我试图创建一个Web方法谁会得到一个Json字符串。我需要获取该json字符串并将其解析为java对象。
当我试图做到这一点时,我得到了例外:
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:
@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);
}
慕容3067478
相关分类