Question

我试图插入一个简单的基础，其中Spring JdbcTemplate使用MapSqlParamaterSource映射参数查询，并将错误视为以下数据：

public void adiciona(Conta conta) {
  String sql = "insert into contas (descricao, paga, valor, tipo) values (:descricao,:paga,:valor,:tipo)";
  MapSqlParameterSource pss = new MapSqlParameterSource();
  pss.addValue("descricao", conta.getDescricao());
  pss.addValue("paga", "true");
  pss.addValue("valor", conta.getValor());
  pss.addValue("tipo", conta.getTipo());
  getJdbcTemplate().update(sql, pss);     
}

错误日志：

mar 22, 2015 12:16:00 PM org.apache.catalina.core.StandardWrapperValve invoke
GRAVE: Servlet.service() for servlet [spring mvc] in context with path [/contas] threw exception [Request processing failed; nested exception is org.springframework.dao.TransientDataAccessResourceException: PreparedStatementCallback; SQL [insert into contas (descricao, paga, valor, tipo) values (:descricao,:paga,:valor,:tipo)]; Invalid argument value: java.io.NotSerializableException; nested exception is java.sql.SQLException: Invalid argument value: java.io.NotSerializableException] with root cause
java.io.NotSerializableException: org.springframework.jdbc.core.namedparam.MapSqlParameterSource

Java Bean：

public class Conta implements Serializable {

  private static final long serialVersionUID = 4678852901357132238L;

  private Long id;
  private String descricao;
  private boolean paga;
  private double valor;
  private Calendar dataPagamento;
  private TipoDaConta tipo;

  // getters and settes

有人可以告诉我如何解决这个问题吗？

Answer 1

我也面临同样的问题，我通过改变它来解决它

this.getJdbcTemplate().update(query,mapSqlParameterSource);

到

this.getNamedParameterJdbcTemplate().update(query,mapSqlParameterSource);

Answer 2

尝试类似：

 this.getJdbcTemplate().update(updateStatement, new Object[] {conta.getDescricao(), "true", conta.getValor(), conta.getTipo()});

Spring：java.io.NotSerializableException：MapSqlParameterSource

2 个答案: