blocks|key|955435|text|只需尝试将其转换为int类型，然后在不起作用时退出。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|955436|try:
++++val+=+int(userInput)
except+ValueError:
++++print("That's+not+an+int!")|code-block|syntax|javascript|955437|请参见Handling+Exceptions+https://docs.python.org/3/tutorial/errors.html#handling-exceptions|offset|length|955438|entityMap|0|LINK|mutability|MUTABLE|url|https://docs.python.org/3/tutorial/errors.html#handling-exceptions^0|0|0|N|1U|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|U|8|@]|9|@$I|V|J|W|1|X]]|A|$]]|$1|K|3|-4|5|6|7|Y|8|@]|9|@]|A|$]]]|L|$M|$5|N|O|P|A|$Q|R]]]]

Simply try converting it to an int and then bailing out if it doesn't work.
<pre><code>try:
 val = int(userInput)
except ValueError:
 print(&quot;That's not an int!&quot;)
</code></pre>
See Handling Exceptions <a href="https://docs.python.org/3/tutorial/errors.html#handling-exceptions" rel="noreferrer">https://docs.python.org/3/tutorial/errors.html#handling-exceptions</a>

blocks|key|955473|text|显然，这不适用于负值，但适用于正数。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|955474|使用isdigit()|offset|length|style|CODE|955475|if+userinput.isdigit():
++++#do+stuff|code-block|syntax|javascript|955476|entityMap^0|0|2|9|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|P|8|@$D|Q|E|R|F|G]]|9|@]|A|$]]|$1|H|3|I|5|J|7|S|8|@]|9|@]|A|$K|L]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

Apparently this will not work for negative values, but it will for positive numbers. 

Use <code>isdigit()</code>

<pre><code>if userinput.isdigit():
 #do stuff
</code></pre>

blocks|key|955908|text|方法isnumeric()将完成这项工作(Documentation+for+python3.x)：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|955909|>>>a+=+'123'
>>>a.isnumeric()
True|code-block|syntax|javascript|955910|但请记住：|955911|>>>a+=+'-1'
>>>a.isnumeric()
False|955912|如果字符串中的所有字符都是数字字符，并且至少有一个字符，则isnumeric()返回True。|955913|所以负数是不被接受的。|955914|entityMap|0|LINK|mutability|MUTABLE|url|https://docs.python.org/3/library/stdtypes.html#str.isnumeric^0|2|B|L|R|0|0|0|0|0|T|B|0|0^^$0|@$1|2|3|4|5|6|7|10|8|@$9|11|A|12|B|C]]|D|@$9|13|A|14|1|15]]|E|$]]|$1|F|3|G|5|H|7|16|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|17|8|@]|D|@]|E|$]]|$1|M|3|N|5|H|7|18|8|@]|D|@]|E|$I|J]]|$1|O|3|P|5|6|7|19|8|@$9|1A|A|1B|B|C]]|D|@]|E|$]]|$1|Q|3|R|5|6|7|1C|8|@]|D|@]|E|$]]|$1|S|3|-4|5|6|7|1D|8|@]|D|@]|E|$]]]|T|$U|$5|V|W|X|E|$Y|Z]]]]

The method <code>isnumeric()</code> will do the job (<a href="https://docs.python.org/3/library/stdtypes.html#str.isnumeric" rel="noreferrer">Documentation for python3.x</a>):

<pre><code>&gt;&gt;&gt;a = '123'
&gt;&gt;&gt;a.isnumeric()
True
</code></pre>

But remember:

<pre><code>&gt;&gt;&gt;a = '-1'
&gt;&gt;&gt;a.isnumeric()
False
</code></pre>

<code>isnumeric()</code> returns True if all characters in the string are numeric characters, and there is at least one character.

So negative numbers are not accepted.

blocks|key|739631|text|对于Python+3，以下内容将起作用。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|739632|userInput+=+0
while+True:
++try:
+++++userInput+=+int(input("Enter+something:+"))+++++++
++except+ValueError:
+++++print("Not+an+integer!")
+++++continue
++else:
+++++print("Yes+an+integer!")
+++++break+|code-block|syntax|javascript|739633|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|$]]

For Python 3 the following will work.

<pre><code>userInput = 0
while True:
 try:
 userInput = int(input("Enter something: ")) 
 except ValueError:
 print("Not an integer!")
 continue
 else:
 print("Yes an integer!")
 break 
</code></pre>

blocks|key|955600|text|编辑:你也可以使用下面的代码来找出它是一个数字还是负数|type|unstyled|depth|inlineStyleRanges|entityRanges|data|955601|import+re
num_format+=+re.compile("%5E[\-]?[1-9][0-9]*\.?[0-9]%2B$")
isnumber+=+re.match(num_format,givennumber)
if+isnumber:
++++print+"given+string+is+number"|code-block|syntax|javascript|955602|你也可以改变你的格式以满足你的特定需求。我看到这篇文章也有点late.but，希望这能帮助其他正在寻找答案的人:)。如果给定代码中有任何错误，请让我知道。|955603|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

EDITED:
You could also use this below code to find out if its a number or also a negative

<pre><code>import re
num_format = re.compile("^[\-]?[1-9][0-9]*\.?[0-9]+$")
isnumber = re.match(num_format,givennumber)
if isnumber:
 print "given string is number"
</code></pre>

you could also change your format to your specific requirement.
I am seeing this post a little too late.but hope this helps other persons who are looking for answers :) . let me know if anythings wrong in the given code.

blocks|key|956304|text|如果你特别需要int或float，你可以尝试"is+not+int“或"is+not+float"：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|956305|user_input+=+''
while+user_input+is+not+int:
++++try:
++++++++user_input+=+int(input('Enter+a+number:+'))
++++++++break
++++except+ValueError:
++++++++print('Please+enter+a+valid+number:+')

print('You+entered+{}'.format(a))|code-block|syntax|javascript|956306|如果您只需要使用int，那么我所见过的最优雅的解决方案是".isdigit()“方法：|956307|a+=+''
while+a.isdigit()+==+False:
++++a+=+input('Enter+a+number:+')

print('You+entered+{}'.format(a))|956308|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|$]]

If you specifically need an int or float, you could try "is not int" or "is not float":

<pre><code>user_input = ''
while user_input is not int:
 try:
 user_input = int(input('Enter a number: '))
 break
 except ValueError:
 print('Please enter a valid number: ')

print('You entered {}'.format(a))
</code></pre>

If you only need to work with ints, then the most elegant solution I've seen is the ".isdigit()" method:

<pre><code>a = ''
while a.isdigit() == False:
 a = input('Enter a number: ')

print('You entered {}'.format(a))
</code></pre>

blocks|key|955657|text|对于负数，我建议使用@karthik27|type|unstyled|depth|inlineStyleRanges|entityRanges|data|955658|import+re
num_format+=+re.compile(r'%5E\-?[1-9][0-9]*\.?[0-9]*')|code-block|syntax|javascript|955659|然后对该正则表达式执行任意操作，match()、findall()等|955660|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

I would recommend this, @karthik27, for negative numbers 

<pre><code>import re
num_format = re.compile(r'^\-?[1-9][0-9]*\.?[0-9]*')
</code></pre>

Then do whatever you want with that regular expression, match(), findall() etc

blocks|key|956099|text|最优雅的解决方案是已经提出的，|type|unstyled|depth|inlineStyleRanges|entityRanges|data|956100|a=123
bool_a+=+a.isnumeric()|code-block|syntax|javascript|956101|不幸的是，它对负整数和a的一般浮点值都不起作用。如果你的重点是检查'a‘是否是整数以外的泛型数字，我建议使用以下方法，它适用于所有类型的浮点数和整型数:)。下面是测试结果：|956102|def+isanumber(a):

++++try:
++++++++float(repr(a))
++++++++bool_a+=+True
++++except:
++++++++bool_a+=+False

++++return+bool_a


a+=+1+#+integer
isanumber(a)
>>>+True

a+=+-2.5982347892+#+general+float
isanumber(a)
>>>+True

a+=+'1'+#+actually+a+string
isanumber(a)
>>>+False|956103|我希望它对您有用:)|956104|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 elegant solutions would be the already proposed,

<pre><code>a=123
bool_a = a.isnumeric()
</code></pre>

Unfortunatelly it doesn't work both for negative integers and for general float values of a. If your point is to check if 'a' is a generic number beyond integers i'd suggest the following one, which works for every kind of float and integer :). Here is the test:

<pre><code>def isanumber(a):

 try:
 float(repr(a))
 bool_a = True
 except:
 bool_a = False

 return bool_a


a = 1 # integer
isanumber(a)
&gt;&gt;&gt; True

a = -2.5982347892 # general float
isanumber(a)
&gt;&gt;&gt; True

a = '1' # actually a string
isanumber(a)
&gt;&gt;&gt; False
</code></pre>

I hope you find it useful :)

blocks|key|740278|text|可以对字符串使用isdigit()方法。在本例中，正如您所说，输入始终是一个字符串：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|740279|++++user_input+=+input("Enter+something:")
++++if+user_input.isdigit():
++++++++print("Is+a+number")
++++else:
++++++++print("Not+a+number")|code-block|syntax|javascript|740280|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 the isdigit() method for strings.
In this case, as you said the input is always a string: 

<pre><code> user_input = input("Enter something:")
 if user_input.isdigit():
 print("Is a number")
 else:
 print("Not a number")
</code></pre>

blocks|key|740455|text|此解决方案将只接受整数，只接受整数。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|740456|def+is_number(s):
++++while+s.isdigit()+==+False:
++++++++s+=+raw_input("Enter+only+numbers:+")
++++return+int(s)


#+Your+program+starts+here++++
user_input+=+is_number(raw_input("Enter+a+number:+"))|code-block|syntax|javascript|740457|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|$]]

This solution will accept only integers and nothing but integers.

<pre><code>def is_number(s):
 while s.isdigit() == False:
 s = raw_input("Enter only numbers: ")
 return int(s)


# Your program starts here 
user_input = is_number(raw_input("Enter a number: "))
</code></pre>

blocks|key|955543|text|这适用于任何数字，包括小数：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|955544|import+fractions

def+isnumber(s):
+++try:
+++++float(s)
+++++return+True
+++except+ValueError:
+++++try:+
+++++++Fraction(s)
+++++++return+True
+++++except+ValueError:+
+++++++return+False|code-block|syntax|javascript|955545|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|$]]

This works with any number, including a fraction:

<pre><code>import fractions

def isnumber(s):
 try:
 float(s)
 return True
 except ValueError:
 try: 
 Fraction(s)
 return True
 except ValueError: 
 return False
</code></pre>

blocks|key|740404|text|为什么不将输入除以一个数字呢？这种方式适用于所有东西。负数、浮点数和负数浮点数。也可以使用空格和零。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|740405|numList+=+[499,+-486,+0.1255468,+-0.21554,+'a',+"this",+"long+string+here",+"455+street+area",+0,+""]

for+item+in+numList:

++++try:
++++++++print+(item+/+2)+#You+can+divide+by+any+number+really,+except+zero
++++except:
++++++++print+"Not+A+Number:+"+%2B+item|code-block|syntax|javascript|740406|结果：|740407|249
-243
0.0627734
-0.10777
Not+A+Number:+a
Not+A+Number:+this
Not+A+Number:+long+string+here
Not+A+Number:+455+street+area
0
Not+A+Number:+|740408|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|$]]

Why not divide the input by a number? This way works with everything. Negatives, floats, and negative floats. Also Blank spaces and zero.

<pre><code>numList = [499, -486, 0.1255468, -0.21554, 'a', "this", "long string here", "455 street area", 0, ""]

for item in numList:

 try:
 print (item / 2) #You can divide by any number really, except zero
 except:
 print "Not A Number: " + item
</code></pre>

Result:

<pre><code>249
-243
0.0627734
-0.10777
Not A Number: a
Not A Number: this
Not A Number: long string here
Not A Number: 455 street area
0
Not A Number: 
</code></pre>

blocks|key|955861|text|我知道这已经很晚了，但这是为了帮助那些不得不花6个小时试图弄清楚这个问题的人。(这就是我所做的)：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|955862|这可以完美地工作：(检查输入中是否有字母/检查输入是否为整型或浮点型)|955863|a=(raw_input("Amount:"))

try:
++++int(a)
except+ValueError:
++++try:
++++++++float(a)
++++except+ValueError:
++++++++print+"This+is+not+a+number"
++++++++a=0


if+a==0:
++++a=0
else:
++++print+a
++++#Do+stuff|code-block|syntax|javascript|955864|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|L|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|M|8|@]|9|@]|A|$G|H]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

I know this is pretty late but its to help anyone else that had to spend 6 hours trying to figure this out. (thats what I did):

This works flawlessly: (checks if any letter is in the input/checks if input is either integer or float)

<pre><code>a=(raw_input("Amount:"))

try:
 int(a)
except ValueError:
 try:
 float(a)
 except ValueError:
 print "This is not a number"
 a=0


if a==0:
 a=0
else:
 print a
 #Do stuff
</code></pre>

blocks|key|740051|text|下面是一个简单的函数，它检查输入的INT和RANGE。此处，如果输入是1-100之间的整数，则返回“True”，否则返回“False”|type|unstyled|depth|inlineStyleRanges|entityRanges|data|740052|def+validate(userInput):

++++try:
++++++++val+=+int(userInput)
++++++++if+val+>+0+and+val+<+101:
++++++++++++valid+=+True
++++++++else:
++++++++++++valid+=+False

++++except+Exception:
++++++++valid+=+False

++++return+valid|code-block|syntax|javascript|740053|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|$]]

Here is a simple function that checks input for INT and RANGE. Here, returns 'True' if input is integer between 1-100, 'False' otherwise

<pre><code>def validate(userInput):

 try:
 val = int(userInput)
 if val &gt; 0 and val &lt; 101:
 valid = True
 else:
 valid = False

 except Exception:
 valid = False

 return valid
</code></pre>

blocks|key|956338|text|我一直在使用一种不同的方法，我认为我应该分享。首先创建一个有效的范围：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|956339|valid+=+[str(i)+for+i+in+range(-10,11)]+#++["-10","-9...."10"]+|code-block|syntax|javascript|956340|现在问一个号码，如果不在列表中，继续问：|956341|p+=+input("Enter+a+number:+")

while+p+not+in+valid:
++++p+=+input("Not+valid.+Try+to+enter+a+number+again:+")|956342|最后，转换为int+(这是可行的，因为list只包含整数作为字符串：|956343|p+=+int(p)|956344|entityMap^0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|S|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|U|8|@]|9|@]|A|$]]|$1|M|3|N|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|O|3|-4|5|6|7|W|8|@]|9|@]|A|$]]]|P|$]]

I've been using a different approach I thought I'd share. Start with creating a valid range:

<pre><code>valid = [str(i) for i in range(-10,11)] # ["-10","-9...."10"] 
</code></pre>

Now ask for a number and if not in list continue asking:

<pre><code>p = input("Enter a number: ")

while p not in valid:
 p = input("Not valid. Try to enter a number again: ")
</code></pre>

Lastly convert to int (which will work because list only contains integers as strings:

<pre><code>p = int(p)
</code></pre>

blocks|key|955780|text|今天早上我还遇到了一些问题，用户可以输入非整数响应来响应我对整数的特定请求。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|955781|这就是最终对我来说效果很好的解决方案，可以强迫我得到我想要的答案：|955782|player_number+=+0
while+player_number+!=+1+and+player_number+!=2:
++++player_number+=+raw_input("Are+you+Player+1+or+2?+")
++++try:
++++++++player_number+=+int(player_number)
++++except+ValueError:
++++++++print+"Please+enter+'1'+or+'2'..."|code-block|syntax|javascript|955783|我甚至在到达try:语句之前就会得到异常。|955784|player_number+=+int(raw_input("Are+you+Player+1+or+2?+")+|955785|并且用户输入"J“或任何其他非整数字符。最好将其作为原始输入，检查原始输入是否可以转换为整数，然后再进行转换。|955786|entityMap^0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|R|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|S|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|T|8|@]|9|@]|A|$]]|$1|K|3|L|5|F|7|U|8|@]|9|@]|A|$G|H]]|$1|M|3|N|5|6|7|V|8|@]|9|@]|A|$]]|$1|O|3|-4|5|6|7|W|8|@]|9|@]|A|$]]]|P|$]]

I also ran into problems this morning with users being able to enter non-integer responses to my specific request for an integer.

This was the solution that ended up working well for me to force an answer I wanted:

<pre><code>player_number = 0
while player_number != 1 and player_number !=2:
 player_number = raw_input("Are you Player 1 or 2? ")
 try:
 player_number = int(player_number)
 except ValueError:
 print "Please enter '1' or '2'..."
</code></pre>

I would get exceptions before even reaching the try: statement when I used 

<pre><code>player_number = int(raw_input("Are you Player 1 or 2? ") 
</code></pre>

and the user entered "J" or any other non-integer character. It worked out best to take it as raw input, check to see if that raw input could be converted to an integer, and then convert it afterward.

blocks|key|740066|text|以下是最简单的解决方案：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|740067|a=+input("Choose+the+option\n")

if(int(a)):
++++print+(a);
else:
++++print("Try+Again")|code-block|syntax|javascript|740068|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|$]]

Here is the simplest solution:

<pre><code>a= input("Choose the option\n")

if(int(a)):
 print (a);
else:
 print("Try Again")
</code></pre>

blocks|key|956174|text|while+True:
++++b1=input('Type+a+number:')
++++try:
++++++++a1=int(b1)
++++except+ValueError:
++++++++print+('"%25(a1)s"+is+not+a+number.+Try+again.'+%25{'a1':b1})+++++++
++++else:
++++++++print+('You+typed+"{}".'.format(a1))
++++++++break|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|956175|这会产生一个循环来检查input是否为整数，结果将如下所示：|unstyled|956176|>>>+%25Run+1.1.py
Type+a+number:d
"d"+is+not+a+number.+Try+again.
Type+a+number:
>>>+%25Run+1.1.py
Type+a+number:4
You+typed+4.
>>>+|956177|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|L|8|@]|9|@]|A|$]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$B|C]]|$1|I|3|-4|5|F|7|N|8|@]|9|@]|A|$]]]|J|$]]

<pre><code>while True:
 b1=input('Type a number:')
 try:
 a1=int(b1)
 except ValueError:
 print ('"%(a1)s" is not a number. Try again.' %{'a1':b1}) 
 else:
 print ('You typed "{}".'.format(a1))
 break
</code></pre>

This makes a loop to check whether input is an integer or not, result would look like below:

<pre><code>&gt;&gt;&gt; %Run 1.1.py
Type a number:d
"d" is not a number. Try again.
Type a number:
&gt;&gt;&gt; %Run 1.1.py
Type a number:4
You typed 4.
&gt;&gt;&gt; 
</code></pre>

blocks|key|740215|text|如果您希望计算浮点数，并且希望接受NaNs作为输入，而不接受'abc'等其他字符串，则可以执行以下操作：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|740216|def+isnumber(x):
++++import+numpy
++++try:
++++++++return+type(numpy.float(x))+==+float
++++except+ValueError:
++++++++return+False|code-block|syntax|javascript|740217|entityMap^0|H|3|U|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|$]]

If you wanted to evaluate floats, and you wanted to accept <code>NaN</code>s as input but not other strings like <code>'abc'</code>, you could do the following:

<pre><code>def isnumber(x):
 import numpy
 try:
 return type(numpy.float(x)) == float
 except ValueError:
 return False
</code></pre>

blocks|key|740505|text|尝尝这个!即使我输入了负数，它对我也是有效的。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|740506|++def+length(s):
++++return+len(s)

+++s+=+input("Enter+the+String:+")
++++try:
++++++++if+(type(int(s)))==int+:
++++++++++++print("You+input+an+integer")

++++except+ValueError:++++++
++++++++print("it+is+a+string+with+length+"+%2B+str(length(s)))+++|code-block|syntax|javascript|740507|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|$]]

try this! it worked for me even if I input negative numbers.

<pre><code> def length(s):
 return len(s)

 s = input("Enter the String: ")
 try:
 if (type(int(s)))==int :
 print("You input an integer")

 except ValueError: 
 print("it is a string with length " + str(length(s))) 
</code></pre>

blocks|key|740555|text|您可以键入：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|740556|user_input+=+input("Enter+something:+")

if+type(user_input)+==+int:
++++print(user_input,+"Is+a+number")
else:
++++print("Not+a+number")
++
try:
++++val+=+int(user_input)
except+ValueError:
++++print("That's+not+an+int!")|code-block|syntax|javascript|740557|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 Type:
<pre><code>user_input = input(&quot;Enter something: &quot;)

if type(user_input) == int:
 print(user_input, &quot;Is a number&quot;)
else:
 print(&quot;Not a number&quot;)
 
try:
 val = int(user_input)
except ValueError:
 print(&quot;That's not an int!&quot;)
</code></pre>

blocks|key|956552|text|正在检查Decimal类型：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|956553|import+decimal
isinstance(x,+decimal.Decimal)|code-block|syntax|javascript|956554|entityMap^0|4|7|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|P|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Q|8|@]|D|@]|E|$]]]|L|$]]

Checking for <code>Decimal</code> type:
<pre><code>import decimal
isinstance(x, decimal.Decimal)
</code></pre>

blocks|key|956121|text|基于答案中的灵感。我定义了一个如下的函数。看起来它工作得很好。如果您发现任何问题，请让我知道|type|unstyled|depth|inlineStyleRanges|entityRanges|data|956122|def+isanumber(inp):
++++try:
++++++++val+=+int(inp)
++++++++return+True
++++except+ValueError:
++++++++try:
++++++++++++val+=+float(inp)
++++++++++++return+True
++++++++except+ValueError:
++++++++++++return+False|code-block|syntax|javascript|956123|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|$]]

Based on inspiration from answer. I defined a function as below. Looks like its working fine. Please let me know if you find any issue

<pre><code>def isanumber(inp):
 try:
 val = int(inp)
 return True
 except ValueError:
 try:
 val = float(inp)
 return True
 except ValueError:
 return False
</code></pre>

blocks|key|739997|text|a=10

isinstance(a,int)++#True

b='abc'

isinstance(b,int)++#False|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|739998|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>a=10

isinstance(a,int) #True

b='abc'

isinstance(b,int) #False
</code></pre>

How do I check if a user's string input is a number (e.g. <code>-1</code>, <code>0</code>, <code>1</code>, etc.)?

<pre><code>user_input = input("Enter something:")

if type(user_input) == int:
 print("Is a number")
else:
 print("Not a number")
</code></pre>

The above won't work since <code>input</code> always returns a string.

How to check if string input is a number?

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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如何检查用户输入的字符串是否为数字(例如-1、0、1等)？user_input = input("Enter something:")if type(user_input) == int:    print("Is a number")else:    print("Not a number")上面的方法不起作用，因为input总是返回一个字符串。

问如何检查字符串输入是否为数字？
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问如何检查字符串输入是否为数字？
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