Question

我使用以下PHP代码从服务器上的表中获取员工列表：

DB_Functions.php中的

函数：

public function getEmployeeList($name) {
    $stmt = $this->con->prepare("SELECT employee_name FROM employees WHERE name = ?");

    $stmt->bind_param('s', $name);

    if ($stmt->execute()) {
        $result = $stmt->get_result();
        $stmt->close();

        if ($result->num_rows < 1) {
            return NULL;
        } else {
            return $result;
        }
    }
}

这是我的employees.php文件，用于获取商店的员工列表：

<?php

require_once 'include/DB_Functions.php';
$db = new DB_Functions();

$response = array('error' => FALSE);

if (isset($_POST['name'])) {
    $name = $_POST['name'];

    $result = $db->getEmployeeList($name);

    if (is_null($result) || !$result) {
        $response['error'] = TRUE;
        $response['error_msg'] = 'No employees have been added to this profile.';

        echo json_encode($response);
    } else {
        $response['error'] = FALSE;
        while ($row = $result->fetch_assoc()) {
            $response['employee_list'][] = $row;
        }

        echo json_encode($response);
    }
} else {
    $response['error'] = TRUE;
    $response['error_msg'] = 'You have not logged in to your store\'s account, please log in first.';

    echo json_encode($response);
}

?>

最后，这是我的Android代码，用于获取此列表：

private void loadEmployeeList(final String name) {
    String tag_string_req = "req_employees";

    pDialog.setMessage("Fetching employees...");
    showDialog();

    StringRequest stringRequest = new StringRequest(Method.POST, AppConfig.URL_EMPLOYEES, new Response.Listener<String>() {
        @Override
        public void onResponse(String response) {
            Log.d(TAG, "Employee Response: " + response);
            hideDialog();

            try {
                JSONObject jsonObject = new JSONObject(response);
                boolean error = jsonObject.getBoolean("error");

                if (!error) {

                } else {
                    String errorMsg = jsonObject.getString("error_msg");
                    Toast.makeText(getApplicationContext(), errorMsg, Toast.LENGTH_SHORT).show();
                }
            } catch (JSONException e) {
                e.printStackTrace();
                Toast.makeText(getApplicationContext(), "Connection error", Toast.LENGTH_SHORT).show();
            }
        }
    }, new Response.ErrorListener() {
        @Override
        public void onErrorResponse(VolleyError error) {
            Log.e(TAG, "Employee Error: " + error.getMessage());
            Toast.makeText(getApplicationContext(), error.getMessage(), Toast.LENGTH_SHORT).show();
            hideDialog();
        }
    }) {
        @Override
        protected Map<String, String> getParams() {
            Map<String, String> params = new HashMap<>();
            params.put("name", name);

            return params;
        }
    };
    AppController.getInstance().addToRequestQueue(stringRequest, tag_string_req);
}

所以，当我收到回复并将其打印到日志时，这就是我得到的：

员工响应：{＆＃34;错误＆＃34;：false，＆＃34; employee_list＆＃34;：[{＆＃34; employee_name＆＃34;：＆＃34; employee1＆＃34;}，{＆＃34; employee_name＆＃34;：＆＃34;和Employee2＆＃34;}，{＆＃34; employee_name＆＃34;：＆＃34; employee3＆＃34;}]}

我的问题是，如何提取内容并将其放入列表视图？

Answer 1

首先，您将从php代码获取JSONObject响应。 “employee_list”对象引用一个Array，您可以解析此结果并将其存储在Arraylist中，以便在Listview实现中使用。

List<String> Employees = new ArrayList<String>();

JSONObject result = new JSONObject(response);
JSONArray json = result.getJSONArray("employee_list");
for (int i=0; i<json.length(); i++) {
    JSONObject data= json.getJSONObject(i);
    String name = data.getString("employee_name");
    Employees.add(name);
}

试试这个...

Answer 2

您应该在编码之前展平数组。

e.g。

变化：

$response['employee_list'][] = $row;

要：

$employees[] = $row;

编辑：

var_dump $ row以确保其有效并且......

if( isset( $row[ 'employee_name' ] ) )
{
$employees[] = $row[ 'employee_name' ];
}

并在while循环之后：

$response[ 'employees' ] = $employees;

这应该允许你将它作为json数组而不是json对象（employee_list键添加的额外级别）接收它。

Answer 3

试试这个：

$result = '{"error":false,"employee_list":[{"employee_name":"employee1"},{"employee_name":"employee2"},{"employee_name":"employee3"}]}';

$arr = array();

$t = objToArray($result, $arr);

print_r($t);


public function objToArray($obj, &$arr) {

        if (!is_object($obj) && !is_array($obj)) {
            $arr = $obj;
            return $arr;
        }
        foreach ($obj as $key => $value) {
            if (!empty($value)) {
                $arr[$key] = array();
                $this->objToArray($value, $arr[$key]);
            } else {
                $arr[$key] = $value;
            }
        }
        return $arr;
    }

如何使用相同的密钥从json对象中获取项目？

3 个答案: