如何使用Slim Framework将响应自定义状态代码和消息作为JSON返回给响应主体
问题描述:
I'm new with Slim framework. I want to know how to return response with the body and status code as I wish. For now, it only performs as I wish if the query returns records, but not if the query failed (e.g. select * from notexiststable) or returns nothing and with Firefox I see the status is 204 in both cases.
So here's the code:
$app->get('/models', function (Request $request, Response $response) {
$conn = getConnection();
$result = pg_query($conn, "SELECT * FROM model where m_id = 2;"); //this will return 0 record!
if (!$result) {
$data = array("Error Message" => 'Query did not execute successfully');
$newResponse = $response->withJson($data, 500);
}
else {
if (pg_num_rows($result) == 0) {
$data = array("Warning Message" => "There's no existent Model!");
$newResponse = $response->withJson($data, 204);
}
else {
$data = array();
while ($row = pg_fetch_assoc($result)) {
$data['Models'][] = $row;
}
$newResponse = $response->withJson($data, 200);
}
}
return $newResponse;
});
Thank you in advanced!
答
Status code 204
is No Content
and the HTTP specification notes that there cannot be a body in the message.
Change the status code for your error to an error code, such as 400
.
ie. Change:
if (pg_num_rows($result) == 0) {
$data = array("Warning Message" => "There's no existent Model!");
$newResponse = $response->withJson($data, 204);
}
to:
if (pg_num_rows($result) == 0) {
$data = array("Warning Message" => "There's no existent Model!");
$newResponse = $response->withJson($data, 400);
}
and the JSON will be sent to the client.