java - 来自服务器的 JSON 数组获取两个值

Question

我是编程新手，请原谅我的问题。我目前正在从远程服务器获取列officer_name。将其解析为 JSON 并发送到应用程序。但我想添加另一个具有 INT 数据类型的列数据 reference 并将其与 officer_name 一起发送到应用程序。我怎样才能实现这样的目标？

sql查询

$sql = "select officer_name from user_tbl where user_name like '".$user_name."' and password like '".$password."' and activate = 1;";
$result = mysqli_query($con,$sql);
$response = array();
if (mysqli_num_rows($result)>0)
{
$row = mysqli_fetch_row($result);
$officer_name = $row[0];
$code = "Login_Success";
array_push($response,array("code"=>$code,"officer_name"=>$officer_name));
echo json_encode($response);

获取 JSON 的 Android Java 代码

public void onResponse(String response) {
                                try {
                                    JSONArray jsonArray = new JSONArray(response);
                                    JSONObject jsonObject = jsonArray.getJSONObject(0);
                                    String code = jsonObject.getString("code");
                                    if (code.equals("Login_Failed"))
                                    {
                                        builder.setTitle("Login Error");
                                        displayAlert(jsonObject.getString("message"));
                                    }
                                    else
                                    {
                                        Intent intent = new Intent(Login.this,MainActivity.class);
                                        Bundle bundle = new Bundle();
                                        bundle.putString("officer_name", jsonObject.getString("officer_name"));
                                        intent.putExtras(bundle);
                                        startActivity(intent);
                                    }

到目前为止，我只能从服务器获取officer_name。我想从我的数据库添加一个具有 INT 值的数据引用列。

Answer 1

看起来是这样，因为您仅选择 OFFICER_NAME 而没有其他列。 当您想选择更多时，只需输入(SELECTOfficer_name、data_u_want2、data_u_need3 FROM 等..)

我获取用户数据的脚本通常如下所示:

$username = $_POST["username"];

                                  // SELECT what you want
$statement = mysqli_prepare($con,"SELECT * FROM user WHERE username = ?");

mysqli_stmt_bind_param($statement,"s",$username);
mysqli_stmt_execute($statement);

mysqli_stmt_store_result($statement);

mysqli_stmt_bind_result($statement, 
                        $userID, 
                        $username, 
                        $password, 
                        $email, 
                        $age, 
                        $realname, 
                        $streetname, 
                        $postcode, 
                        $city,
                        $state);

$response = array();

$response["success"] = false;

while(mysqli_stmt_fetch($statement)){
    $response["success"] = true;        
    $response["username"] = $username;
    $response["password"] = $password;
    $response["email"] = $email;
    $response["age"] = $age;
    $response["realname"] = $realname;
    $response["streetname"] = $streetname;
    $response["postcode"] = $postcode;
    $response["city"] = $city;
    $response["state"] = $state;        
}

echo json_encode($response);

您将来应该使用准备好的语句来防止 sql 注入(inject)。

您可以将 $username 更改为 $officer_name 或其他名称，并将您想要的列放入 $response