blocks|key|313950|text|如果您正在编写兼容2.x和3.x的代码，则可能需要使用six|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|313951|from+six+import+string_types
isinstance(s,+string_types)|code-block|syntax|javascript|313952|entityMap|0|LINK|mutability|MUTABLE|url|http://packages.python.org/six/#six.string_types^0|R|3|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@$A|R|B|S|1|T]]|C|$]]|$1|D|3|E|5|F|7|U|8|@]|9|@]|C|$G|H]]|$1|I|3|-4|5|6|7|V|8|@]|9|@]|C|$]]]|J|$K|$5|L|M|N|C|$O|P]]]]

If you're writing 2.x-and-3.x-compatible code, you'll probably want to use <a href="http://packages.python.org/six/#six.string_types" rel="noreferrer">six</a>:

<pre><code>from six import string_types
isinstance(s, string_types)
</code></pre>

blocks|key|314135|text|我发现的最简洁的方法不依赖于six这样的包，是：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|314136|try:
++basestring
except+NameError:
++basestring+=+str|code-block|syntax|javascript|314137|然后，假设您一直在以最通用的方式检查Python+2中的字符串，|314138|isinstance(s,+basestring)|314139|现在也适用于Python+3%2B。|314140|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|S|8|@]|9|@]|A|$]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

The most terse approach I've found without relying on packages like six, is:

<pre><code>try:
 basestring
except NameError:
 basestring = str
</code></pre>

then, assuming you've been checking for strings in Python 2 in the most generic manner,

<pre><code>isinstance(s, basestring)
</code></pre>

will now also work for Python 3+.

blocks|key|314171|text|这个怎么样，在所有情况下都有效？|type|unstyled|depth|inlineStyleRanges|entityRanges|data|314172|isinstance(x,+("".__class__,+u"".__class__))|code-block|syntax|javascript|314173|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

What about this, works in all cases?

<pre><code>isinstance(x, ("".__class__, u"".__class__))
</code></pre>

blocks|key|314002|text|这是@Lev+Levitsky的答案，稍微重写了一下。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|314003|try:
++++isinstance("",+basestring)
++++def+isstr(s):
++++++++return+isinstance(s,+basestring)
except+NameError:
++++def+isstr(s):
++++++++return+isinstance(s,+str)|code-block|syntax|javascript|314004|try/except测试完成一次，然后定义一个始终有效且尽可能快的函数。|offset|length|style|CODE|314005|编辑:实际上，我们甚至不需要调用isinstance()；我们只需要评估basestring，看看是否得到了NameError|314006|try:
++++basestring++#+attempt+to+evaluate+basestring
++++def+isstr(s):
++++++++return+isinstance(s,+basestring)
except+NameError:
++++def+isstr(s):
++++++++return+isinstance(s,+str)|314007|不过，我认为调用isinstance()会更容易一些。|314008|entityMap^0|0|0|0|3|4|6|0|G|C|10|A|1I|9|0|0|8|C|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|W|8|@$I|X|J|Y|K|L]|$I|Z|J|10|K|L]]|9|@]|A|$]]|$1|M|3|N|5|6|7|11|8|@$I|12|J|13|K|L]|$I|14|J|15|K|L]|$I|16|J|17|K|L]]|9|@]|A|$]]|$1|O|3|P|5|D|7|18|8|@]|9|@]|A|$E|F]]|$1|Q|3|R|5|6|7|19|8|@$I|1A|J|1B|K|L]]|9|@]|A|$]]|$1|S|3|-4|5|6|7|1C|8|@]|9|@]|A|$]]]|T|$]]

This is @Lev Levitsky's answer, re-written a bit.

<pre><code>try:
 isinstance("", basestring)
 def isstr(s):
 return isinstance(s, basestring)
except NameError:
 def isstr(s):
 return isinstance(s, str)
</code></pre>

The <code>try</code>/<code>except</code> test is done once, and then defines a function that always works and is as fast as possible.

