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[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况下调用我的Api时,它应该返回一个json响应,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);
}
在这种情况下,当我没有使用邮递员在标头中发送User-Token字段时,我会正确地获得响应,但由于该响应,我收到了未定义索引“ User-token”的错误,我不希望显示此错误
只需稍微修改一下代码即可。
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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