Question

如何在slim框架中同时读取Json和File？我正在尝试这个：

$entityManager = $this -> get(‘em’);
try{
        //Salvar foto
        $files = $request->getUploadedFiles();
        $newimage = $files['foto'];
        if ($newimage->getError() === UPLOAD_ERR_OK) {
            $uploadFileName = $newimage->getClientFilename();
            $type = $newimage->getClientMediaType();
            $name = uniqid('img-' . date('d-m-y') . '-');
            $name .= $newimage->getClientFilename();
            //  $imgs[] = array('url' => '/Photos/' . $name);
            //local server
            $newimage->moveTo("/home/citycare/imgs/$name");#/home/citycare/Imgs/User/$name
            //localdev
            $photoURL = "/home/citycare/public_html/Imgs/user/$name";#/home/citycare/Imgs/User/$name
        }

    //Instância da entidade Login
    $login = new Login();
    //recuperando o parâmetro objeto login do json


    $fk_login_cidadao = $request->getParam('fk_login_cidadao');
    //setando valores do objeto login
    $login ->setLogin($fk_login_cidadao['login']);
    $login ->setEmail($fk_login_cidadao['email']);
    $login ->setSenha($fk_login_cidadao['senha']);
    $login ->setStatus_login($fk_login_cidadao['status_login']);
    $login ->setAsAdministrador($fk_login_cidadao['administrador']);
    //salvando login
    $entityManager->persist($login);
    $entityManager->flush();
    //buscando login recém salvo
    $loginRepository = $entityManager->getRepository('App\Models\Entity\Login');
    //pegando login
    $loginCidadao = $loginRepository->find($login->getId_login());

    //Instância da entidade Cidadao
    $cidadao = new Cidadao();
    //setando valores do objeto cidadao
    $cidadao ->setFk_login_cidadao($loginCidadao);
    $cidadao->setNome($request->getParam('nome'));
    $cidadao->setSexo($request->getParam('sexo'));
    $cidadao ->setSobrenome($request->getParam('sobrenome'));
    $cidadao ->setEstado($request->getParam('estado'));
    $cidadao ->setCidade($request->getParam('cidade'));
    $cidadao ->setDir_foto_usuario($request->getParam($photoURL));
    $entityManager->persist($cidadao);
    $entityManager->flush();
    //retornando confirmação do evento completo
    $return = $response->withJson(["result" => true],201)->withHeader('Content-type', 'application/json');


} catch (Exception $ex){
    //código e mensagem do erro
    throw new Exception($ex->getMessage(), $ex->getCode());
}
return $return;

}

but I’m getting this {“message”:"An exception occurred while executing ‘INSERT INTO login (email, login, senha, status_login, administrador) VALUES (?, ?, ?, ?, ?)’ with params [null, null, “”, null, null]:\n\nSQLSTATE[23000]:

我正在从Android应用程序中收到Json和File，现在我正在尝试使用内容类型的json帖子;

内容 - 处置：表单数据;名称= “cidadao”

这是json：

内容类型：multipart / form-data;字符集= utf-8的 { “CIDADE”： “艾瓦巴”， “国家体制”： “塞阿拉州”， “fk_login_cidadao”：{ “administrador”：假的， “电子邮件”： “jvictor@jvictor.com”， “登陆”： “kikokiko”，“senha “：” 123123123" ， “status_login”：真}， “诺姆”： “jijo”， “sexo”： “Masculino”， “sobrenome”： “jdiasjdia”}

Answer 1

你的json密钥是cidadao。 List<string> urls = xmldoc2.Descendants("PID").Select(x => x.Attribute("5678").Value).ToList(); <Tools> <Tools> <Tool> <ID>1234</ID> <PID>5678</PID> <Name>Test</Name> </Tool> </Tools> <Type> <Parent> <ID>5678</ID> <PID>9999</PID> <Name>Test2</Name> </Parent> </Type> </Tools>，您需要对其进行解码。
解码后的字符串被解析为对象，您可以像这样访问它：$fk_login_cidadao = json_decode( $request->getParam('cidadao') );

阅读Json和File Slim Framework

1 个答案: