Question

下面我用postman测试它返回响应但是当我检查PHP时它没有返回任何响应。请帮助我。

cURL文件创建：

$curl_file = curl_file_create(\Yii::$app->basePath.'/web/uploads/'.$uploadedFile->name,'pbix',$uploadedFile->baseName);
            $params = ['file' => $curl_file];

cURL流程：

$curl = curl_init();
        curl_setopt_array($curl, array(
          CURLOPT_URL => "https://***.***.**.***/v1.0/collections/**/workspaces/**/***?datasetDisplayName=new test",
          CURLOPT_RETURNTRANSFER => true,
          CURLOPT_ENCODING => "",
          CURLOPT_HTTP_VERSION => CURL_HTTP_VERSION_1_1,
          CURLOPT_CUSTOMREQUEST => 'POST',
          CURLOPT_POSTFIELDS => $params,
          CURLOPT_HTTPHEADER => array(
            "authorization: key",
          ),
        ));
        curl_setopt($curl, CURLOPT_SSL_VERIFYHOST, 0);
        curl_setopt($curl, CURLOPT_SSL_VERIFYPEER, 0);

        $response = curl_exec($curl);
        $err = curl_error($curl);

        curl_close($curl);
        if ($err) {
          echo "cURL Error #:" . $err;
        } else {
          echo $response;
        }

Answer 1

问题是参数中的空格。space应替换为%20。你需要做的如下：

$url="https://***.***.**.***/v1.0/collections/**/workspaces/3b09eff6-69ae-4932-9419-14eeb69a9dbd/***?datasetDisplayName=new test";
$curl_url=str_replace(" ","%20",$url);

或者您必须使用urlencode。在编码要在URL的查询部分中使用的字符串时使用。

$curl_url="https://***.***.**.***/v1.0/collections/**/workspaces/3b09eff6-69ae-4932-9419-14eeb69a9dbd/***?datasetDisplayName=urlencode(new test)";

然后

curl_setopt_array($curl, array(
          CURLOPT_URL => $curl_url,
          CURLOPT_RETURNTRANSFER => true,
          CURLOPT_ENCODING => "",
          CURLOPT_HTTP_VERSION => CURL_HTTP_VERSION_1_1,
          CURLOPT_CUSTOMREQUEST => 'POST',
          CURLOPT_POSTFIELDS => $params,
          CURLOPT_HTTPHEADER => array(
            "authorization: key",
            ),
        ));

卷曲操作无效，但邮递员成功

1 个答案: