

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());

仅当 getJdbcTemplate()返回类型为

This only works if getJdbcTemplate() returns an instance of type NamedParameterJdbcTemplate

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