Android/php：使用 MySqli 数据库表中的 JSONArray 行填充 AutoCompleteTextView 列表

Question

我试图在我的应用程序中实现一个数据库搜索功能，它搜索一个用户表并返回一个用户数组，这些用户喜欢搜索字符串。我需要这个数组来填充 acTextView 下方下拉的列表，但我无法让 JSONArray 正确传递，而且我不确定如何将数据传递给 acTextView。对此的任何帮助将不胜感激！

应用端搜索活动：

public class PopupAddContact extends AppCompatActivity {

@Override
protected void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    setContentView(R.layout.popup_add_contact);

    getWindow().setLayout(ConstraintLayout.LayoutParams.WRAP_CONTENT, ConstraintLayout.LayoutParams.WRAP_CONTENT);
    getWindow().setBackgroundDrawable(new ColorDrawable(Color.TRANSPARENT));

    final AutoCompleteTextView searchInput = (AutoCompleteTextView) findViewById(R.id.searchText);

    searchInput.addTextChangedListener(new TextWatcher() {
        @Override
        public void beforeTextChanged(CharSequence charSequence, int i, int i1, int i2) {

        }

        @Override
        public void onTextChanged(CharSequence charSequence, int i, int i1, int i2) {
            final String name = searchInput.getText().toString();

            Response.Listener<String> responseListener = new Response.Listener<String>() {
                @Override
                public void onResponse(String response) {
                    try {
                        System.out.println(response);
                        //ArrayList<String> array = new ArrayList<String>();
                        JSONArray jsonResponse = new JSONArray(response);

                        System.out.println(jsonResponse);
                        System.out.println("working");


                        /*
                        if (jsonResponse != null) {

                            Integer contact_user_id = jsonResponse.getInt();
                            String contact_name = jsonResponse.getString("name");
                            String contact_email = jsonResponse.getString("email");



                        } else if(response != null){
                            Toast.makeText(PopupAddContact.this, "No users found", Toast.LENGTH_SHORT).show();
                        } */

                    } catch (JSONException e) {
                        e.printStackTrace();
                    }
                }
            };

            PopupContactRequest AddContactRequest = new PopupContactRequest(name, responseListener);
            RequestQueue queue = Volley.newRequestQueue(PopupAddContact.this);
            queue.add(AddContactRequest);

        }

        @Override
        public void afterTextChanged(Editable editable) {

        }
    });
}

用于检索用户数组的 PHP 脚本：

if(!empty($_POST["name"])){ // check post data and then start
$mysqli = new mysqli("db creds"); 

$name = $_POST["name"];
$response=array(); 

if (mysqli_connect_errno()) {
    printf("Connect failed: %s\n", mysqli_connect_error());
    exit();
}

$query = "SELECT user_id, name, email FROM users WHERE name LIKE %?%"; 

if ($stmt = $mysqli->prepare($query)) { // prepare query

    $stmt->bind_param("s", $name); // bind parameters

    $stmt->execute();

    $stmt->bind_result($data);

    while ($stmt->fetch()) {
        $response[] = $data; //assign each data to response array
    }

}

echo json_encode($response);

$mysqli->close();

}else{
    echo "no users found";
}

Answer 1

我认为您的 php 文件没有回显有效的 json，请尝试更改以下行：

if ($stmt = $mysqli->prepare($query)) { // prepare query

    $stmt->bind_param("s", $name); // bind parameters

    $stmt->execute();

    $stmt->bind_result($user_id, $name, $email);
    while ($stmt->fetch()) {
        $response[] = [
            "user_id" => $user_id,
            "name" => $name,
            "email" => $email,
        ]; //assign each data to response array
    }
}

echo json_encode(array_values($response));

首先您应该调用 php 网站并在 json 验证器中验证结果，例如 https://jsonlint.com/，然后您可以继续使用 java 代码。

在回显 json 响应之前编写 json 标头也是一个好习惯：

header('Content-Type: application/json');
echo json_encode($response);