EDIT: Actually, we don't even need to call <code>isinstance()</code>; we just need to evaluate <code>basestring</code> and see if we get a <code>NameError</code>:

<pre><code>try:
 basestring # attempt to evaluate basestring
 def isstr(s):
 return isinstance(s, basestring)
except NameError:
 def isstr(s):
 return isinstance(s, str)
</code></pre>

I think it is easier to follow with the call to <code>isinstance()</code>, though.

blocks|key|313856|text|也许可以使用像这样的变通方法|type|unstyled|depth|inlineStyleRanges|entityRanges|data|313857|def+isstr(s):
++++try:
++++++++return+isinstance(s,+basestring)
++++except+NameError:
++++++++return+isinstance(s,+str)|code-block|syntax|javascript|313858|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Maybe use a workaround like

<pre><code>def isstr(s):
 try:
 return isinstance(s, basestring)
 except NameError:
 return isinstance(s, str)
</code></pre>

blocks|key|314054|text|您可以通过调用object.__class__来获取对象的类，因此为了检查object是否为默认字符串类型：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|314055|++++isinstance(object,"".__class__)|code-block|syntax|javascript|314056|您可以将以下代码放在代码的顶部，以便在python+2中用引号括起的字符串是unicode：|314057|++++from+__future__+import+unicode_literals|314058|entityMap^0|7|G|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@$9|R|A|S|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|T|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|U|8|@]|D|@]|E|$]]|$1|M|3|N|5|H|7|V|8|@]|D|@]|E|$I|J]]|$1|O|3|-4|5|6|7|W|8|@]|D|@]|E|$]]]|P|$]]

You can get the class of an object by calling <code>object.__class__</code>, so in order to check if object is the default string type:

<pre><code> isinstance(object,"".__class__)
</code></pre>

And You can place the following in the top of Your code so that strings enclosed by quotes are in unicode in python 2:

<pre><code> from __future__ import unicode_literals
</code></pre>

blocks|key|531260|text|您可以在代码的开头尝试执行以下操作：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|531261|from+__future__+import+print_function
import+sys
if+sys.version[0]+==+"2":
++++py3+=+False
else:
++++py3+=+True
if+py3:+
++++basstring+=+str
else:
++++basstring+=+basestring|code-block|syntax|javascript|531262|在后面的代码中：|531263|anystring+=+"test"
#+anystring+=+1
if+isinstance(anystring,+basstring):
++++print("This+is+a+string")
else:
++++print("No+string")|531264|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

You can try this at the beginning of your code:

<pre><code>from __future__ import print_function
import sys
if sys.version[0] == "2":
 py3 = False
else:
 py3 = True
if py3: 
 basstring = str
else:
 basstring = basestring
</code></pre>

and later in the code:

<pre><code>anystring = "test"
# anystring = 1
if isinstance(anystring, basstring):
 print("This is a string")
else:
 print("No string")
</code></pre>

blocks|key|314281|text|注意!在Python2中，str和bytes本质上是相同的。如果您试图区分这两者，这可能会导致错误。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|314282|>>>+size+=+5++++
>>>+byte_arr+=+bytes(size)
>>>+isinstance(byte_arr,+bytes)
True
>>>+isinstance(byte_arr,+str)
True|code-block|syntax|javascript|314283|entityMap^0|D|3|H|5|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]|$9|P|A|Q|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|R|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|S|8|@]|D|@]|E|$]]]|L|$]]

Be careful! In python 2, <code>str</code> and <code>bytes</code> are essentially the same. This can cause a bug if you are trying to distinguish between the two. 

<pre><code>&gt;&gt;&gt; size = 5 
&gt;&gt;&gt; byte_arr = bytes(size)
&gt;&gt;&gt; isinstance(byte_arr, bytes)
True
&gt;&gt;&gt; isinstance(byte_arr, str)
True
</code></pre>

blocks|key|531207|text|类型(字符串)+==字符串|type|unstyled|depth|inlineStyleRanges|entityRanges|data|531208|如果是字符串，则返回true；如果不是，则返回false|531209|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|F|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|G|8|@]|9|@]|A|$]]|$1|D|3|-4|5|6|7|H|8|@]|9|@]|A|$]]]|E|$]]

type(string) == str

returns true if its a string, and false if not

I'm aware that I can use: <code>isinstance(x, str)</code> in python-3.x but I need to check if something is a string in python-2.x as well. Will <code>isinstance(x, str)</code> work as expected in python-2.x? Or will I need to check the version and use <code>isinstance(x, basestr)</code>?

Specifically, in python-2.x:

<pre><code>&gt;&gt;&gt;isinstance(u"test", str)
False
</code></pre>

and python-3.x does not have <code>u"foo"</code>

How to check if variable is string with python 2 and 3 compatibility

Python

我知道我可以在python-3.x中使用：isinstance(x, str)，但我也需要检查在python-2.x中是否也是字符串。在isinstance(x, str) -2.x中，python会像预期的那样工作吗？或者我需要检查版本并使用isinstance(x, basestr)具体来说，在python-2.x中：>>>isinstance(u"test", str)False并且pyth

问如何检查变量是否为兼容python 2和3的字符串
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