Question

我已经看过两个或更多使用改装2.0来应用帖子的教程。所有这些都和我一样实现。但我得到如下错误

onFailure：com.google.gson.JsonSyntaxException：java.lang.IllegalStateException预期为BEGIN_OBJECT但在第1行第1行为STRING

我想传递我从编辑文本中获取的简单字符串。然后将它发布在我的服务器数据库中...但它只是返回此异常。如何使它正确我想在服务器端接收数据作为字符串，或者我想以字符串

发送数据

Api服务界面

import retrofit.Call;
import retrofit.http.Body;
import retrofit.http.GET;
import retrofit.http.POST;

/**
 * Created by Shaon on 8/14/2016.
 */
public interface APIService {
    @GET("my_json")
    Call<List<People>> getPeopleDetails();

    @POST("my_json/insert.php")
    Call<People> setPeopleDetails(@Body People people);
}

人类

import com.google.gson.annotations.Expose;
import com.google.gson.annotations.SerializedName;

/**
 * Created by Shaon on 8/14/2016.
 */
public class People {

    private String id = "";

    @SerializedName("name")
    @Expose
    private String name = "";

    public String getId() {
        return id;
    }


    public String getName() {
        return name;
    }

    public void setName(String name) {
        this.name = name;
    }

}

我的帖子方法

Retrofit retrofit = new Retrofit.Builder()
                .baseUrl("http://w...content-available-to-author-only...e.org/").
                        addConverterFactory(GsonConverterFactory.create())
                .build();

        APIService service = retrofit.create(APIService.class);
        People people = new People();
        people.setName(editName.getText().toString());

        Call<People> peopleCall = service.setPeopleDetails(people);

        peopleCall.enqueue(new Callback<People>() {
            @Override
            public void onResponse(Response<People> response, Retrofit retrofit) {
                hidepDialog();
                Log.d("onResponse", "There is an error");
            }

            @Override
            public void onFailure(Throwable t) {
                hidepDialog();
                Log.d("onFailure", t.toString());
            }
        });

我的服务器端代码

    $id = $_POST["id"];
$name = $_POST["name"];

try {
    $conn = new PDO("mysql:host=$servername;dbname=$dbname", $username, $password);
    // set the PDO error mode to exception
    $conn->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
    $sql = "INSERT INTO mytable (name)
    VALUES ('$name')";
    // use exec() because no results are returned
    $conn->exec($sql);


    echo "New record created successfully";
    }
catch(PDOException $e)
    {
    echo $sql . "<br>" . $e->getMessage();
    }

$conn = null;
?>

在失败时获取错误：com.google.gson.JsonSyntaxException：java.lang.IllegalStateException

0 个答案: