blocks|key|4669402|text|使用str.isdigit|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|4669403|>>>+"12345".isdigit()
True
>>>+"12345a".isdigit()
False
>>>|code-block|syntax|javascript|4669404|entityMap|0|LINK|mutability|MUTABLE|url|http://docs.python.org/2.7/library/stdtypes.html#str.isdigit^0|2|B|2|B|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@$9|T|A|U|B|C]]|D|@$9|V|A|W|1|X]]|E|$]]|$1|F|3|G|5|H|7|Y|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Z|8|@]|D|@]|E|$]]]|L|$M|$5|N|O|P|E|$Q|R]]]]

Use <a href="http://docs.python.org/2.7/library/stdtypes.html#str.isdigit"><code>str.isdigit</code></a>:

<pre><code>&gt;&gt;&gt; "12345".isdigit()
True
&gt;&gt;&gt; "12345a".isdigit()
False
&gt;&gt;&gt;
</code></pre>

blocks|key|4669418|text|使用string+isdigit函数：|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|4669419|>>>+s+=+'12345'
>>>+s.isdigit()
True
>>>+s+=+'1abc'
>>>+s.isdigit()
False|code-block|syntax|javascript|4669420|entityMap|0|LINK|mutability|MUTABLE|url|https://docs.python.org/3/library/stdtypes.html#str.isdigit^0|9|7|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]]]]

Use string <a href="https://docs.python.org/3/library/stdtypes.html#str.isdigit" rel="nofollow noreferrer">isdigit</a> function:

<pre><code>&gt;&gt;&gt; s = '12345'
&gt;&gt;&gt; s.isdigit()
True
&gt;&gt;&gt; s = '1abc'
&gt;&gt;&gt; s.isdigit()
False
</code></pre>

blocks|key|4669426|text|你可以在这里使用try+catch块：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4669427|s="1234"
try:
++++num=int(s)
++++print+"S+contains+only+digits"
except:
++++print+"S+doesn't+contain+digits+ONLY"|code-block|syntax|javascript|4669428|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|$]]

You can use try catch block here:

<pre><code>s="1234"
try:
 num=int(s)
 print "S contains only digits"
except:
 print "S doesn't contain digits ONLY"
</code></pre>

blocks|key|163303|text|您也可以使用正则表达式，|type|unstyled|depth|inlineStyleRanges|entityRanges|data|163304|import+re|code-block|syntax|javascript|163305|例如：-1)+word+=+"3487954“|163306|re.match('%5E[0-9]*$',word)|163307|例如：-2)+word+=+"3487.954“|163308|re.match('%5E[0-9\.]*$',word)|163309|例如：-3)+word+=+"3487.954+328“|163310|re.match('%5E[0-9\.\+]*$',word)|163311|如你所见，所有的3个，例如，意味着你的字符串中只有no。因此，您可以按照它们提供的相应解决方案进行操作。|163312|entityMap^0|0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|X|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Y|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|Z|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|10|8|@]|9|@]|A|$]]|$1|M|3|N|5|D|7|11|8|@]|9|@]|A|$E|F]]|$1|O|3|P|5|6|7|12|8|@]|9|@]|A|$]]|$1|Q|3|R|5|D|7|13|8|@]|9|@]|A|$E|F]]|$1|S|3|T|5|6|7|14|8|@]|9|@]|A|$]]|$1|U|3|-4|5|6|7|15|8|@]|9|@]|A|$]]]|V|$]]

You can also use the regex, 

<pre><code>import re
</code></pre>

eg:-1) word = "3487954"

<pre><code>re.match('^[0-9]*$',word)
</code></pre>

eg:-2) word = "3487.954"

<pre><code>re.match('^[0-9\.]*$',word)
</code></pre>

eg:-3) word = "3487.954 328"

<pre><code>re.match('^[0-9\.\ ]*$',word)
</code></pre>

As you can see all 3 eg means that there is only no in your string. So you can follow the respective solutions given with them.

blocks|key|163349|text|正如这篇评论中指出的那样，isdigit()方法对于这个用例并不是完全准确的，因为它为一些类似数字的字符返回True：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|163350|>>>+"\u2070".isdigit()+#+unicode+escaped+'superscript+zero'+
True|code-block|syntax|javascript|163351|如果需要避免这种情况，下面的简单函数会检查字符串中的所有字符是否都是介于"0“和”9“之间的数字：|163352|import+string

def+contains_only_digits(s):
++++#+True+for+"",+"0",+"123"
++++#+False+for+"1.2",+"1,2",+"-1",+"a",+"a1"
++++for+ch+in+s:
++++++++if+not+ch+in+string.digits:
++++++++++++return+False
++++return+True|163353|在问题中的示例中使用：|163354|if+len(isbn)+==+10+and+contains_only_digits(isbn):
++++print+("Works")|163355|entityMap^0|D|9|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@$9|V|A|W|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|X|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|Y|8|@]|D|@]|E|$]]|$1|M|3|N|5|H|7|Z|8|@]|D|@]|E|$I|J]]|$1|O|3|P|5|6|7|10|8|@]|D|@]|E|$]]|$1|Q|3|R|5|H|7|11|8|@]|D|@]|E|$I|J]]|$1|S|3|-4|5|6|7|12|8|@]|D|@]|E|$]]]|T|$]]

As pointed out in this comment <a href="https://stackoverflow.com/questions/21388541/how-do-you-check-in-python-whether-a-string-contains-only-numbers#comment96723571_21388567">How do you check in python whether a string contains only numbers?</a> the <code>isdigit()</code> method is not totally accurate for this use case, because it returns True for some digit-like characters:

<pre><code>&gt;&gt;&gt; "\u2070".isdigit() # unicode escaped 'superscript zero' 
True
</code></pre>

If this needs to be avoided, the following simple function checks, if all characters in a string are a digit between "0" and "9":

<pre><code>import string

def contains_only_digits(s):
 # True for "", "0", "123"
 # False for "1.2", "1,2", "-1", "a", "a1"
 for ch in s:
 if not ch in string.digits:
 return False
 return True
</code></pre>

Used in the example from the question:

<pre><code>if len(isbn) == 10 and contains_only_digits(isbn):
 print ("Works")
</code></pre>

blocks|key|4669495|text|我可以想到两种方法来检查一个字符串是否都有数字|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4669496|方法1(使用python中内置的isdigit()函数)：-|4669497|>>>st+=+'12345'
>>>st.isdigit()
True
>>>st+=+'1abcd'
>>>st.isdigit()
False|code-block|syntax|javascript|4669498|方法2(在字符串顶部执行异常处理)：-|4669499|st="1abcd"
try:
++++number=int(st)
++++print("String+has+all+digits+in+it")
except:
++++print("String+does+not+have+all+digits+in+it")|4669500|上述代码的输出将是：|4669501|String+does+not+have+all+digits+in+it|4669502|entityMap^0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|T|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|U|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|V|8|@]|9|@]|A|$]]|$1|K|3|L|5|F|7|W|8|@]|9|@]|A|$G|H]]|$1|M|3|N|5|6|7|X|8|@]|9|@]|A|$]]|$1|O|3|P|5|F|7|Y|8|@]|9|@]|A|$G|H]]|$1|Q|3|-4|5|6|7|Z|8|@]|9|@]|A|$]]]|R|$]]

There are 2 methods that I can think of to check whether a string has all digits of not

Method 1(Using the built-in isdigit() function in python):-

<pre><code>&gt;&gt;&gt;st = '12345'
&gt;&gt;&gt;st.isdigit()
True
&gt;&gt;&gt;st = '1abcd'
&gt;&gt;&gt;st.isdigit()
False
</code></pre>

Method 2(Performing Exception Handling on top of the string):-

<pre><code>st="1abcd"
try:
 number=int(st)
 print("String has all digits in it")
except:
 print("String does not have all digits in it")
</code></pre>

The output of the above code will be:

<pre><code>String does not have all digits in it
</code></pre>

blocks|key|4669546|text|您可以使用str.isdigit()方法或str.isnumeric()方法|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4669547|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

you can use str.isdigit() method or str.isnumeric() method

How do you check whether a string contains only numbers?

I've given it a go here. I'd like to see the simplest way to accomplish this.

<pre><code>import string

def main():
 isbn = input("Enter your 10 digit ISBN number: ")
 if len(isbn) == 10 and string.digits == True:
 print ("Works")
 else:
 print("Error, 10 digit number was not inputted and/or letters were inputted.")
 main()

if __name__ == "__main__":
 main()
 input("Press enter to exit: ")
</code></pre>

How do you check in python whether a string contains only numbers?

如何检查字符串是否只包含数字？我在这里试过了。我希望看到实现这一点的最简单的方法。import stringdef main():    isbn = input("Enter your 10 digit ISBN number: ")    if len(isbn) == 10 and string.digits =...

问如何在python中检查字符串是否只包含数字？
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