[英]What if I don't want to do composer update, I just want to use what ever in composer.json and composer.lock
[英]I just want to return JSON error response when ever the token field is missing in headers in codeigniter
只是只要想在缺少令牌的情況下調用我的Api時,它應該返回一個json響應,提示“令牌是必需的”。
$headers = apache_request_headers();
$token = $headers['User-Token'];
if(isset($token)){
$user_details = $this->user_basic_model->get_user_by_token($token);
if (!empty($user_details))
{
$data = $this->fare_model->get_all_fare();
if (empty($data)) {
$arr = array('msg' => 'Error!', 'status_code' => 404);
echo json_encode($arr);
}
else {
$success_array = array('msg' => 'Success!', 'status_code' => 200);
echo json_encode(array_merge($data, $success_array));
}
}
else
{
$arr = array('success' => 'false', 'msg' => 'Unauthorize Request',
'status_code' => 401);
echo json_encode($arr);
}
}
else
{
$arr = array('success' => 'false', 'msg' => 'Token is required',
'status_code' => 401);
echo json_encode($arr);
}
在這種情況下,當我沒有使用郵遞員在標頭中發送User-Token字段時,我會正確地獲得響應,但由於該響應,我收到了未定義索引“ User-token”的錯誤,我不希望顯示此錯誤
只需稍微修改一下代碼即可。
if (isset($headers['User-Token'])) {
$token = $headers['User-Token'];
if (empty($token)) {
//your code
} else {
$arr = array('success' => 'false', 'msg' => 'Token is required',
'status_code' => 401);
echo json_encode($arr);
exit();
}
} else {
$arr = array('success' => 'false', 'msg' => 'Token is required',
'status_code' => 401);
echo json_encode($arr);
exit();
}
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