Question

我这里有这段代码可通过AWS部署。和我在AWS上进行测试的问题如下：

当我按如下方式将值传递给x时：

“ x”：1.0或“ x”：100.0 .... etc（小数部分为零）它被视为整数，错误表示“ errorMessage”：此行中的“ java.lang.Integer无法转换为java.lang.Double” x_Ptr [i] = str.get（i）.doubleValue（）;

如果x通过分数等于“ x”的分数：1.5或“ x”的分数：0.444，则它有效

这里有什么建议吗？

public class Handler implements RequestHandler<Map<String, Object>, ApiGatewayResponse> {
    private static final Logger LOG = Logger.getLogger(Handler.class);

    @Override
    public ApiGatewayResponse handleRequest(Map<String, Object> event, Context context) {
        LOG.info("received: " + event);

        List<Double> str = new ArrayList<Double>();

        str = (ArrayList<Double>)(event.get("x"));
        double[] x_Ptr = new double[str.size()];

        for (int i = 0; i < x_Ptr.length; i++) {
            x_Ptr[i] = str.get(i).doubleValue();
        }

        System.out.println(x_Ptr[0]);

        Map<String, String> headers = new HashMap<>();
        headers.put("Content-Type", "application/json");

        Output bodyout = new Output(x_Ptr);

        return ApiGatewayResponse.builder()
                .setStatusCode(200)
                .setObjectBody(bodyout)
                .setHeaders(headers)
                .build();
    }
}

Answer 1

为输入类型定义一个POJO，问题是反序列化时类型信息不可用，并且可以从数据中推断出来。

public partial class admin_category : System.Web.UI.Page
{
    SqlConnection con = new SqlConnection(@"Data Source=(LocalDB)\v11.0;AttachDbFilename=|DataDirectory|\Database.mdf;Integrated Security=True");
    SqlDataSource sd = new SqlDataSource();

    protected void Page_Load(object sender, EventArgs e)
    {

        this.UnobtrusiveValidationMode = System.Web.UI.UnobtrusiveValidationMode.None;

        con.Open();

    }
    protected void Button1_Click(object sender, EventArgs e)
    {
        SqlCommand cmd = new SqlCommand(" insert into Men Category (CategoryName) values (@CategoryName)",con);
        cmd.Parameters.AddWithValue("@CategoryName",Cat_nm.Text);

        cmd.ExecuteNonQuery();
        con.Close();

        Response.Write("<script>alert('Category Added Sucessfully');</script>");

    }
}

“ errorMessage”：“ java.lang.Integer无法转换为java.lang.Double”

1 个答案: