当前位置:Gxlcms > PHP教程 > postman上传图片到服务器方法

postman上传图片到服务器方法

时间:2021-07-01 10:21:17 帮助过:89人阅读

本文主要和大家分享postman上传图片到服务器方法,结合图文的形式,希望能帮助到大家。

1. postman的设置:

这里写图片描述

2. PHP代码:

public function byte(){
    $base_path = "./uploads/"; //存放目录
    if(!is_dir($base_path)){
        mkdir($base_path,0777,true);
    }
    $target_path = $base_path . basename ( $_FILES ['upload'] ['name'] );
    if (move_uploaded_file ( $_FILES ['upload'] ['tmp_name'], $target_path )) {
        $info['mess'] = 'ok';
        //$info['flag'] = 0;
        exit(json_encode($info));
    } else {
        $array = array (
            "flag" => 0,
            "mess" => "There was an error uploading the file, please try again!" . $_FILES ['upload'] ['error']
        );
        exit(json_encode ( $array ));
    }
}

3. api说明:

请求方式参数是否必须参数说明
POSTuploadY所要上传的文件

这里写图片描述

1. postman的设置:

这里写图片描述

2. PHP代码:

public function byte(){
    $base_path = "./uploads/"; //存放目录
    if(!is_dir($base_path)){
        mkdir($base_path,0777,true);
    }
    $target_path = $base_path . basename ( $_FILES ['upload'] ['name'] );
    if (move_uploaded_file ( $_FILES ['upload'] ['tmp_name'], $target_path )) {
        $info['mess'] = 'ok';
        //$info['flag'] = 0;
        exit(json_encode($info));
    } else {
        $array = array (
            "flag" => 0,
            "mess" => "There was an error uploading the file, please try again!" . $_FILES ['upload'] ['error']
        );
        exit(json_encode ( $array ));
    }
}

3. api说明:

请求方式参数是否必须参数说明
POSTuploadY所要上传的文件

这里写图片描述

以上就是postman上传图片到服务器方法的详细内容,更多请关注Gxl网其它相关文章!

人气教程排行