blocks|key|2905456|text|那个？限定词是你的朋友。生成尽可能短的匹配而不是贪婪的匹配。使用它作为名字，如下所示：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2905457|%5E(.%2B?)+(.%2B)$|code-block|syntax|javascript|2905458|第一组获取第一个空间之前的所有内容，第二组获取其余空间。|2905459|当然，如果名字中包含空格，现在该怎么办？|2905460|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|P|8|@]|9|@]|A|$]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

The ? qualifier is your friend. Makes a shortest-possible match instead of a greedy one. Use it for the first name, as in:

<pre><code>^(.+?) (.+)$
</code></pre>

Group 1 grabs everything up to the first space, group 2 gets the rest.

Of course, now what do you do if the first name contains spaces?

blocks|key|2069581|text|尝试如下所示：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2069582|%5E[%5E\s][\w\s]*[%5E\s]$|code-block|syntax|javascript|2069583|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|$]]

Try something like this:

<pre><code>^[^\s][\w\s]*[^\s]$
</code></pre>

blocks|key|2075043|text|这应该可以完成这项工作：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2075044|%5E[a-zA-Z][a-zA-Z+]*[a-zA-Z]$|code-block|syntax|javascript|2075045|编辑：这是一个小小的改进，允许在名称中使用后一个名称和连字符/撇号：|offset|length|style|BOLD|2075046|%5E[a-zA-Z'][a-zA-Z'-+]*[a-zA-Z']?$|2075047|entityMap^0|0|0|0|3|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|S|8|@$I|T|J|U|K|L]]|9|@]|A|$]]|$1|M|3|N|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|O|3|-4|5|6|7|W|8|@]|9|@]|A|$]]]|P|$]]

This should do the job:

<pre><code>^[a-zA-Z][a-zA-Z ]*[a-zA-Z]$
</code></pre>

Edit: Here's a slight improvement that allows one-latter names and hyphens/apostrophes in the name:

<pre><code>^[a-zA-Z'][a-zA-Z'- ]*[a-zA-Z']?$
</code></pre>

blocks|key|163682|text|%5E\p{L}%2B(\s%2B\p{L}%2B)*$|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|163683|此正则表达式具有以下功能：|unstyled|163684|163685|将匹配一个字母的姓氏(例如，马尔科姆X的姓氏)|unordered-list-item|163686|将不匹配包含数字的姓氏(类似于任何带有\w或[%5E+]的姓氏)|offset|length|style|CODE|163687|匹配unicode字母|163688|163689|但是像“O‘’Connor”这样的姓氏或者用连字符连接的姓氏呢？嗯..。|163690|entityMap^0|0|0|0|0|J|2|M|4|0|0|0|0^^$0|@$1|2|3|4|5|6|7|X|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|Y|8|@]|9|@]|A|$]]|$1|G|3|-4|5|F|7|Z|8|@]|9|@]|A|$]]|$1|H|3|I|5|J|7|10|8|@]|9|@]|A|$]]|$1|K|3|L|5|J|7|11|8|@$M|12|N|13|O|P]|$M|14|N|15|O|P]]|9|@]|A|$]]|$1|Q|3|R|5|J|7|16|8|@]|9|@]|A|$]]|$1|S|3|-4|5|F|7|17|8|@]|9|@]|A|$]]|$1|T|3|U|5|F|7|18|8|@]|9|@]|A|$]]|$1|V|3|-4|5|F|7|19|8|@]|9|@]|A|$]]]|W|$]]

<pre><code>^\p{L}+(\s+\p{L}+)*$
</code></pre>

This regex has the following features:

<ul>
<li>Will match a one letter last name (e.g. Malcolm X's last name)</li>
<li>Will not match last names containing numbers (like anything with a <code>\w</code> or a <code>[^ ]</code> will)</li>
<li>Matches unicode letters</li>
</ul>

But what about last names like "O'Connor" or hyphenated last names ... hmm ...

blocks|key|2075070|text|我想这更像是你想要的：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2075071|%5E[%5E+][a-zA-Z+]%2B[%5E+]$|code-block|syntax|javascript|2075072|这应该与行的开头匹配，没有空格、字母字符或空格，末尾没有空格。|2075073|这在irb中有效，但上次我使用C#时，我使用了类似的正则表达式：|2075074|(0表示好，0表示失败)|2075075|>>+"Di+Giorno"+=~+/%5E[%5E+][a-zA-Z+]%2B[%5E+]$/
=>+0
>>+"DiGiorno"+=~+/%5E[%5E+][a-zA-Z+]%2B[%5E+]$/
=>+0
>>+"+DiGiorno"+=~+/%5E[%5E+][a-zA-Z+]%2B[%5E+]$/
=>+nil
>>+"DiGiorno+"+=~+/%5E[%5E+][a-zA-Z+]%2B[%5E+]$/
=>+nil
>>+"Di+Gior+no"+=~+/%5E[%5E+][a-zA-Z+]%2B[%5E+]$/
=>+0|2075076|entityMap^0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|S|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|T|8|@]|9|@]|A|$]]|$1|K|3|L|5|6|7|U|8|@]|9|@]|A|$]]|$1|M|3|N|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|O|3|-4|5|6|7|W|8|@]|9|@]|A|$]]]|P|$]]

