blocks|key|3509830|text|您的代码不工作，因为您添加了您目前的字母，在结尾-因此，它将是相同的。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3509831|相反，您可以在循环一次字符串时计数'a'的数量(for循环更适合，但也可以使用)-然后打印剩余的'b'。|offset|length|style|CODE|3509832|word+=+"ababababaabab"
lw+=+len(word)
i+=+0+
a_s=+0++
while+i+<+lw:
++++if+word[i]+==+'a':
++++++++print("a",+end="")++#+dont+end+the+line+quite+yet
++++++++a_s+%2B=+1+
++++i+%2B=+1

#+print+b's+for+remaining+wordlength+minus+printed+a's
print('b'+*+(lw-a_s))|code-block|syntax|javascript|3509833|或者更好地使用for循环：|3509834|a_s+=+0
for+letter+in+word:
++++if+letter+==+"a":
++++++++a_s+%2B=1
++++++++print(letter,+end="")
print('b'+*+(lw-a_s))|3509835|entityMap^0|0|H|3|1C|3|0|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|T|8|@$D|U|E|V|F|G]|$D|W|E|X|F|G]]|9|@]|A|$]]|$1|H|3|I|5|J|7|Y|8|@]|9|@]|A|$K|L]]|$1|M|3|N|5|6|7|Z|8|@]|9|@]|A|$]]|$1|O|3|P|5|J|7|10|8|@]|9|@]|A|$K|L]]|$1|Q|3|-4|5|6|7|11|8|@]|9|@]|A|$]]]|R|$]]

Your code does not work because you add the letter that you currently have at the end - so it will be the same afterwards.
You can instead count amount of <code>'a'</code>'s while looping through the string once (for loop would be better suited but while can be used as well) - then print the remaining <code>'b'</code>:
<pre><code>word = &quot;ababababaabab&quot;
lw = len(word)
i = 0 
a_s= 0 
while i &lt; lw:
 if word[i] == 'a':
 print(&quot;a&quot;, end=&quot;&quot;) # dont end the line quite yet
 a_s += 1 
 i += 1

# print b's for remaining wordlength minus printed a's
print('b' * (lw-a_s))
</code></pre>
or better with for loop:
<pre><code>a_s = 0
for letter in word:
 if letter == &quot;a&quot;:
 a_s +=1
 print(letter, end=&quot;&quot;)
print('b' * (lw-a_s))
</code></pre>

blocks|key|604896|text|根据您当前的想法，您只需根据字母的值将其附加到一端或另一端即可。这将允许a在开始时积累，而b可以在结束时进行构建。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|604897|word+=+"ababababaabab"
wordfixed+=+""
i+=+0

while+i+<+len(word):
++++if+word[i]+==+"a":
++++++++wordfixed+=+word[i]+%2B+wordfixed+
++++elif+word[i]+==+"b":
++++++++wordfixed+=+wordfixed+%2B+word[i]
++++i+%2B=+1


print(wordfixed)
#+aaaaaaabbbbbb|code-block|syntax|javascript|604898|entityMap^0|10|1|19|1|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|$]]

Using your current idea, you could just append to one end or the other depending on the value of the letter. This would allow <code>a</code> to accumulate at the start and <code>b</code> to build off the end.
<pre><code>word = &quot;ababababaabab&quot;
wordfixed = &quot;&quot;
i = 0

while i &lt; len(word):
 if word[i] == &quot;a&quot;:
 wordfixed = word[i] + wordfixed 
 elif word[i] == &quot;b&quot;:
 wordfixed = wordfixed + word[i]
 i += 1


print(wordfixed)
# aaaaaaabbbbbb
</code></pre>

blocks|key|269170|text|您可以创建两个“助手”变量:一个用于积累a，另一个用于b。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|269171|word+=+"ababababaabab"
wordfixed+=+""
wordfixed2+=+""
wordfixed3+=+""
i+=+0

while+i+<+len(word):
++++if+word[i]+==+"a":
++++++++wordfixed2+%2B=+"a"
++++elif+word[i]+==+"b":
++++++++wordfixed3+%2B=+"b"
++++i+%2B=+1

wordfixed+=+wordfixed2+%2B+wordfixed3
print(wordfixed)|code-block|syntax|javascript|269172|我认为这样做是正确的。|269173|entityMap^0|K|1|R|1|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@$9|P|A|Q|B|C]|$9|R|A|S|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|T|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|U|8|@]|D|@]|E|$]]|$1|M|3|-4|5|6|7|V|8|@]|D|@]|E|$]]]|N|$]]

You can create two &quot;helper&quot; variables: one to accumulate the <code>a</code>s, and one for the <code>b</code>s. Then in the end just concatenate them in the right order:
<pre><code>word = &quot;ababababaabab&quot;
wordfixed = &quot;&quot;
wordfixed2 = &quot;&quot;
wordfixed3 = &quot;&quot;
i = 0

while i &lt; len(word):
 if word[i] == &quot;a&quot;:
 wordfixed2 += &quot;a&quot;
 elif word[i] == &quot;b&quot;:
 wordfixed3 += &quot;b&quot;
 i += 1

wordfixed = wordfixed2 + wordfixed3
print(wordfixed)
</code></pre>
I think that's the right way to do it.

blocks|key|3509897|text|我将使用@Mark_Meyer解决方案，但是如果您真的不需要只使用while循环，那么下面是一个简单的方法：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3509898|word+=+"ababababaabab"
print(word.count('a')*'a'+%2B+word.count('b')*'b')
#+aaaaaaabbbbbb|code-block|syntax|javascript|3509899|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|$]]

I'll go with @Mark_Meyer solution but if you don't really need to use only the while loop, here is an easy way to go :
<pre><code>word = &quot;ababababaabab&quot;
print(word.count('a')*'a' + word.count('b')*'b')
# aaaaaaabbbbbb
</code></pre>

Given the String <code>&quot;aaababababab&quot;</code> (Only <code>'a'</code> or <code>'b'</code> allowed):
How do I order the string alphabetically to be like <code>&quot;aaaaaaabbbbb&quot;</code> by only using a while loop?
This is my current attempt:
<pre><code>word = &quot;ababababaabab&quot;
wordfixed = &quot;&quot;
i = 0

while i &lt; len(word):
 if word[i] == &quot;a&quot;:
 wordfixed += &quot;a&quot;
 i += 1

 elif word[i] == &quot;b&quot;:
 wordfixed += (word[i])
 i += 1

print(wordfixed)
</code></pre>
But it simply prints the original word back. How can I get it to print the desired output?

How can I arrange a string by its alphabetical order using only While loop and conditions?

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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给定字符串"aaababababab" (只有'a'或'b'允许)：如何通过只使用like循环将字符串按字母顺序排序为"aaaaaaabbbbb"？这是我目前的尝试：word = "ababababaabab"wordfixed = ""i = 0while i < len(word): if word[i] =...

问如何使用While循环和条件按字母顺序排列字符串？
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问如何使用While循环和条件按字母顺序排列字符串？EN