如何开通淘宝客

 简单的正则表达式： 
 
 \w+ 
 
 这匹配一串“单词”字符。 这几乎是你想要的。 
 
 这稍微更准确： 
 
 \w(?<!\d)[\w'-]* 
 
 它匹配任意数量的单词字符，确保第一个字符不是数字。 
 
 这是我的比赛： 
 
 
 
 1 LOLOLOL 
 2你是 
 3 BEEN 
 4 PWN3D 
 5 einszwei 
 6 drei 
 
 
 现在，这更像是它。 
 
 编辑： 
 负面后视的原因是，一些正则表达式支持Unicode字符。 使用[a-zA-Z]会错过很多需要的“单词”字符。 允许\w和不允许\d包含所有可能会在任何文本块中启动单词的Unicode字符。 
 
 编辑2： 
 我找到了一个更简洁的方式来获得负面后视效果：双负面字符类与一个负排除。 
 
 [^\W\d][\w'-]*(?<=\w) 
 
 这与上述内容相同，除了它还确保该单词以单词字符结束 。 最后是： 
 
 [^\W\d](\w|[-']{1,2}(?=\w))* 
 
 确保连续不超过两个非单词字符。 又名，它匹配“单词”，但不匹配“单词”，这是有道理的。 如果你希望它匹配“单词”，而不是“单词---上”，你可以将2更改为3 。 

Simple Regex:
 
\w+
 
This matches a string of "word" characters. That is almost what you want.
 
This is slightly more accurate:
 
\w(?<!\d)[\w'-]*
 
It matches any number of word characters, ensuring that the first character was not a digit.
 
Here are my matches:
 
 
 
1 LOLOLOL
 2 YOU'VE
 3 BEEN
 4 PWN3D
 5 einszwei
 6 drei
 
 
Now, that's more like it.
 
EDIT:
 The reason for the negative look-behind, is that some regex flavors support Unicode characters. Using [a-zA-Z] would miss quite a few "word" characters that are desirable. Allowing \w and disallowing \d includes all Unicode characters that would conceivably start a word in any block of text.
 
EDIT 2:
 I have found a more concise way to get the effect of the negative lookbehind: Double negative character class with a single negative exclusion.
 
[^\W\d][\w'-]*(?<=\w)
 
This is the same as the above with the exception that it also ensures that the word ends with a word character. And, finally, there is:
 
[^\W\d](\w|[-']{1,2}(?=\w))*
 
Ensuring that there are no more than two non-word-characters in a row. Aka, It matches "word-up" but not "word--up", which makes sense. If you want it to match "word--up", but not "word---up", you can change the 2 to a 3.