Question

尝试使用Spring-JDBC。我使用this作为参考。我正在尝试获得具有相同姓氏的演员列表。运行此代码给了我想要的结果：

public List<String> getActorsWithSameLastName(String lastName,
            NamedParameterJdbcTemplate template) {
        String query = "SELECT FIRSTNAME FROM ACTORS WHERE LASTNAME=:LASTNAME";
        Map<String, String> paramMap = new HashMap<String, String>();
        paramMap.put("LASTNAME", lastName);
        return template.queryForList(query, paramMap, String.class);
    }

我有List<String>个姓氏。如何获得具有列表的演员列表？我是否遍历姓氏列表并每次调用getActorsWithSameLastName()或者弹簧是否提供了迭代的方式并为我提取结果？请指教。

Answer 1

使用IN子句..

How to use SELECT IN clause in JDBCTemplates?

List<String> lastnames= new ArrayList<>();

Map namedParameters = Collections.singletonMap("lastnamevalues", lastnames);

StringBuffer recordQueryString = new StringBuffer();

recordQueryString.append("select FIRSTNAME, LASTNAME from ACTORS where lastname in (:lastnamevalues)");

List nameInvolvements = this.namedparameterJdbcTemplate.query(recordQueryString.toString(), namedParameters, new MyMapper());

Answer 2

您也可以使用MapSqlParameterSource

String query = "SELECT FIRSTNAME FROM ACTORS WHERE LASTNAME in (:LASTNAME)";
Set<String> ids = ....;

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

this.namedparameterJdbcTemplate.query(query.toString(), parameters);

如何使用NamedParameterJDBCTemplate中的字符串列表来获取结果

2 个答案: