blocks|key|3961210|text|你可以简单的做|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3961211|if+(number+%25+3+==+0)+and+(number+%25+5+==+0):
++++your+code+here...|code-block|syntax|javascript|3961212|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 simply do
<pre><code>if (number % 3 == 0) and (number % 5 == 0):
 your code here...
</code></pre>

blocks|key|3525082|text|使用Python语法，您需要使用and而不是&。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|3525083|如果a大于b，如果c大于a，则|3525084|测试：|blockquote|3525085|3525086|A=+200+b=+33+c=+500如果a>b和c>+a:+print(“两种条件均为真”)|3525087|3525088|来源：https://www.w3schools.com/python/python_conditions.asp#:~:text=The%2520and%2520keyword%2520is%2520a%2520logical%2520operator%252C%2520and%2520is%2520used%2520to%2520combine%2520conditional%2520statements|3525089|entityMap|0|LINK|mutability|MUTABLE|url|https://www.w3schools.com/python/python_conditions.asp#:%257E:text=The%2520and%2520keyword%2520is%2520a%2520logical%2520operator%252C%2520and%2520is%2520used%2520to%2520combine%2520conditional%2520statements^0|G|3|M|1|0|0|0|0|0|0|3|4V|0|0^^$0|@$1|2|3|4|5|6|7|Y|8|@$9|Z|A|10|B|C]|$9|11|A|12|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|13|8|@]|D|@]|E|$]]|$1|H|3|I|5|J|7|14|8|@]|D|@]|E|$]]|$1|K|3|-4|5|6|7|15|8|@]|D|@]|E|$]]|$1|L|3|M|5|6|7|16|8|@]|D|@]|E|$]]|$1|N|3|-4|5|6|7|17|8|@]|D|@]|E|$]]|$1|O|3|P|5|6|7|18|8|@]|D|@$9|19|A|1A|1|1B]]|E|$]]|$1|Q|3|-4|5|6|7|1C|8|@]|D|@]|E|$]]]|R|$S|$5|T|U|V|E|$W|X]]]]

With the Python syntax, you need to use an <code>and</code> instead of <code>&amp;</code>.
<blockquote>
Test if a is greater than b, AND if c is greater than a:
<pre><code>a = 200
b = 33
c = 500
if a &gt; b and c &gt; a:
 print(&quot;Both conditions are True&quot;)
</code></pre>
</blockquote>
Source: <a href="https://www.w3schools.com/python/python_conditions.asp#:%7E:text=The%20and%20keyword%20is%20a%20logical%20operator%2C%20and%20is%20used%20to%20combine%20conditional%20statements" rel="nofollow noreferrer">https://www.w3schools.com/python/python_conditions.asp#:~:text=The%20and%20keyword%20is%20a%20logical%20operator%2C%20and%20is%20used%20to%20combine%20conditional%20statements</a>

blocks|key|3961334|text|你的程序完全出错了吗？Python3允许关键字&和and用作“和”条件语句。不管用哪种方式都行..。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|3961335|entityMap^0|N|1|P|3|0^^$0|@$1|2|3|4|5|6|7|H|8|@$9|I|A|J|B|C]|$9|K|A|L|B|C]]|D|@]|E|$]]|$1|F|3|-4|5|6|7|M|8|@]|D|@]|E|$]]]|G|$]]

Does your program error at all? Python 3 allows keyword <code>&amp;</code> as well as <code>and</code> to be used as &quot;and&quot; conditional statements. It should work either way...

blocks|key|4056352|text|您可以使用“和”或“或”运算符，它们之间的区别是“和”应该调用True如果所有条件都是True，或者-如果其中一个条件是True。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4056353|示例：|4056354|a+=+1
b+=+2
c+=+'d'
if+type(a)+==+int+and+type(b)+==+int:
+++#+that+will+be+printed
+++print("a,+b+are+numbers")
if+type(a)+==+int+and+type(c)+==+int:
+++#+that+will+not+be+printed
+++print("a,+c+are+numbers")
if+type(a)+==+int+or+type(c)+==+int:
+++#+that+will+be+printed
+++print("a+or+c+is+number")|code-block|syntax|javascript|4056355|您的代码应该如下所示：|4056356|Starting_Num+=+int(input('Please+enter+intial+number:+'))

Ending_Num+=+int(input('Please+enter+ending+number:+'))

for+number+in+range(Starting_Num,+(Ending_Num%2B1)):

++++if+(number+%25+3+==+0):
++++++++print(number,+'--',++3)
++++elif+(number+%25+5+==+0):
++++++++print(number,+'--',+5)
++++elif+(number+%25+3+==+0)+and+(number+%25+5+==+0):
++++++++print(number,+'--+both')
++++elif+(number+%25+3+!=+0)+and+(number+%25+5+!=+0):
++++++++print(number)|4056357|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|P|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|Q|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|R|8|@]|9|@]|A|$]]|$1|K|3|L|5|F|7|S|8|@]|9|@]|A|$G|H]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

