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Question

website = str(raw_input('Website: '))
palavras_chaves = ['joomla', 'Joomla']
palavras_chaves2 = ['wordpress', 'Wordpress', 'WordPress', 'wp']    
abrindo = urllib2.urlopen(website)
abrindo = abrindo.read()
if palavras_chaves[0] and palavras_chaves[1] in abrindo:
    print '\nÉ um joomla!'

elif palavras_chaves2[0] and palavras_chaves2[1] and palavras_chaves2[2] and palavras_chaves2[3] in abrindo:        
    print '\nÉ um wordpress!'

else:
    print '\nÉ um CMS desconhecido!'

我想知道如何检查所有字符串而不指定它。似乎很简单，但我尝试了一些我想到的并且没有工作的东西。

Answer 1

正如@fourtheye指出的那样，你需要内置的all()函数，你可以利用列表理解来完成这项工作，如下所示。

所以试试这个：

if all(i in abrindo for i in palavras_chaves):
    print '\nÉ um joomla!'

elif all(i in abrindo for i in palavras_chaves2):
    print '\nÉ um wordpress!'
else:
    print '\nÉ um CMS desconhecido!'

编辑：

我认为如果您手动执行此操作：

if palavras_chaves[0] and palavras_chaves[1] in abrindo:

可行，因为palavras_chaves[1]位于abrindo但palavras_chaves[0]可能不在abrindo，但在all()使用palavras_chaves abrindo时位于abrindo，以防一个字符串不在'Works as expected'， 它失败！ < / p>

请参阅此示例，您尝试执行此类操作，并且您希望它返回>>> if 'Hey' and 'Bye' in 'Bye': ... print 'Not expected right?' ... else: ... print 'Works as expected' ... #Not expected right?但是它会返回相反的

all()

但是使用if all(i in 'Bye'for i in ['Hey','Bye']): print 'Not expected right?' else: print 'Works as expected' #Works as expected：

{{1}}

Answer 2

使用python的内置集，它具有issubset方法。 http://docs.python.org/2/library/stdtypes.html#set

如何检查列表中的所有字符串？

2 个答案:

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