Python-urllib.request和flask

Question

我有這個P2P python代碼，並且我試圖從中將POST請求發送到flask ：

在我的P2P方面，我有：

...

for reply in con:
    jsonData = json.loads(reply)
    print(jsonData)
    print(type(jsonData) is dict, tuple)
    data = urlencode(jsonData)
    print(data + " : urlencode")
    data = data.encode()
    print(data.__class__)
    req = urllib.request.Request("http://0.0.0.0:5000/validate", data)
    response = urllib.request.urlopen(req)
    #res = response.read()
    print(response.read().decode('utf8') + " : response in alice")

對於我的flask代碼，我有：

@app.route('/validate', methods=['POST'])
def validate():

    print(request.args)

request args始終輸出為：

ImmutableMultiDict([])

P2P端的輸出為：

{'index': 2043, 'message': 'New Block Forged', 'previous_hash': 'fa4a49cd092869db788490e79a933e7a45107ce513523500e5cd9c85e25426de', 'proof': 168158, 'transactions': [{'amount': 1, 'recipient': '6760d061731c493e94897164c2362476', 'sender': '0'}]}
True <class 'tuple'>
index=2043&message=New+Block+Forged&previous_hash=fa4a49cd092869db788490e79a933e7a45107ce513523500e5cd9c85e25426de&proof=168158&transactions=%5B%7B%27amount%27%3A+1%2C+%27recipient%27%3A+%276760d061731c493e94897164c2362476%27%2C+%27sender%27%3A+%270%27%7D%5D : urlencode
<class 'bytes'>
{
  "add": true
}
 : response in alice

如您所見， urllib.request.urlopen的data看起來正確。 為什么不通過燒瓶側？

Answer 1

您正在發出POST請求以validate端點。 並且request.args僅返回url查詢字符串。 因此，實際數據將在request.form可用。

回答

請發出GET請求...，以便可以在request.args獲得data

data = urllib.parse.urlencode(json_data)

url = 'http://localhost:5000?{}'.format(data)
with urllib.request.urlopen(url) as response:
    pass