I think this is more what you were looking for:

<pre><code>^[^ ][a-zA-Z ]+[^ ]$
</code></pre>

This should match the beginning of the line with no space, alpha characters or a space, and no space at the end.

This works in irb, but last time I worked with C#, I've used similar regexes:

(zero is good, nil means failed)

<pre><code>&gt;&gt; "Di Giorno" =~ /^[^ ][a-zA-Z ]+[^ ]$/
=&gt; 0
&gt;&gt; "DiGiorno" =~ /^[^ ][a-zA-Z ]+[^ ]$/
=&gt; 0
&gt;&gt; " DiGiorno" =~ /^[^ ][a-zA-Z ]+[^ ]$/
=&gt; nil
&gt;&gt; "DiGiorno " =~ /^[^ ][a-zA-Z ]+[^ ]$/
=&gt; nil
&gt;&gt; "Di Gior no" =~ /^[^ ][a-zA-Z ]+[^ ]$/
=&gt; 0
</code></pre>

blocks|key|2905528|text|在名字"Ṣalāḥad-Dīn+Yūsuf+ibn+Ayyūb“(见http://en.wikipedia.org/wiki/Saladdin)中，哪个是名字，哪个是最后？"Roberto+Garcia+y+Vega“(发明的)这个名字是怎么回事？“蒋介石”(见http://en.wikipedia.org/wiki/Chang_Kai-shek)？|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|2905529|名称中的空格是您最不关心的问题！参见Personal+names+in+a+global+application:+What+to+store。|2905530|entityMap|0|LINK|mutability|MUTABLE|url|http://en.wikipedia.org/wiki/Saladdin|1|http://en.wikipedia.org/wiki/Chang_Kai-shek|2|https://stackoverflow.com/questions/620118/personal-names-in-a-global-application-what-to-store^0|Y|11|0|3N|17|1|0|I|1H|2|0^^$0|@$1|2|3|4|5|6|7|R|8|@]|9|@$A|S|B|T|1|U]|$A|V|B|W|1|X]]|C|$]]|$1|D|3|E|5|6|7|Y|8|@]|9|@$A|Z|B|10|1|11]]|C|$]]|$1|F|3|-4|5|6|7|12|8|@]|9|@]|C|$]]]|G|$H|$5|I|J|K|C|$L|M]]|N|$5|I|J|K|C|$L|O]]|P|$5|I|J|K|C|$L|Q]]]]

In the name "Ṣalāḥ ad-Dīn Yūsuf ibn Ayyūb" (see <a href="http://en.wikipedia.org/wiki/Saladdin" rel="nofollow noreferrer">http://en.wikipedia.org/wiki/Saladdin</a>), which is the first name, and which is the last? What about in the name "Roberto Garcia y Vega" (invented)? "Chiang Kai-shek" (see <a href="http://en.wikipedia.org/wiki/Chang_Kai-shek" rel="nofollow noreferrer">http://en.wikipedia.org/wiki/Chang_Kai-shek</a>)?

Spaces in names are the least of your problems! See <a href="https://stackoverflow.com/questions/620118/personal-names-in-a-global-application-what-to-store">Personal names in a global application: What to store</a>.

blocks|key|5020120|text|这里有一个更好的方法：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|5020121|/%5E[a-zA-Z]%2B(([\'\,\.\-+][a-zA-Z+])?[a-zA-Z]*)*$/|code-block|syntax|javascript|5020122|允许使用标准标点符号和空格，但不能以标点符号开头。|5020123|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|$]]

Here's a better one:

<pre><code>/^[a-zA-Z]+(([\'\,\.\- ][a-zA-Z ])?[a-zA-Z]*)*$/
</code></pre>

Allows standard punctuation and spaces, but cannot start with punctuation.

I'd like a Regular Expression for C# that matches "Johnson", "Del Sol", or "Del La Range"; in other words, it should match words with spaces in the middle but no space at the start or at the end.

Regular Expression in C# for Last Name that includes internal space

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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我希望C#的正则表达式能够匹配"Johnson“、"Del Sol”或"Del La Range"；换句话说，它应该匹配中间有空格但开头或结尾没有空格的单词。

问C#中包含内部空格的姓氏正则表达式
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