blocks|key|1511244|text|您在顶部定义的是派、咖啡和冰淇淋的变量名。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1511245|您从raw_input获得的是文本字符串。|1511246|它们不会仅仅因为相同而匹配，你必须以某种方式匹配它们。我建议更多的是：|1511247|pie+=+2.75
coffee+=+1.50
icecream+=+2.00
cost+=+0

while+True:

++++choice+=+raw_input("What+would+you+like+to+buy?+pie,+coffee,+icecream,+or+nothing?")

++++if+choice+==+"nothing":
++++++++break
++++if+choice+==+"pie":
++++++++cost+=+cost+%2B+pie
++++if+choice+=+"coffee":
++++++++cost+=+cost+%2B+coffee

++++++++#...+etc.

print+"Your+total+is:+${0}".format(cost)|code-block|syntax|javascript|1511248|如果你想避免大量的if语句，看看@Evert建议的一个保存价格的字典。|offset|length|style|CODE|1511249|entityMap^0|0|0|0|0|9|2|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|T|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|U|8|@]|9|@]|A|$]]|$1|F|3|G|5|H|7|V|8|@]|9|@]|A|$I|J]]|$1|K|3|L|5|6|7|W|8|@$M|X|N|Y|O|P]]|9|@]|A|$]]|$1|Q|3|-4|5|6|7|Z|8|@]|9|@]|A|$]]]|R|$]]

What you define at the top are variable names for pie, coffee and ice cream.

What you get from raw_input are strings of text.

They don't match up just by being the same, you have to match them up somehow. I suggest more like:

<pre><code>pie = 2.75
coffee = 1.50
icecream = 2.00
cost = 0

while True:

 choice = raw_input("What would you like to buy? pie, coffee, icecream, or nothing?")

 if choice == "nothing":
 break
 if choice == "pie":
 cost = cost + pie
 if choice = "coffee":
 cost = cost + coffee

 #... etc.

print "Your total is: ${0}".format(cost)
</code></pre>

And if you want to avoid lots of <code>if</code> statements, look at @Evert's suggestion of a dictionary holding the prices.

blocks|key|1829694|text|看起来你根本不需要三个选择。考虑以下代码|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1829695|pie+=+2.75
coffee+=+1.50
icecream+=+2.00
cost+=+0
while+True:
++++choice01+=+0
++++choice01+=+raw_input("What+would+you+like+to+buy?+pie,+coffee,+icecream,+or+nothing?")

++++if+choice01+==+"nothing":
++++++++break
++++elif+choice01=="coffee":
++++++++choice01+=+coffee
++++elif+choice01=="pie":
++++++++choice01+=+pie
++++elif+choice01=="icecream":
++++++++choice01==icecream
++++else:
++++++++choice01=0
++++cost%2B=choice01

print+"Your+total+is:+${0}"+.format(cost)|code-block|syntax|javascript|1829696|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|$]]

It seems you don't need three choices at all. consider the following code

<pre><code>pie = 2.75
coffee = 1.50
icecream = 2.00
cost = 0
while True:
 choice01 = 0
 choice01 = raw_input("What would you like to buy? pie, coffee, icecream, or nothing?")

 if choice01 == "nothing":
 break
 elif choice01=="coffee":
 choice01 = coffee
 elif choice01=="pie":
 choice01 = pie
 elif choice01=="icecream":
 choice01==icecream
 else:
 choice01=0
 cost+=choice01

print "Your total is: ${0}" .format(cost)
</code></pre>

blocks|key|1829774|text|让我们来关注一下你的代码在做什么。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1829775|choice01+=+raw_input("What+would+you+like+to+buy?+pie,+coffee,+icecream,+or+nothing?")|code-block|syntax|javascript|1829776|当用户回答这个问题时，他们的答案是一个字符串。它现在在choice01中。对于这个例子，让我们假设他们输入"pie“(没有引号)。|offset|length|style|CODE|1829777|我们对choice02+=行重复此操作，用户选择"coffee“。|1829778|让我们看一下只有这两个选项的cost+=代码行。|1829779|cost+=+choice01+%2B+choice02|1829780|我们刚刚确定choice01是字符串值"pie“，而choice02是字符串值"coffee”，因此cost+=+"piecoffee"|1829781|1829782|你怎么解决这个问题呢？|1829783|您希望在顶部利用这些变量。其中一种方法是创建一个字典：|1829784|prices+=+{"pie":+2.75,
++++"coffee":+1.50,
++++"icecream":+2.00,
++++"nothing":+0.00
}

...

cost+=+prices[choice01]%2Bprices[choice02]%2Bprices[choice03]|1829785|1829786|我做错什么了？|1829787|在字典中，我设置了你的4个可能的值和相关的价格。"nothing“的值为0.00，因为您在成本计算中使用了该值。它使数学变得简单，而且它是有效的，因为你假设总是会有3个选择。|1829788|重要的是要注意，使用这种方法，如果用户键入他们的答案(即，"cofee“而不是”cofee“)，它将抛出一个异常。这是一项由您决定如何处理此类错误的活动。有很多点你可以添加这样的检查。|1829789|1829790|其他修复|1829791|您还需要解决其他一些问题：|1829792|按照您的代码原样，|1829793|while+choice:将无法正常工作。你没有定义choice。另一种方法是简单地执行while+True，然后在循环结束时中断。|unordered-list-item|1829794|您可以将整个输入循环压缩为一个简单的ask，并将其添加到运行总数中。这将允许您有3个以上的选择。|1829795|1829796|示例：|1829797|prices+=+{"pie":+2.75,
++++++++"coffee":+1.50,
++++++++"icecream":+2.00
++++}

cost+=+0.00
while+True:
++++choice+=+raw_input("What+would+you+like+to+buy?+pie,+coffee,+icecream,+or+nothing?")
++++if+choice+==+"nothing":
++++++++break
++++cost+=+cost+%2B+prices[choice]

print+"Your+total+is:+${0}"+.format(cost)|1829798|输出：|1829799|What+would+you+like+to+buy?+pie,+coffee,+icecream,+or+nothing?+pie
What+would+you+like+to+buy?+pie,+coffee,+icecream,+or+nothing?+nothing
Your+total+is:+$2.75|1829800|注意，我们只有一次用户输入问题，并且我们在循环开始之前定义了cost。然后我们只需在每次循环中添加它。我还从字典中删除了"nothing“键，因为您在将选择添加到成本之前打破了循环。|1829801|entityMap^0|0|0|R|8|0|3|A|0|E|6|0|0|6|8|Q|8|1E|I|0|0|0|0|0|0|0|0|0|0|0|0|0|0|D|Q|6|18|A|0|0|0|0|0|0|0|U|4|0^^$0|@$1|2|3|4|5|6|7|1X|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|1Y|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|1Z|8|@$I|20|J|21|K|L]]|9|@]|A|$]]|$1|M|3|N|5|6|7|22|8|@$I|23|J|24|K|L]]|9|@]|A|$]]|$1|O|3|P|5|6|7|25|8|@$I|26|J|27|K|L]]|9|@]|A|$]]|$1|Q|3|R|5|D|7|28|8|@]|9|@]|A|$E|F]]|$1|S|3|T|5|6|7|29|8|@$I|2A|J|2B|K|L]|$I|2C|J|2D|K|L]|$I|2E|J|2F|K|L]]|9|@]|A|$]]|$1|U|3|-4|5|6|7|2G|8|@]|9|@]|A|$]]|$1|V|3|W|5|6|7|2H|8|@]|9|@]|A|$]]|$1|X|3|Y|5|6|7|2I|8|@]|9|@]|A|$]]|$1|Z|3|10|5|D|7|2J|8|@]|9|@]|A|$E|F]]|$1|11|3|-4|5|6|7|2K|8|@]|9|@]|A|$]]|$1|12|3|13|5|6|7|2L|8|@]|9|@]|A|$]]|$1|14|3|15|5|6|7|2M|8|@]|9|@]|A|$]]|$1|16|3|17|5|6|7|2N|8|@]|9|@]|A|$]]|$1|18|3|-4|5|6|7|2O|8|@]|9|@]|A|$]]|$1|19|3|1A|5|6|7|2P|8|@]|9|@]|A|$]]|$1|1B|3|1C|5|6|7|2Q|8|@]|9|@]|A|$]]|$1|1D|3|1E|5|6|7|2R|8|@]|9|@]|A|$]]|$1|1F|3|1G|5|1H|7|2S|8|@$I|2T|J|2U|K|L]|$I|2V|J|2W|K|L]|$I|2X|J|2Y|K|L]]|9|@]|A|$]]|$1|1I|3|1J|5|1H|7|2Z|8|@]|9|@]|A|$]]|$1|1K|3|-4|5|6|7|30|8|@]|9|@]|A|$]]|$1|1L|3|1M|5|6|7|31|8|@]|9|@]|A|$]]|$1|1N|3|1O|5|D|7|32|8|@]|9|@]|A|$E|F]]|$1|1P|3|1Q|5|6|7|33|8|@]|9|@]|A|$]]|$1|1R|3|1S|5|D|7|34|8|@]|9|@]|A|$E|F]]|$1|1T|3|1U|5|6|7|35|8|@$I|36|J|37|K|L]]|9|@]|A|$]]|$1|1V|3|-4|5|6|7|38|8|@]|9|@]|A|$]]]|1W|$]]

