java.lang.IllegalStateException：应为 BEGIN_OBJECT 但为 STRING
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【中文标题】java.lang.IllegalStateException：应为 BEGIN_OBJECT 但为 STRING【英文标题】：java.lang.IllegalStateException: Expected BEGIN_OBJECT but was STRING 【发布时间】：2016-06-01 20:57:49 【问题描述】：我编写了一个Login 功能并与web service 进行通信 我正在使用Retrofit，但总是得到“失败”
获取：
retrofit.RetrofitError: com.google.gson.JsonSyntaxException: java.lang.IllegalStateException: Expected BEGIN_OBJECT but was STRING at line 1 column 1 path $
LoginAPI.java：
public interface LoginApi
@FormUrlEncoded
@POST("/retrofit_user/login.php")
public void getlogin(
@Field("username") String username,
@Field("password") String password,
Callback<User> response);
LoginActivity.java：
restAdapter = new RestAdapter.Builder().setEndpoint(ROOT_LOGIN).build();
restAdapter.setLogLevel(RestAdapter.LogLevel.FULL);
loginApi = restAdapter.create(LoginApi.class);
buttonlogin.setOnClickListener(new View.OnClickListener()
@Override
public void onClick(View v)
loginApi.getlogin(editTextUsername.getText().toString(),
editTextPassword.getText().toString(), new Callback<User>()
@Override
public void success(User s, Response response)
Toast.makeText(LoginActivity.this,"Logged In",Toast.LENGTH_LONG).show();
if(s != null)
Log.d("name:", s.getName());
Log.d("email:", s.getEmail());
Log.d("id:", String.valueOf(s.getId()));
@Override
public void failure(RetrofitError error)
Log.d("error:", error.toString());
Toast.makeText(LoginActivity.this,"Failed",Toast.LENGTH_LONG).show();
);
);
User.java：
public class User
String name;
String username;
String password;
String email;
int id;
public String getName()
return name;
public void setName(String name)
this.name = name;
public String getUsername()
return username;
public void setUsername(String username)
this.username = username;
public String getPassword()
return password;
public void setPassword(String password)
this.password = password;
public String getEmail()
return email;
public void setEmail(String email)
this.email = email;
public int getId()
return id;
public void setId(int id)
this.id = id;
日志说
D/Retrofit: <--- HTTP 200 http://domain.info/retrofit_user/login.php (942ms)
D/Retrofit: : HTTP/1.1 200 OK
D/Retrofit: Connection: Keep-Alive
D/Retrofit: Content-Type: text/html; charset=UTF-8
D/Retrofit: Date: Sat, 20 Feb 2016 06:00:39 GMT
D/Retrofit: Keep-Alive: timeout=5, max=100
D/Retrofit: Server: Apache/2.4.16 (Unix) OpenSSL/1.0.1e-fips mod_bwlimited/1.4
D/Retrofit: Transfer-Encoding: chunked
D/Retrofit: X-android-Received-Millis: 1455948038590
D/Retrofit: X-Android-Response-Source: NETWORK 200
D/Retrofit: X-Android-Selected-Transport: http/1.1
D/Retrofit: X-Android-Sent-Millis: 1455948037988
D/Retrofit: X-Powered-By: PHP/5.6.14
D/Retrofit: Done
D/Retrofit: <--- END HTTP (4-byte body)
D/error:: retrofit.RetrofitError: com.google.gson.JsonSyntaxException: java.lang.IllegalStateException: Expected BEGIN_OBJECT but was STRING at line 1 column 1 path $
D/GraphicBuffer: create handle(0x55bb2e98) (w:256, h:66, f:1)
D/GraphicBuffer: close handle(0x55bb2e98) (w:256 h:66 f:1)
这是我的login.php 脚本：
<?php
$objConnect = mysql_connect("localhost","username","password");
$objDB = mysql_select_db("database");
/*** for sample */
// $_POST["username"] = "sun";
// $_POST["password"] = "live";
$username = $_REQUEST['username'];
$password = $_REQUEST['password'];
$strSQL = "select * from test_users where username = '".$username."' and password = '".$password."' ";
$objQuery = mysql_query($strSQL);
$intNumRows = mysql_num_rows($objQuery);
if($intNumRows==0)
echo "Not Done" ;
else
echo "Done" ;
mysql_close($objConnect);
?>
【问题讨论】：
您的请求正在返回 HTTP/1.1 200 OK。 200 代码用于成功请求。我认为改造没有任何问题 是的，但我的问题是为什么我登录失败 在failure 方法中记录error 变量。它说什么？
@KNeerajLal 刚刚检查得到：D/error:: retrofit.RetrofitError: com.google.gson.JsonSyntaxException: java.lang.IllegalStateException: 预期 BEGIN_OBJECT 但在第 1 列第 1 行路径 $
我认为你的接口方法不正确。如果我是你，我会这样写 Callback问题是，retrofit 需要一个 JSON 输出，它可以适合 login.php 的 User 类，但所有 login.php 给出的是“未完成”或“完成”。
你必须让你的php脚本输出json格式如下，
"name": "name",
"username": "username",
"password": "password",
"email": "email",
"id": id
您必须编辑 php 代码。会是这样的，
<?php
$objConnect = mysql_connect("localhost","username","password");
$objDB = mysql_select_db("database");
/*** for sample */
// $_POST["username"] = "sun";
// $_POST["password"] = "live";
$username = $_REQUEST['username'];
$password = $_REQUEST['password'];
$strSQL = "select * from test_users where username = '".$username."' and password = '".$password."' ";
$objQuery = mysql_query($strSQL);
$intNumRows = mysql_num_rows($objQuery);
if($intNumRows==0)
$response["success"] = 0;
$response["name"] = "";
$response["username"] = "";
$response["password"] = "";
$response["email"] = "";
$response["id"] = 0;
else
$row = mysql_fetch_array($objQuery);
$response["success"] = 1;
$response["name"] = $row["name"];
$response["username"] = $row["username"];
$response["password"] = $row["password"];
$response["email"] = $row["email"];
$response["id"] = $row["id"];
mysql_close($objConnect);
echo json_encode($response, JSON_NUMERIC_CHECK);
?>
还将int success; 字段添加到User 类并使用它来验证输出。所以你的成功方法是，
@Override
public void success(User s, Response response)
if(s != null && s.getSuccess() == 1)
Toast.makeText(LoginActivity.this,"Logged In",Toast.LENGTH_LONG).show();
Log.d("name:", s.getName());
Log.d("email:", s.getEmail());
Log.d("id:", String.valueOf(s.getId()));
else
Toast.makeText(LoginActivity.this,"Invalid!",Toast.LENGTH_LONG).show();
更新
id 字段是数据库中的int，但它以字符串形式返回 id。请参阅this 线程。在 php 中将 JSON_NUMERIC_CHECK 添加到 json_encode() 会有所帮助。
【讨论】：
我试过了，它可以工作，但是在 " " 双引号中获取 id，这是我的 php 的输出： "success":1,"name":"Sun Live","username" :"sun","password":"live","email":"sun@live.ifno","id":"1" 整数字段中没有 bro id 字段：id Primary int(11)以上是关于java.lang.IllegalStateException：应为 BEGIN_OBJECT 但为 STRING的主要内容，如果未能解决你的问题，请参考以下文章