我正在使用一个连接到数据库并输出是否登录的servlet，当使用printWriter.write(JsonObject)时，我得到了其余的错误(意外 token L)。我正在使用tomcat服务器托管数据。
public class Login extends HttpServlet {
static final String JDBC_DRIVER = "com.mysql.jdbc.Driver";
static final String DB_URL = "jdbc:mysql://localhost/employeedatabase";
// Database credentials
static final String USER = "root";
static final String PASS = "admin";
static Connection conn = null;
static Statement stmt = null;
static ResultSet rs;
static PrintWriter out;
private static final long serialVersionUID = 1L;
/**
* @see HttpServlet#HttpServlet()
*/
public Login() {
super();
}
/**
* @see HttpServlet#doGet(HttpServletRequest request, HttpServletResponse response)
*/
protected void doGet(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
}
/**
* @see HttpServlet#doPost(HttpServletRequest request, HttpServletResponse response)
*/
protected void doPost(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
response.setContentType("application/json; charset=UTF-8");
out = response.getWriter();
String email = request.getParameter("username");
String password = request.getParameter("password");
String result = "";
if(Validation.validateNull(email) && Validation.validateNull(password)){
if(!Validation.validateEmail(email))
{
result = "Invalid email";
}
if(databaseFuctions.Login(email,password))
{
result = "Login accepted";
try {
Class.forName("com.mysql.jdbc.Driver").newInstance();
conn = DriverManager.getConnection(DB_URL, USER, PASS);
stmt = conn.createStatement();
getFromDatabase("manager",email,request,response);
// getFromDatabase("qa",email,request,response);
// getFromDatabase("developer",email,request,response);
} catch (Exception e1) {
// TODO Auto-generated catch block
System.out.println(e1.getMessage());
}
}
else if(!databaseFuctions.Login(email, password))
{
result = "Login invalid";
}}
else{
result = "No login/password entered";
}
out.write(result);
}
public static void getFromDatabase(String table,String email,HttpServletRequest request, HttpServletResponse response){
JSONObject JObject = new JSONObject();
ResultSet ds;
try {
ds = stmt.executeQuery("SELECT * from "+table+" where email = '"+email+"'");
while(ds.next())
{
int id = ds.getInt("id");
int salary = ds.getInt("salary");
String name = ds.getString("name");
String role = ds.getString("role");
String emailAddress = ds.getString("email");
String phone = ds.getString("phone");
JObject.put("id", id);
JObject.put("salary", salary);
JObject.put("name", name);
JObject.put("role", role);
JObject.put("email", emailAddress);
JObject.put("phone", phone);
}
}
catch (Exception e)
{
System.out.println(e.getMessage());
}
out.print(JObject.toString());
out.flush();
System.out.println(JObject.toString());
}
在系统中打印时，我会获取所有正确的数据，或者从静止状态检查原始数据后会得到正确的数据。但是我不太明白为什么打印机会抛出异常，但任何帮助都是惊人的
最佳答案
好的，如果由于客户端错误而导致，则返回的是格式错误的JSON值，因此返回的内容如下:{ id: 13, name: "Name"}Login invalid，则第一个字符L对于JSON语法无效。
这是因为您在响应中编写了getFromDatabase out.print(JObject.toString());方法的json字符串，并且在方法调用之后，将响应字符串result = "Login invalid"; out.write(result);添加到响应中，导致您的JSON无效。
解决此问题的一种方法是从方法getFromDatabase返回JSONObject，然后将result方法放入此对象JObject.put("result", result)中，然后将该对象写入响应中。