我正在使用php和在BOX_API中上传一个文件。但我错过了一些东西。这是我第一次使用像这样的接口--我读过手册。但这对我来说是一种困惑。我的问题有两个:
-First,为什么您必须只传递给post调用文件的名称,而不是文件上传的整个文件头?在表单中上传文件不像通过post调用传递文件的名称;
-secondly,因此,我是否必须为文件上传创建一个表单,或者仅仅是一个文本区域,在其中写入要传递给BOX的文件的名称?
更新:这是我上传表格的代码:
$form = new Zend_Form;
$form->setAction('/imball-reagens/public/upload')
->setMethod('post');
$file = new Zend_Form_Element_File('file');
$file->setLabel('Choose a file to upload:');
$file->addValidator('alnum');
$file->setRequired(true);
$form->addElement($file);
$access_token = new Zend_Form_Element_Hidden(array('name' => 'access_token', 'value' => $result->access_token));
$form->addElement($access_token);
$refresh_token = new Zend_Form_Element_Hidden(array('name' => 'refresh_token', 'value' => $result->refresh_token));
$form->addElement($refresh_token);
$form->addElement('submit', 'upload', array('label' => 'Upload File'));
echo $form;这是到方框API的POST,该API在表单之后:
$access_token= $this->getRequest()->getParam('access_token');
$client = new Zend_Http_Client('https://upload.box.com/api/2.0/files/content');
$client->setMethod(Zend_Http_Client::POST);
$client->setHeaders('Authorization: Bearer '.$access_token);
$data = $_FILES["file"]["name"];
$client->setParameterPost(array(
'filename' => '@'.$data,
'parent_id' => '0'
));
$response = $client->request()->getBody();
$this->view->response= $response;
$result = json_decode($response);它引发的错误如下:
{"type":"error","status":400,"code":"invalid_request_parameters","help_url":"http:\/\/developers.box.com\/docs\/#errors","message":"Invalid input parameters in request","request_id":"172518183652dcf2a16af73"}发布于 2014-01-20 19:26:34
在没有看到所有代码的情况下调试是很棘手的,但在粘贴的位中,它看起来像是将$_FILES["file"]["name"]传递给API --这只包含用户上传的文件的原始名称--您需要将位置传递到服务器上的文件,该文件将数据发送到Box API客户端,以便它能够抓取数据并将其发送到服务器--这应该存储在$_FILES["file"]["tmp_name"]中。
我建议将代码更改为以下代码,然后再试一次:
$access_token= $this->getRequest()->getParam('access_token');
$client = new Zend_Http_Client('https://upload.box.com/api/2.0/files/content');
$client->setMethod(Zend_Http_Client::POST);
$client->setHeaders('Authorization: Bearer '.$access_token);
$data = $_FILES["file"]["tmp_name"];
$client->setParameterPost(array(
'parent_id' => '0'
));
$client->setFileUpload($data, 'filename');
$response = $client->request()->getBody();
$this->view->response= $response;
$result = json_decode($response);https://stackoverflow.com/questions/21230295
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