需要一个字符串,但在第 3 行第 4 列路径 $.SUCCESS 处是 BEGIN_OBJECT
Expected a string but was BEGIN_OBJECT at line 3 column 4 path $.SUCCESS
我用 Retrofit2 做了一个 post 请求方法,但我在响应中遇到了这个问题。
Expected a string but was BEGIN_OBJECT at line 3 column 4 path $.SUCCESS
响应应该是
{
"SUCCESS" :
{
"200" : "access granted",
"ra" : "approved",
"la" : "approved",
"ch" : "approved"
}
}
我将此代码用于 post 请求
@POST("login")
Call<Post> createPost(@Body Post post);
对于 POJO class
public class Post {
private String anthony;
private String SUCCESS;
public Post(String name) {
this.anthony = name;
}
public String getSUCCESS() {
return SUCCESS;
}
}
方法我使用下面的代码
private void createPost() {
Post post = new Post("mypassword");
Call<Post> call = jsonPlaceHolderApi.createPost(post);
call.enqueue(new Callback<Post>() {
@Override
public void onResponse(Call<Post> call, Response<Post> response) {
if (!response.isSuccessful()) {
textViewResult.setText("Code: " + response.code());
return;
}
Post postResponse = response.body();
String content = "";
content += "Code: " + response.code() + "\n";
content += "S" + postResponse.getSUCCESS();
textViewResult.setText(content);
}
@Override
public void onFailure(Call<Post> call, Throwable t) {
textViewResult.setText(t.getMessage());
}
});
}
有人知道我的代码有什么问题吗?我希望在“SUCCESS”json 对象中得到响应。
您希望 SUCCESS 成为您想要的响应中的对象,但您已将其定义为 Post class 中的字符串。您应该使用对象来代替 SUCCESS。
public class Post {
private String anthony;
private PostSuccess SUCCESS;
public Post(String name) {
this.anthony = name;
}
public PostSuccess getSUCCESS() {
return SUCCESS;
}
}
public class PostSuccess {
@JsonProperty("200")
private String _200;
private String ra;
private String la;
private String ch;
}
我用 Retrofit2 做了一个 post 请求方法,但我在响应中遇到了这个问题。
Expected a string but was BEGIN_OBJECT at line 3 column 4 path $.SUCCESS
响应应该是
{
"SUCCESS" :
{
"200" : "access granted",
"ra" : "approved",
"la" : "approved",
"ch" : "approved"
}
}
我将此代码用于 post 请求
@POST("login")
Call<Post> createPost(@Body Post post);
对于 POJO class
public class Post {
private String anthony;
private String SUCCESS;
public Post(String name) {
this.anthony = name;
}
public String getSUCCESS() {
return SUCCESS;
}
}
方法我使用下面的代码
private void createPost() {
Post post = new Post("mypassword");
Call<Post> call = jsonPlaceHolderApi.createPost(post);
call.enqueue(new Callback<Post>() {
@Override
public void onResponse(Call<Post> call, Response<Post> response) {
if (!response.isSuccessful()) {
textViewResult.setText("Code: " + response.code());
return;
}
Post postResponse = response.body();
String content = "";
content += "Code: " + response.code() + "\n";
content += "S" + postResponse.getSUCCESS();
textViewResult.setText(content);
}
@Override
public void onFailure(Call<Post> call, Throwable t) {
textViewResult.setText(t.getMessage());
}
});
}
有人知道我的代码有什么问题吗?我希望在“SUCCESS”json 对象中得到响应。
您希望 SUCCESS 成为您想要的响应中的对象,但您已将其定义为 Post class 中的字符串。您应该使用对象来代替 SUCCESS。
public class Post {
private String anthony;
private PostSuccess SUCCESS;
public Post(String name) {
this.anthony = name;
}
public PostSuccess getSUCCESS() {
return SUCCESS;
}
}
public class PostSuccess {
@JsonProperty("200")
private String _200;
private String ra;
private String la;
private String ch;
}