使用PHP中的卷曲上载图片上载数据

Question

我有一个API的URL。我的任务是在形式数据库中存储的数据。但是，当使用卷发工作时，抛出错误URL未设置。请看代码。

 $full_name=$_REQUEST['full_name'];
          $email=$_REQUEST['email'];
          $mobile_number=$_REQUEST['mobile_number'];
          $pass=$_REQUEST['pass'];
          $work_address=$_REQUEST['work_address'];
          $file0=$_FILES['doc1']['name'];
          $file1=$_FILES['doc2']['name'];
          $file2=$_FILES['doc3']['name'];
        $url="http://34.195.215.247/sudzero/api/washer-register.php?fname='" . $full_name . "'&email='" . $email . "'&pass='" . $pass . "'&address='" . $work_address . "'&phone='".$mobile_number."'&file0='".$file0."'&file1='".$file1."'&file2='".$file2."'"; echo $url;
        $curl    = curl_init(); 
          $headers = array("Content-Type:multipart/form-data");
            curl_setopt_array($curl, array(
            curlopt_returntransfer => 1,
            curlopt_url => $url,
            CURLOPT_HEADER => true,
            curlopt_useragent => 'cubewires sample curl request',
            curlopt_post => 1,
            CURLOPT_HTTPHEADER => $headers,
            curlopt_postfields => array(
                'fname' => $full_name,
                'email' => $email,
                'pass' => $pass,
                'address' => $work_address,
                'phone' => $mobile_number,
                'file0' => $file0,
                'file1' => $file1,
                'file2' => $file2
            )
        ));
        $resp = curl_exec($curl);
        if (curl_errno($curl)) {
        $msg = curl_error($curl);
            }
            else {
                $msg = 'data uploaded successfully.';
            }
        curl_close($curl);
        $return = array('msg' => $msg);
        echo json_encode($return);

请帮助我的朋友。我失去了两个小时。谢谢

Answer 1

大写卷曲选项

curlopt_url => $url,
//change the above to 
CURLOPT_URL => $url,

也是以下字段

CURLOPT_RETURNTRANSFER => 1,
CURLOPT_USERAGENT => 'cubewires sample curl request',
CURLOPT_POST => 1,

Answer 2

$ file0 = $ _files ['doc1'] ['tmp_name'];

$ file1 = $ _ files ['doc2'] ['tmp_name'];

$ file2 = $ _ files ['doc3'] ['tmp_name'];