blocks|key|5118804|text|如果年份是闰年，则isleap()方法返回True，否则返回False。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|5118805|import+calendar
print(calendar.isleap(2016))|code-block|syntax|javascript|5118806|entityMap^0|9|8|L|4|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]|$9|R|A|S|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|T|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|U|8|@]|D|@]|E|$]]]|L|$]]

The <code>isleap()</code> method returns <code>True</code> if the year is a leap year, otherwise <code>False</code>.

<pre><code>import calendar
print(calendar.isleap(2016))
</code></pre>

blocks|key|1269738|text|有几处更改，您需要将年份传递给函数，并将"+year“作为函数中的变量，然后在打印时将年份转换为字符串。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1269739|import+calendar

def+leap_year(year):
++++if+calendar.isleap(year)+==+True:
++++++++print(str(year)+%2B+'+is+a+leap+year')
++++else:
++++++++print(str(year)+%2B+'+is+not+a+leap+year')

leap_year(2016)|code-block|syntax|javascript|1269740|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|$]]

A couple of changes, you need to pass in the year to the function and have "year" as a variable in your function, and then cast the year to a string when printing.

<pre><code>import calendar

def leap_year(year):
 if calendar.isleap(year) == True:
 print(str(year) + ' is a leap year')
 else:
 print(str(year) + ' is not a leap year')

leap_year(2016)
</code></pre>

blocks|key|1334809|text|代码可能是|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1334810|def+leap_year(check_year):
++++if+((check_year%254==0)and+not(check_year%25100==0))or+(check_year%25400==0)++:
++++++++print(str(year)+%2B+'is+a+leap+year')
++++else:
++++++++print(str(year)+%2B+'is+not+a+leap+year')

year+=+2016

leap_year(year)|code-block|syntax|javascript|1334811|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|$]]

The code could have been 

<pre><code>def leap_year(check_year):
 if ((check_year%4==0)and not(check_year%100==0))or (check_year%400==0) :
 print(str(year) + 'is a leap year')
 else:
 print(str(year) + 'is not a leap year')

year = 2016

leap_year(year)
</code></pre>

blocks|key|1269779|text|你的代码有几个问题。首先，您的函数创建了一个名为year的新局部变量，它隐藏了全局变量。由于名称查找的工作方式，您不能将全局变量year传递给calendar.isleap并将结果赋给局部变量。对局部变量使用不同的名称。其次，您希望查看calendar.isleap的结果是否为True，而不是函数本身。在表达式中使用新的局部变量。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1269780|def+leap_year():
++++result+=+calendar.isleap(year)
++++if+result+==+True:
++++++++print(str(year)+%2B+'is+a+leap+year')
++++else:
++++++++print(str(year)+%2B+'is+not+a+leap+year')|code-block|syntax|javascript|1269781|就风格而言，不要将布尔变量与布尔常量进行比较；只需直接测试变量：|1269782|if+result:
++++print(year+%2B+'is+a+leap+year')|1269783|这意味着您甚至不需要本地变量；只需在if语句中直接使用函数调用：|1269784|def+leap_year():
++++if+calendar.isleap(year):
++++++++print(year+%2B+'is+a+leap+year')
++++else:
++++++++print(year+%2B+'is+not+a+leap+year')|1269785|最后，在这里使用全局变量是一种糟糕的风格。取而代之的是，将它们的年份作为参数传递给leap_year。|1269786|def+leap_year(y):
++++if+calendar.isleap(y):
++++++++print(str(y)+%2B+'is+a+leap+year')
++++else:
++++++++print(str(y)+%2B+'is+not+a+leap+year')

year+=+2016
leap_year(year)++#+Or+simply+leap_year(2016)|1269787|entityMap^0|O|4|1S|4|1Z|F|3A|F|3V|4|0|0|0|0|I|2|0|0|15|9|0|0^^$0|@$1|2|3|4|5|6|7|Y|8|@$9|Z|A|10|B|C]|$9|11|A|12|B|C]|$9|13|A|14|B|C]|$9|15|A|16|B|C]|$9|17|A|18|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|19|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|1A|8|@]|D|@]|E|$]]|$1|M|3|N|5|H|7|1B|8|@]|D|@]|E|$I|J]]|$1|O|3|P|5|6|7|1C|8|@$9|1D|A|1E|B|C]]|D|@]|E|$]]|$1|Q|3|R|5|H|7|1F|8|@]|D|@]|E|$I|J]]|$1|S|3|T|5|6|7|1G|8|@$9|1H|A|1I|B|C]]|D|@]|E|$]]|$1|U|3|V|5|H|7|1J|8|@]|D|@]|E|$I|J]]|$1|W|3|-4|5|6|7|1K|8|@]|D|@]|E|$]]]|X|$]]

There are several problems with your code. First, your function creates a new local variable named <code>year</code> that shadows the global variable. Due to how name lookup works, you can't pass the global variable <code>year</code> to <code>calendar.isleap</code> and assign the result to the local variable. Use a different name for the local variable. Second, you want to see if the result of <code>calendar.isleap</code> is <code>True</code>, not the function itself. Use your new local variable in the expression.

<pre><code>def leap_year():
 result = calendar.isleap(year)
 if result == True:
 print(str(year) + 'is a leap year')
 else:
 print(str(year) + 'is not a leap year')
</code></pre>

As a matter of style, don't compare a boolean variable to a boolean constant; just testing the variable directly:

<pre><code>if result:
 print(year + 'is a leap year')
</code></pre>

This means that you don't even need the local variable; just use the function call directly in the <code>if</code> statement:

<pre><code>def leap_year():
 if calendar.isleap(year):
 print(year + 'is a leap year')
 else:
 print(year + 'is not a leap year')
</code></pre>

Finally, using a global variable here is bad style. Instead, pass they year to check as an argument to <code>leap_year</code>.

<pre><code>def leap_year(y):
 if calendar.isleap(y):
 print(str(y) + 'is a leap year')
 else:
 print(str(y) + 'is not a leap year')

year = 2016
leap_year(year) # Or simply leap_year(2016)
</code></pre>

So I am currently trying to create a function whereby when you input a specific year and calendar.isleap(year) will then process whether that year is a leap year with True or False. 

My issue is I can't seem to figure out a function whereby it will decide whether the year input is True of False and will then print out that year along with a string. The example I have created and tried to use is below:

<pre><code>import calendar

def leap_year():
 year = calendar.isleap(year)
 if calendar.isleap == True:
 print(year + 'is a leap year')
 else:
 print(year + 'is not a leap year')

year = 2016

leap_year()
</code></pre>

Any help on this would be appreciated. Thanks

Python Calendar Leap Year Function Code Not Working

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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因此，我目前正在尝试创建一个函数，当您输入特定年份时，calendar.isleap( year )将处理该年是否是True或False的闰年。我的问题是，我似乎找不出一个函数，它可以判断年份输入是True还是False，然后将该年份与字符串一起打印出来。下面是我创建并尝试使用的示例：import calendardef leap_year():    year = calendar.isleap

问Python日历闰年函数代码不起作用
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问Python日历闰年函数代码不起作用
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