a = '1000' # also tested for 'abc100', 'a100bc', '100abc'

isinstance(a, str) or isinstance(a, unicode)

返回True

type(a) in [str, unicode]

返回True

我看过:

hasattr(s, 'endswith') 

Python 2的另一种方法，不使用basestring:

isinstance(s, (str, unicode))

但仍然不能在Python 3中工作，因为unicode没有定义(在Python 3中)。

因为在Python3中没有定义basestring，这个小技巧可能有助于使代码兼容:

try: # check whether python knows about 'basestring'
   basestring
except NameError: # no, it doesn't (it's Python3); use 'str' instead
   basestring=str

之后，您可以在Python2和Python3上运行以下测试

isinstance(myvar, basestring)

如果检查的不是整型和字符串，类型模块也存在。 http://docs.python.org/library/types.html

如果你不想依赖于外部库，这适用于Python 2.7+和Python 3 (http://ideone.com/uB4Kdc):

# your code goes here
s = ["test"];
#s = "test";
isString = False;

if(isinstance(s, str)):
    isString = True;
try:
    if(isinstance(s, basestring)):
        isString = True;
except NameError:
    pass;

if(isString):
    print("String");
else:
    print("Not String");