php - Slim\Http\Request 类的对象无法转换为 double

Question

我正在尝试发送一个友谊请求，将用户 ID 组合在一个表中，但似乎我在某个地方犯了一个错误。这是代码片段:

public function sendRequest($user_one_id, $user_two_id, $status, $action_user_id) {
    $stmt = $this->conn->prepare("INSERT INTO relationship(user_one_id, user_two_id, status, action_user_id) VALUES (?, ?, ?, ?)");
    $stmt->bind_param("iiii", $user_one_id, $user_two_id, $status, $action_user_id);
    $result = $stmt->execute(); // HERE IS AN ERROR
    $stmt->close();

    if (!$result) {
        return NULL;
    } else {
        return TRUE;
    }
}

我得到的错误:

Object of class Slim\Http\Request could not be converted to double

编辑:

这就是我调用方法的方式:

$app->post('/send_request', function() use ($app) {

// check for required params
verifyRequiredParams(array('user_one_id', 'user_two_id', 'status', 'action_user_id'));

// reading post params
$user_one_id = $app->request('user_one_id');
$user_two_id = $app->request('user_two_id');
$status = $app->request('status');
$action_user_id = $app->request('action_user_id');
$response = array();

$db = new DbHandler();

$relationship = $db->sendRequest($user_one_id, $user_two_id, $status, $action_user_id);

if ($relationship != NULL) {
    $response["error"] = false;
    $response["request"]["user_one_id"] = $relationship["user_one_id"];
    $response["request"]["user_two_id"] = $relationship["user_two_id"];
    $response["request"]["status"] = $relationship["status"];
    $response["request"]["action_user_id"] = $relationship["action_user_id"];
} else {
    $response["error"] = true;
    $response["message"] = "An error occurred. Please try again later.";
}
echoRespnse(200, $response);
});

Answer 1

verifyRequiredParams 函数中发生了什么？在您发布的代码中，我没有看到 Slim\Http\Request 对象的任何转换。

$user_one_id = $app->request('user_one_id');
$user_two_id = $app->request('user_two_id');
$status = $app->request('status');
$action_user_id = $app->request('action_user_id');

据我所知，这些是您对 Slim 请求对象执行任何操作的唯一代码行。您能否验证您设置的这些变量中的值是否正确？喜欢 var_dump() 它们并在您的代码中放置一个 exit 并查看其内容。