Question

我试图创建一个获取Json字符串的Web方法。我需要获取json字符串并将其解析为java对象。

当我试图这样做时，我得到了异常：

javax.servlet.ServletException: com.google.gson.JsonSyntaxException: 
java.io.EOFException: End of input at line 1 column 2

根本原因

com.google.gson.JsonSyntaxException: java.io.EOFException: End of input at 
line 1 column 2

根本原因

 java.io.EOFException: End of input at line 1 column 2

那是我的代码吗？

@GET
@Produces("application/text")
@Path("checkuser/{user}")
public String checkUser(@PathParam("user") String mu) throws SQLException, ClassNotFoundException {

    gson = new Gson();
    modelUserGet = gson.fromJson(mu, ModelUser.class);

    StringBuilder query = new StringBuilder();

    query.append("SELECT user, password, email,");
    query.append(" telephone, creation_data, last_update_data ");
    query.append("FROM user ");
    query.append("WHERE user ='");
    query.append(modelUserGet.getUser());
    query.append("' ");

    datamysql = new DataMySqlAccess();
    Statement st = datamysql.getConnection().createStatement();
    ResultSet rs = st.executeQuery(query.toString());



    String result = "";

    if(rs.next() == true){

        modelUserSend = new ModelUser(
                rs.getString("user"),
                rs.getString("password"),
                rs.getString("email"),
                rs.getString("telephone"),
                rs.getString("creation_data"),
                rs.getString("last_update_data")
        );

        if(!modelUserSend.getPassword().equals(modelUserGet.getPassword()))
            result = "INVALID_PASSWORD";
        else
            result = "OK";
    } else 
        result = "INVALID_USER";


    modelUserSend.setCheckUserReponse(result);

    return gson.toJson(modelUserSend);
}

这是我的模型课

private String user;
private String password;
private String email;
private String telephone;
private String creationData;
private String lastUpdateData;
private String checkUserResponse;

public ModelUser(String user, String password, String email, String telephone, String creationData, String lastUpdateData){
    this.user = user;
    this.password = password;
    this.email = email;
    this.telephone = telephone;
    this.creationData = creationData;
    this.lastUpdateData = lastUpdateData;
}

然后我发送该参数json字符串

{
    "user":"admin",
    "password": "admin",
    "email":"admin@admin.com",
    "telephone":"(11) 3761-5292",
    "creationData":"2018-07-08",
    "lastUpdateData":"2018-07-08"
}

我做错什么了吗？

Thx。 o /

Answer 1

@PathParam（“ user”）字符串mu

这是您要获取的PathParameter，它是一个字符串。这就是例外的原因。无法转换为模型。

在您的情况下，请将其作为单独的路径参数变量并在方法中使用它。

编辑1

我认为，用户名是您从参数中唯一需要的项目。因此，最好将用户名作为Path参数来获取。还有其他细节，您就是从数据库中获得的。

def f(arr):
    best_val = calc_x(arr)
    used = set(arr)
    for i, v in enumerate(arr):
        for replacement in used:
            if replacement == v:
                continue

            arr2 = arr[0:i] + replacement + arr[i:]

            y = abs(replacement - v)
            x = calc_x(arr2)

            best_val = min(best_val, x + y)
    return best_val

}

Answer 2

使用该代码，我可以发送json，获取该json并将其转换为JavaObject，以比String参数更好地工作。

@GET
@Produces(MediaType.APPLICATION_JSON)
@Consumes(MediaType.APPLICATION_JSON)
@Path("checkuser/{user}")
public String checkUser(@PathParam("user") String mu) throws SQLException, 
ClassNotFoundException {

    gson = new Gson();
    modelUserGet = gson.fromJson(mu, ModelUser.class);

    StringBuilder query = new StringBuilder();

    query.append("SELECT user, password, email,");
    query.append(" telephone, creation_data, last_update_data ");
    query.append("FROM user ");
    query.append("WHERE user ='");
    query.append(modelUserGet.getUser());
    query.append("' ");

    datamysql = new DataMySqlAccess();
    Statement st = datamysql.getConnection().createStatement();
    ResultSet rs = st.executeQuery(query.toString());

    String result = "";

    if(rs.next() == true){

        modelUserSend = new ModelUser(
                rs.getString("user"),
                rs.getString("password"),
                rs.getString("email"),
                rs.getString("telephone"),
                rs.getString("creation_data"),
                rs.getString("last_update_data")
        );

        if(!modelUserSend.getPassword().equals(modelUserGet.getPassword()))
            result = "INVALID_PASSWORD";
        else
            result = "OK";
    } else 
        result = "INVALID_USER";


    modelUserSend.setCheckUserReponse(result);

    return gson.toJson(modelUserSend);
}

尝试使用Gson库将Json解析为Java对象时抛出异常

2 个答案: