Retrofit2 java.lang.IllegalStateException:预期BEGIN_OBJECT但在线串1列1路$
问题描述:
我创建/使用retrofit2我的服务器端注册新用户
<?php
$name= $_POST['name'];
$email = $_POST['email'];
$password= $_POST['password'];
$gender= $_POST['gender'];
$con = mysqli_connect("localhost", "root", "qwerty", "db");
$query= mysqli_query($con, "INSERT INTO users(name,email, password, gender) VALUES('$name','$email', '$password', '$gender')");
if($query){
echo "You are sucessfully Registered";
}
else{
echo "your details could not be registered";
}
mysqli_close($con);
?>
我的模型类是
public class User {
@SerializedName("name")
@Expose
public String name;
@SerializedName("email")
@Expose
public String email;
@SerializedName("password")
@Expose
public String password;
@SerializedName("gender")
@Expose
public String gender;
public User(String name, String email, String password, String gender) {
this.name = name;
this.email = email;
this.password = password;
this.gender = gender;
}
邮政服务
@POST("register.php")
Call<User> createUser(@Body User user);
发布数据
mRegisterButton.setOnClickListener(new View.OnClickListener() {
@Override
public void onClick(View view) {
if (validateForm()){
mUserCall = mRestManager.getJobService()
.createUser(new User(mNameEditText.getText().toString(),
mEmailEditText.getText().toString(),
mPasswordEditText.getText().toString(),
mGenders[position]));
mUserCall.enqueue(new Callback<User>() {
@Override
public void onResponse(Call<User> call, Response<User> response) {
User user1 = response.body();
Toast.makeText(getApplicationContext(), user1.name , Toast.LENGTH_SHORT).show();
}
@Override
public void onFailure(Call<User> call, Throwable t) {
Log.e("REGISTER_ERROR", "Message is " + t.getMessage());
}
});
}
}
});
我收到提示
java.lang.IllegalStateException: Expected BEGIN_OBJECT but was STRING at line 1 column 1 path $
我已经找到了一些解决方案,并作出
@FormUrlEncoded
@POST("register.php")
Call<User> createUser(@Field("name") String name,
@Field("email") String email,
@Field("password") String password,
@Field("gender") String gender);
发生相应的变化,但它也不起作用,同样的错误信息,如何解决这个错误。
答
检查你的回应,它是不同于你的期望。
Json必须从“{”也有响应是不同于你的Pojo 用户所以从新的响应使新的Pojo。
使用调用 –
添加你的json响应 –
你没有返回一个User对象,你正在为PHP返回一个成功/失败的字符串。 –