如何 return 使用自定义状态代码和消息作为 JSON 响应主体响应 Slim Framework

Question

我是 Slim 框架的新手。我想知道如何 return 根据需要使用正文和状态代码进行响应。目前，它仅在查询 returns 记录时按我的意愿执行，但如果查询失败（例如 select * 来自 notexiststable）或 returns 什么都没有，我看到了 Firefox在这两种情况下，状态都是 204。

代码如下：

$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;

});

提前致谢！

Answer 1

状态代码 204 是 No Content 并且 HTTP 规范指出消息中不能有正文。

将错误的状态代码更改为错误代码，例如 400。

即。变化：

if (pg_num_rows($result) == 0) {
    $data = array("Warning Message" => "There's no existent Model!");
    $newResponse = $response->withJson($data, 204);
}

至：

if (pg_num_rows($result) == 0) {
    $data = array("Warning Message" => "There's no existent Model!");
    $newResponse = $response->withJson($data, 400);
}

并且 JSON 将发送给客户端。

如何 return 使用自定义状态代码和消息作为 JSON 响应主体响应 Slim Framework

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