UnboundLocalError:局部变量……在赋值前被引用 [英] UnboundLocalError: local variable … referenced before assignment

问题描述

import hmac, base64, hashlib, urllib2
base = 'https://.......'

def makereq(key, secret, path, data):
    hash_data = path + chr(0) + data
    secret = base64.b64decode(secret)
    sha512 = hashlib.sha512
    hmac = str(hmac.new(secret, hash_data, sha512))

    header = {
        'User-Agent': 'My-First-test',
        'Rest-Key': key,
        'Rest-Sign': base64.b64encode(hmac),
        'Accept-encoding': 'GZIP',
        'Content-Type': 'application/x-www-form-urlencoded'
    }

    return urllib2.Request(base + path, data, header)

错误:文件C:/Python27/btctest.py"，第 8 行，在 makereqhmac = str(hmac.new(secret, hash_data, sha512))UnboundLocalError:赋值前引用了局部变量hmac"

Error: File "C:/Python27/btctest.py", line 8, in makereq hmac = str(hmac.new(secret, hash_data, sha512)) UnboundLocalError: local variable 'hmac' referenced before assignment

有人知道为什么吗?谢谢

Somebody knows why? Thanks

推荐答案

如果您在函数中的任何位置分配变量，则该变量将被视为该函数中处处的局部变量.所以你会看到下面的代码出现同样的错误:

If you assign to a variable anywhere in a function, that variable will be treated as a local variable everywhere in that function. So you would see the same error with the following code:

foo = 2
def test():
    print foo
    foo = 3

换句话说，如果函数中存在同名的局部变量，则无法访问全局或外部变量.

In other words, you cannot access the global or external variable if there is a local variable in the function of the same name.

要解决这个问题，只需给你的局部变量 hmac 一个不同的名字:

To fix this, just give your local variable hmac a different name:

def makereq(key, secret, path, data):
    hash_data = path + chr(0) + data
    secret = base64.b64decode(secret)
    sha512 = hashlib.sha512
    my_hmac = str(hmac.new(secret, hash_data, sha512))

    header = {
        'User-Agent': 'My-First-test',
        'Rest-Key': key,
        'Rest-Sign': base64.b64encode(my_hmac),
        'Accept-encoding': 'GZIP',
        'Content-Type': 'application/x-www-form-urlencoded'
    }

    return urllib2.Request(base + path, data, header)

请注意，可以通过使用 global 或 nonlocal 关键字来更改此行为，但您似乎不想在您的情况下使用它们.

Note that this behavior can be changed by using the global or nonlocal keywords, but it doesn't seem like you would want to use those in your case.

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