Let's focus on what your code is doing.

<pre><code>choice01 = raw_input("What would you like to buy? pie, coffee, icecream, or nothing?")
</code></pre>

When the user answers this question, their answer is a string. It is now in <code>choice01</code>. For this example, let's assume they type "pie" (no quotes).

We repeat this with the <code>choice02 =</code> line and the user selects "coffee".

Let's look at your <code>cost =</code> line with just these two choices.

<pre><code>cost = choice01 + choice02
</code></pre>

We just determined that <code>choice01</code> is the string value "pie" and <code>choice02</code> is the string value "coffee" Thus, <code>cost = "piecoffee"</code>

<hr>

<h3>How do you fix this?</h3>

You want to utilize those variables at the top. One way to do this is to create a dictionary:

<pre><code>prices = {"pie": 2.75,
 "coffee": 1.50,
 "icecream": 2.00,
 "nothing": 0.00
}

...

cost = prices[choice01]+prices[choice02]+prices[choice03]
</code></pre>

<hr>

<h3>What did I do?</h3>

In the dictionary, I set up your 4 possible values and associated prices. "nothing" has a value of 0.00, because you utilize that in your cost calculation. It makes the math easy and it works because you are assuming there will always be 3 choices made.

It is important to note that with this method, if the user typos their answer (ie. "cofee" instead of "coffee"), it will throw an exception. It's an activity for you to determine how you want to handle such an error. There are various points you could add such a check. 

<hr>

<h3>Other fixes</h3>

There are a few other things you need to fix:

<ul>
<li><code>while choice:</code> isn't going to work as your code stands. You don't define <code>choice</code>. An alternative is to simply do <code>while True</code> and then at the end of your loop break out. </li>
<li>You can compress the entire input loop to only a simple ask and add to a running total. This would allow you to take more than 3 choices.</li>
</ul>

Example:

<pre><code>prices = {"pie": 2.75,
 "coffee": 1.50,
 "icecream": 2.00
 }

cost = 0.00
while True:
 choice = raw_input("What would you like to buy? pie, coffee, icecream, or nothing?")
 if choice == "nothing":
 break
 cost = cost + prices[choice]

print "Your total is: ${0}" .format(cost)
</code></pre>

Outputs:

<pre><code>What would you like to buy? pie, coffee, icecream, or nothing? pie
What would you like to buy? pie, coffee, icecream, or nothing? nothing
Your total is: $2.75
</code></pre>

Notice that we only have the user input question once and we define <code>cost</code> before the start of our loop. Then we just add to that each time through the loop. I also removed the "nothing" key from the dictionary, because you break out of the loop prior to adding the selection to the cost.

