php文件处理

[lijin_12456]

本地文件怎么通过php代码读取并以文件的形式请求接口

本地文件：D:\note\20190428093404.jpg
想通过php代码直接访问图片，并将图片以文件的形式传到第三方保存图片接口，要怎么实现？求大神指点，万分感谢

[lijin_12456]

$path = 'D:\note\123.mp4'; $data = [ 'advertiser_id' => 1665001408972814, 'video_signature' => '7732573e84c62d7e5e6b25deec3056c3', 'video_file' => '', //video_file 就是要本地选择上传的文件 ]; $url = 'http://yii.local.com/index.php/basic/import-guest/deal'; $ch = curl_init(); curl_setopt($ch, CURLOPT_URL, $url); curl_setopt($ch, CURLOPT_POST, 1); curl_setopt($ch, CURLOPT_RETURNTRANSFER, 1); curl_setopt($ch, CURLOPT_POSTFIELDS, $data); curl_setopt($ch, CURLOPT_SSL_VERIFYPEER, FALSE); curl_setopt($ch, CURLOPT_SSL_VERIFYHOST, FALSE); curl_exec($ch); curl_close($ch);

[chendong_j]

function curl_post($url,$post_data) { $ch = curl_init(); curl_setopt($ch, CURLOPT_URL, $url); curl_setopt($ch, CURLOPT_POST, 1); curl_setopt($ch, CURLOPT_SAFE_UPLOAD, true); curl_setopt($ch, CURLOPT_RETURNTRANSFER, 1); curl_setopt($ch, CURLOPT_POSTFIELDS, $post_data); $output = curl_exec($ch); curl_close($ch); return $output; } $url是上传地址 $post_data是上传的参数 $post_data = [ 　　'name'　　=> 'xxxx', 　　'file'　　=> new \CURLFile('绝对路径'),　　　　 ]; 搬运结束