从010开始依次在代理数组中旋转

Question

在我的python脚本中，我输入了3条数据，称它们为A，B，C。 我想在数组中输入数据，例如，arrayA=[A1，A2。。。A10]（对于arrayB和arrays相同），然后我希望脚本自动运行10次（第一次使用数据A1、B1、C1/第二次使用A2、B2、C2等），每次迭代数据都将更改为每个数组的n索引数据。你知道吗

我用的是硒和铬

def proxyRotate():
    for i in range(len(proxArr)):
        return proxArr[i]
i = 0
while i < len(proxArr):
    print(proxyRotate())
    i += 1

def get_proxies():
    proxies = ['128.199.254.244:3128', '95.85.79.54:53281', '128.199.125.54:2468', '178.45.8.113:53281', '206.189.225.30:3128']
    return proxies

Answer 1

def main(proxArr, ccArr, cvcArr):

    PROXY = proxArr
    creditCard = ccArr
    securityCode = cvcArr



def get_cc():
    cc = ['5136154545452522', '51365445452823', '51361265424522']
    return cc

def get_cvc():
    cvc = ['734', '690', '734']
    return cvc

def get_proxies():
    proxies = ['51.77.545.171:8080', '51.77.254.171:8080', 

'51.77.258.82:8080'] 返回代理

proxArr = get_proxies()
ccArr = get_cc()
cvcArr = get_cvc()

for elem in zip(proxArr, ccArr, cvcArr):
    spotify(elem[0], elem[1], elem[2])

我做错了什么，因为它总是传递我试图输入的第一个值elem[0]等等 但它给了我语法错误

编辑

我已经像你说的那样做了，而且它一直在传递相同的值 打印只是看它是否传递了值

def spotify(elem1, elem2, elem3):

    print("proxy: {}, cc: {}, cvc: {}".format(elem1, elem2, elem3))




    PROXY = elem1
    creditCard = elem2
    securityCode = elem3

    restart=input("Do you wish to start again: ").lower()
    if restart in yeslist:
        spotify(elem1, elem2, elem3)
        print("proxy: {}, cc: {}, cvc: {}".format(elem1, elem2, elem3))
    else:
        exit()

def get_cc():
    cc = ['5136154545452522', '51365445452823', '51361265424522']
    return cc

def get_cvc():
    cvc = ['734', '690', '734']
    return cvc

def get_proxies():
    proxies = ['51.77.545.171:8080', '51.77.254.171:8080', '51.77.258.82:8080']
    return proxies

proxArr = get_proxies()
ccArr = get_cc()
cvcArr = get_cvc()

for elem in zip(proxArr, ccArr, cvcArr):
    spotify(elem[0], elem[1], elem[2])

Answer 2

您可以在代理之间旋转，而无需为其定义单独的函数。在你列完单子之后，你就可以循环一下了。你知道吗

def get_proxies():
    proxies = ['128.199.254.244:3128', '95.85.79.54:53281', '128.199.125.54:2468', '178.45.8.113:53281', '206.189.225.30:3128']
    return proxies

proxArr = get_proxies()

for proxy in proxArr:
    print(proxy)

您可以调用所需的函数，而不是打印出来。你知道吗

编辑

可以使用zip()函数在多个这样的列表上创建迭代器。它的工作原理如下：

def get_cc(): 
    cc = ['56465465465465', '4654654654654', '54654654654654'] 
    return cc

def get_proxies():
    proxies = ['128.199.254.244:3128', '95.85.79.54:53281', '128.199.125.54:2468', '178.45.8.113:53281', '206.189.225.30:3128']
    return proxies

proxArr = get_proxies()
ccArr = get_cc()

for elem in zip(proxArr, ccArr):
    spotify(elem[0], elem[1])

编辑2

spotify函数还应包含3个元素：

def spotify(elem1, elem2, elem3):
    print("proxy: {}, cc: {}, cvc: {}".format(elem1, elem2, elem3))

编辑3

如果我理解正确，这会解决你的问题。您应该将重新运行的逻辑从函数中取出并放入循环中：

import sys

def spotify(elem1, elem2, elem3):

    print("proxy: {}, cc: {}, cvc: {}".format(elem1, elem2, elem3))


def get_cc():
    cc = ['5136154545452522', '51365445452823', '51361265424522']
    return cc

def get_cvc():
    cvc = ['734', '690', '734']
    return cvc

def get_proxies():
    proxies = ['51.77.545.171:8080', '51.77.254.171:8080', '51.77.258.82:8080']
    return proxies

proxArr = get_proxies()
ccArr = get_cc()
cvcArr = get_cvc()
yeslist = ['y','yes']

for elem in zip(proxArr, ccArr, cvcArr):
    spotify(elem[0], elem[1], elem[2])
    restart=input("Do you wish to start again: ").lower()
    if restart not in yeslist:
        sys.exit("Exiting")