blocks|key|4146418|text|Python确保用户只输入pie而不是"pie“，因为将使用|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4146419|：-+choices+=+{'pie'+:+2.75,+'coffee':1.50,+'icecream':+2.00,+'nothing':0}将"pie”作为字典，然后使用for循环。你也可以使用文件循环，但如果没有必要在开始循环之前声明cost=+0，建议使用更简单的循环。这是一个循环：|ordered-list-item|offset|length|style|CODE|4146420|4146421|对于范围(0，3)中的x:+choosed+=raw_input(“你想买什么?派、咖啡、冰淇淋，还是什么都不买？”)成本=成本%2B选择选择|4146422|然后最后(外部循环)用户print+"Your+total+is:+${0}"+.format(cost)获得总金额。|4146423|把所有这些放在一起：|4146424|choices+=+{'pie'+:2.75,+'coffee':1.50,+'icecream':+2.00,+'nothing':+0}
cost+=+0
for+x+in+range(0,+3):
++++++++choosed+=+input("What+would+you+like+to+buy?+pie,+coffee,+icecream,+or+nothing?")
++++++++cost+=+cost+%2B+choices[choosed]
print+"Your+total+is:+${0}"+.format(cost)|code-block|syntax|javascript|4146425|附言:我建议不要要求输入3次，一旦用户进入EOF就停止询问。为此，您可以考虑if+choosed+==+"nothing":+break|4146426|entityMap^0|0|3|1Y|3D|7|0|0|0|C|15|0|0|0|12|U|0^^$0|@$1|2|3|4|5|6|7|Y|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|Z|8|@$E|10|F|11|G|H]|$E|12|F|13|G|H]]|9|@]|A|$]]|$1|I|3|-4|5|6|7|14|8|@]|9|@]|A|$]]|$1|J|3|K|5|6|7|15|8|@]|9|@]|A|$]]|$1|L|3|M|5|6|7|16|8|@$E|17|F|18|G|H]]|9|@]|A|$]]|$1|N|3|O|5|6|7|19|8|@]|9|@]|A|$]]|$1|P|3|Q|5|R|7|1A|8|@]|9|@]|A|$S|T]]|$1|U|3|V|5|6|7|1B|8|@$E|1C|F|1D|G|H]]|9|@]|A|$]]|$1|W|3|-4|5|6|7|1E|8|@]|9|@]|A|$]]]|X|$]]

<ol>
<li>Make sure that user enter only pie and not "pie" because Python will take "pie" as a string.</li>
<li>Consider using dictionary:- <code>choices = {'pie' : 2.75, 'coffee':1.50, 'icecream': 2.00, 'nothing':0}</code> and then use for loop. You can use file loop also, but it is advisable to use simpler loop if not necessary otherwise.</li>
<li>Also declare <code>cost= 0</code> just before starting loop. Here's the loop:-

<pre><code>for x in range(0, 3):
 choosed = raw_input("What would you like to buy? pie, coffee, icecream, or nothing?")
 cost = cost + choices[choosed]
</code></pre>

And then at last (outside loop) user <code>print "Your total is: ${0}" .format(cost)</code> to get total amount.</li>
</ol>

Putting up all this together:-

<pre><code>choices = {'pie' :2.75, 'coffee':1.50, 'icecream': 2.00, 'nothing': 0}
cost = 0
for x in range(0, 3):
 choosed = input("What would you like to buy? pie, coffee, icecream, or nothing?")
 cost = cost + choices[choosed]
print "Your total is: ${0}" .format(cost)
</code></pre>

P.S.:- I'd advice rather than asking input 3 times, stop asking once user enters EOF. For that you can consider <code>if choosed == "nothing": break</code>

This code needs to allow up to 3 inputs for items wanted and print a total cost of all the items. I am VERY new to all this and need all the advice I can get. I can not get the total to print out.

<pre><code>pie = 2.75
coffee = 1.50
icecream = 2.00

while choice:

 choice01 = raw_input("What would you like to buy? pie, coffee, icecream, or nothing?")

 if choice01 == "nothing":
 break

 choice02 = raw_input("What would you like to buy? pie, coffee, icecream, or nothing?")

 if choice02 == "nothing":
 break

 choice03 = raw_input("What would you like to buy? pie, coffee, icecream, or nothing?")

 if choice03 == "nothing":
 break

 cost = choice01+choice02+choice03


 print "Your total is: ${0}" .format(cost)
</code></pre>

What am I doing wrong in my code python that it won't print the totall cost?

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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此代码需要允许输入最多3个想要的物品，并打印所有物品的总成本。我对这一切都是新手，需要我能得到的所有建议。我不能把总数打印出来。pie = 2.75coffee = 1.50icecream = 2.00while choice:    choice01 = raw_input("What would you like...

问我在我的代码python中做错了什么，它不会打印总成本？
EN

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问我在我的代码python中做错了什么，它不会打印总成本？EN