blocks|key|2898480|text|听起来你可以直接使用：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2898481|if+(A.Contains(str[i]))|code-block|syntax|javascript|2898482|但肯定有更有效的方法来做到这一点。特别是，避免在循环中进行字符串连接。|2898483|我的猜测是，有一些Unicode标准化方法也不需要对所有这些数据进行硬编码。我确定我记得在某个地方，关于编码回退，但我不能确定它...编辑:我怀疑它与String.Normalize有关--至少值得一看。|offset|length|style|CODE|2898484|至少，这会更有效率：|2898485|char[]+mutated+=+new+char[str.Length];
for+(int+i+=+0;+i+<+str.Length;+i%2B%2B)
{
++++//+You+could+use+a+local+variable+to+avoid+calling+the+indexer+three
++++//+times+if+you+really+want...
++++mutated[i]+=+A.Contains(str[i])+?+'A'
+++++++++++++++:+a.Contains(str[i])+?+'a'
+++++++++++++++:+str[i];
}
string+strNew+=+new+string(mutated);|2898486|entityMap|0|LINK|mutability|MUTABLE|url|http://msdn.microsoft.com/en-us/library/ebza6ck1.aspx^0|0|0|0|23|G|23|G|0|0|0|0^^$0|@$1|2|3|4|5|6|7|10|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|11|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|12|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|13|8|@$K|14|L|15|M|N]]|9|@$K|16|L|17|1|18]]|A|$]]|$1|O|3|P|5|6|7|19|8|@]|9|@]|A|$]]|$1|Q|3|R|5|D|7|1A|8|@]|9|@]|A|$E|F]]|$1|S|3|-4|5|6|7|1B|8|@]|9|@]|A|$]]]|T|$U|$5|V|W|X|A|$Y|Z]]]]

It sounds like you could just use:

<pre><code>if (A.Contains(str[i]))
</code></pre>

but there are certainly more efficient ways of doing this. In particular, avoid string concatenation in a loop.

My guess is that there are Unicode normalization approaches which don't require you to hard-code all this data, too. I'm sure I remember one somewhere, around encoding fallbacks, but I can't put my finger on it... EDIT: I suspect it's around <a href="http://msdn.microsoft.com/en-us/library/ebza6ck1.aspx" rel="nofollow"><code>String.Normalize</code></a> - worth a look, at least.

At the very least, this would be more efficient:

<pre><code>char[] mutated = new char[str.Length];
for (int i = 0; i &lt; str.Length; i++)
{
 // You could use a local variable to avoid calling the indexer three
 // times if you really want...
 mutated[i] = A.Contains(str[i]) ? 'A'
 : a.Contains(str[i]) ? 'a'
 : str[i];
}
string strNew = new string(mutated);
</code></pre>

blocks|key|2986568|text|这应该是可行的：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2986569|for+(int+i+=+0;+i+<+str.Length;+i%2B%2B)
{+
++++if(A.Contains(str[i]))
++++++++strNew+%2B=+'A'
++++else+if(a.Contains(str[i]))+
++++++++++strNew+%2B='a'
++++else
++++++++strNew+%2B=+str[i];
}|code-block|syntax|javascript|2986570|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 should work: 

<pre><code>for (int i = 0; i &lt; str.Length; i++)
{ 
 if(A.Contains(str[i]))
 strNew += 'A'
 else if(a.Contains(str[i])) 
 strNew +='a'
 else
 strNew += str[i];
}
</code></pre>

blocks|key|2898526|text|尝试使用正则表达式(首先替换为"A“，然后替换为"a"：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2898527|string+result+=+Regex.Replace("Àpple",+"([ÀÁÂÄ])",+"A",+RegexOptions.None);|code-block|syntax|javascript|2898528|然后你也可以对"a“做同样的事情。|2898529|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|$]]

Try with a regex (first replace with "A" and then with "a":

<pre><code>string result = Regex.Replace("Àpple", "([ÀÁÂÄ])", "A", RegexOptions.None);
</code></pre>

And then you can do the same for "a".

I have e.g. 

<pre><code>string str ='Àpple';
string strNew="";
char[] A = {'À','Á','Â','Ä'};
char[] a = {'à','á','â','ä'};
</code></pre>

I want to look through the str and see if found replace with Ascii code 'A' . So the result should be:

<pre><code>strNew = 'Apple';
</code></pre>

Here is my code:

<pre><code>for (int i = 0; i &lt; str.Length; i++)
{ 
 if(str[i].CompareTo(A))
 strNew += 'A'
 else if(str[i].CompareTo(a)) 
 strNew +='a'
 else
 strNew += str[i];
}
</code></pre>

But the compare function doesn't work, so what other function I can use?

C# - Efficient search and replace char array in string

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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我有一个例子。string str ='Àpple';string strNew="";char[] A = {'À','Á','Â','Ä'};char[] a = {'à','á','â','ä'};我想查看一下str，看看是否找到用Ascii代码'A‘替换的内容。所以结果应该是：strNew = 'Apple';...

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