You can use &quot;and&quot; or &quot;or&quot; operator, difference between them is that &quot;and&quot; should call True if all of the condition are True, &quot;or&quot; - if one of it is True.
example:
<pre><code>a = 1
b = 2
c = 'd'
if type(a) == int and type(b) == int:
 # that will be printed
 print(&quot;a, b are numbers&quot;)
if type(a) == int and type(c) == int:
 # that will not be printed
 print(&quot;a, c are numbers&quot;)
if type(a) == int or type(c) == int:
 # that will be printed
 print(&quot;a or c is number&quot;)
</code></pre>
your code should look like:
<pre><code>Starting_Num = int(input('Please enter intial number: '))

Ending_Num = int(input('Please enter ending number: '))

for number in range(Starting_Num, (Ending_Num+1)):

 if (number % 3 == 0):
 print(number, '--', 3)
 elif (number % 5 == 0):
 print(number, '--', 5)
 elif (number % 3 == 0) and (number % 5 == 0):
 print(number, '-- both')
 elif (number % 3 != 0) and (number % 5 != 0):
 print(number)
</code></pre>

blocks|key|4056373|text|问题是你在哪里定位你的‘和’条件。在您的代码中，假设输入符合条件(例如，数字为15)，那么您将输入第一个条件：(if+(number+%25+3+!=+0):)并跳过所有其他条件，因为您使用了if+/+elif语句。您需要做的是将条件elif+(number+%25+3+!=+0)+&+(number+%25+5+!=+0):移动为第一个条件，这将确保在数字可被3和5整除的情况下，您将首先检查这两个条件。此外，在逻辑和使用and运算符的情况下。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|4056374|它应该是这样的：|4056375|Starting_Num+=+int(input('Please+enter+intial+number:+'))

Ending_Num+=+int(input('Please+enter+ending+number:+'))

for+number+in+range(Starting_Num,+(Ending_Num%2B1)):

if+(number+%25+3+==+0)+and+(number+%25+5+==+0):
++++print(number,+'--+both')
elif+(number+%25+3+==+0):
++++print(number,+'--',++3)
elif+(number+%25+5+==+0):
++++print(number,+'--',+5)
else:
++++print(number)++++|code-block|syntax|javascript|4056376|entityMap^0|1K|L|37|17|5R|3|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@$9|P|A|Q|B|C]|$9|R|A|S|B|C]|$9|T|A|U|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|V|8|@]|D|@]|E|$]]|$1|H|3|I|5|J|7|W|8|@]|D|@]|E|$K|L]]|$1|M|3|-4|5|6|7|X|8|@]|D|@]|E|$]]]|N|$]]

The problem is where you position your 'and' condition. In your code, assuming the input fits the condition (e.g. the number is 15) what will happen is that you will enter the first condition: (<code>if (number % 3 != 0):</code>) and skip all the others, because you used [if / elif] statements. What you need to do is move the condition [<code>elif (number % 3 != 0) &amp; (number % 5 != 0):</code>] to be the first condition, which will ensure that in case the number is divisible by both 3 and 5 - you will first check for both conditions.
Also, in case of logical and use the <code>and</code> operator.
It should look something like this:
<pre><code>Starting_Num = int(input('Please enter intial number: '))

Ending_Num = int(input('Please enter ending number: '))

for number in range(Starting_Num, (Ending_Num+1)):

if (number % 3 == 0) and (number % 5 == 0):
 print(number, '-- both')
elif (number % 3 == 0):
 print(number, '--', 3)
elif (number % 5 == 0):
 print(number, '--', 5)
else:
 print(number) 
</code></pre>

<pre class="lang-py prettyprint-override"><code>Starting_Num = int(input('Please enter intial number: '))

Ending_Num = int(input('Please enter ending number: '))

for number in range(Starting_Num, (Ending_Num+1)):

 if (number % 3 == 0):
 print(number, '--', 3)
 elif (number % 5 == 0):
 print(number, '--', 5)
 elif (number % 3 == 0) &amp; (number % 5 == 0):
 print(number, '-- both')
 elif (number % 3 != 0) &amp; (number % 5 != 0):
 print(number)
</code></pre>

How to make the if condition contain more than one condition? i.e.: if (number % 3 == 0) & (number % 5 == 0):

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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Starting_Num = int(input('Please enter intial number: '))Ending_Num = int(input('Please enter ending number: '))for number in range(Starting_Num, (Ending_Num+1)):    if (number % 3 == 0):        print

问如何使if条件包含多个条件？即: if (数字%3 == 0)和(数字%5 == 0)：
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问如何使if条件包含多个条件？即: if (数字%3 == 0)和(数字%5 == 0)：EN

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问如何使if条件包含多个条件？即: if (数字%3 == 0)和(数字%5 == 0)：
EN