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Python Back4app上载对象问题_Python_Python Requests_Parse Server_Back4app - Fatal编程技术网

Python Back4app上载对象问题

Python Back4app上载对象问题,python,python-requests,parse-server,back4app,Python,Python Requests,Parse Server,Back4app,我正试图通过back4app为classsound 我无法上传文件的问题 使用的代码: import requests headers = { "X-Parse-Application-Id": "hidden", "X-Parse-REST-API-Key": "hidden", } data = { "audio": { "__type&

我正试图通过
back4app
为class
sound

我无法上传文件的问题

使用的代码:

import requests

headers = {
    "X-Parse-Application-Id": "hidden",
    "X-Parse-REST-API-Key": "hidden",
}


data = {
    "audio": {
        "__type": open("a.mp3", 'rb'),
        "name": "a.mp3"

    },
    "displayText": "test"
}


def main(url):
    with requests.Session() as req:
        req.headers.update(headers)
        r = req.post(url, data=data)
        print(r.text)


main("https://parseapi.back4app.com/classes/sounds")
输出:

{"code":111,"error":"schema mismatch for sounds.audio; expected File but got Array"}

您首先需要上传文件:

导入json,httplib connection=httplib.HTTPSConnection('parseapi.back4app.com',443) connection.connect() connection.request('POST','/files/a.mp3',open('a.mp3','rb')。read(){ “X-Parse-Application-Id”:“${Application\u Id}”, “X-Parse-REST-API-Key”:“${REST\u-API\u-Key}”, “内容类型”:“音频/mpeg” }) result=json.load(connection.getresponse().read()) 然后创建对象:

connection.request('POST','/classes/sounds',json.dumps({
“显示文本”:“测试”,
“音频”:{
“名称”:结果[“名称”],
“url::结果[“url”],
“类型”:“文件”
}
}), {
“X-Parse-Application-Id”:“${Application\u Id}”,
“X-Parse-REST-API-Key”:“${REST\u-API\u-Key}”,
“内容类型”:“应用程序/json”
})
connection.getresponse().read()

非常感谢您这么做。我正试图通过支持电子邮件与您联系。我只是想知道为什么您还不支持Python。我只是想用一下。现在我明白了。谢谢你,阿加尼刚刚尝试了你的解决方案。请检查,但看起来它无效。请您尝试使用我上面发布的正确编码,好吗?我刚试过,效果很好。我正在使用
httplib
而不是
http.client
http.client
也应该可以工作,但我认为需要相应地修改请求。