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;
}
}
}
这是我的员工.php文件，用于抓取商店的员工列表：
<?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);
}
因此，当我收到响应并将其打印到日志时，这就是我得到的：
员工响应： {"错误"：false，"employee_list"：[{"employee_name"："employee1"}，{"employee_name"："employee2"}，{"employee_name"："employee3"}]}
我的问题是，如何提取内容并将它们放入列表视图中？
首先，你从 php 代码中得到一个 JSONObject 响应。"employee_list"对象引用一个数组，您已经解析了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...
您应该在编码之前平展数组。
例如
改变：
$response['employee_list'][] = $row;
自：
$employees[] = $row;
编辑：
var_dump $row确保这有效并...
if( isset( $row[ 'employee_name' ] ) )
{
$employees[] = $row[ 'employee_name' ];
}
在 while 循环之后：
$response[ 'employees' ] = $employees;
这应该允许您将其作为 json 数组而不是 json 对象接收（employee_list键添加的额外级别）
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;
}