Question

这里这个代码正在运行，然后突然又经过另一次测试后它确实工作并说json包含了像我这样的html元素，我不能让它再次工作我需要帮助

<?php

include("config.php");

if(isset($_POST["email"]) && isset($_POST["username"]) && isset($_POST["password"])){
    $email = $_POST["email"];
    $username = $_POST["username"];
    $password = $_POST["password"];
}   

$con = mysqli_connect(DATABASE_HOST, DATABASE_USERNAME, DATABASE_PASSWORD, DATABASE_NAME);

$response = array();

$statement = mysqli_prepare($con, "SELECT * FROM accounts WHERE email = ? OR username = ?");
mysqli_stmt_bind_param($statement, "ss", $email, $username);
$result = mysqli_stmt_execute($statement);
$rows = $result->num_rows;

if($rows > 0){
    $response["success"] = false;
    $response["message"] = "Email or Username already exists.";
}else{
    mysqli_stmt_close($statement);
    $statement2 = mysqli_prepare($con, "INSERT INTO accounts (email, username, password) VALUES (?, ?, ?)");
    mysqli_stmt_bind_param($statement2, "sss", $email, $username, $password);
    $result2 = mysqli_stmt_execute($statement2);
    $rows2 = $result2->affected_rows;

    if($rows2 > 0){
        $response["success"] = true;
        $response["message"] = "Account created successfuly.";
    }else{
        $response["success"] = false;
        $response["message"] = "Creation error.";
    }
}

echo json_encode($response);

mysqli_close($con);

?>

当我在andoid中解析它时它告诉我它包含的东西thas是没有字符串..这是我的解码json的android代码

private void register() {
        Response.Listener<String> responselistener = new Response.Listener<String>() {
            @Override
            public void onResponse(String response) {
                try {
                    JSONObject jsonresponse = new JSONObject(response);
                    boolean success = jsonresponse.getBoolean("success");
                    String message = jsonresponse.get("message").toString();
                    if(success){
                        Intent gotoContinueRegisterActivityIntent = new Intent(RegisterActivity.this, ContinueRegisterActivity.class);
                        startActivity(gotoContinueRegisterActivityIntent);
                    }else {
                        error.setText(message);
                    }
                } catch (JSONException e) {
                    e.printStackTrace();
                }
            }
        };
        Response.ErrorListener errorListener = new Response.ErrorListener() {
            @Override
            public void onErrorResponse(VolleyError error2) {
                error2.printStackTrace();
            }
        };
        RegisterRequest registerRequest = new RegisterRequest(s_email, s_username, s_password, responselistener, errorListener);
        RequestQueue requestQueue = Volley.newRequestQueue(getBaseContext());
        requestQueue.add(registerRequest);
    }

Answer 1

您最好实际发送标题，让您的Android应用知道它已接收json，因为它可能认为它获取了内容类型：text/html

尝试将此标题放在PHP脚本的顶部：

<?php
header("Content-Type: application/json");

我无法解析json数据

1 个答案: