blocks|key|2903216|text|我不知道你想在这里做什么，\s*%2B，但是你基本上有了最后一部分[0-9][0-9]%2B的答案，它会在最后找到2%2B数字。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2903217|也许可以试试这个(它更简洁)。这将搜索1%2B位数而不是2%2B数。|2903218|\d%2B(\s*-\s*\d%2B)?|code-block|syntax|javascript|2903219|entityMap^0|D|4|V|B|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@$9|P|A|Q|B|C]|$9|R|A|S|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|T|8|@]|D|@]|E|$]]|$1|H|3|I|5|J|7|U|8|@]|D|@]|E|$K|L]]|$1|M|3|-4|5|6|7|V|8|@]|D|@]|E|$]]]|N|$]]

I'm not sure what you're trying to do here <code>\s*+</code> but you basically had the answer with the last part <code>[0-9][0-9]+</code> that would find 2+ digits on the end.

Maybe try this (it's more concise). This searches for 1+ digits instead of 2+

<pre><code>\d+(\s*-\s*\d+)?
</code></pre>

blocks|key|1215733|text|您可以对2-3个数字使用大括号{2,3}，但*%2B也不正确。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1215734|/%5E(([0-9]{1,3}\s-\s)?[0-9]{1,3})\s/|code-block|syntax|javascript|1215735|我嵌套了大括号，所以您只需要正则表达式的第一个结果。|1215736|它就这样破裂了|1215737|([0-9]{1,3}\s-\s)?|1215738|首先，是否有一个1-3位数和空格-空格可选？|1215739|然后..。它的结尾是一个1-3位数，后面是空格吗？|1215740|entityMap^0|M|2|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@$9|X|A|Y|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|Z|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|10|8|@]|D|@]|E|$]]|$1|M|3|N|5|6|7|11|8|@]|D|@]|E|$]]|$1|O|3|P|5|H|7|12|8|@]|D|@]|E|$I|J]]|$1|Q|3|R|5|6|7|13|8|@]|D|@]|E|$]]|$1|S|3|T|5|6|7|14|8|@]|D|@]|E|$]]|$1|U|3|-4|5|6|7|15|8|@]|D|@]|E|$]]]|V|$]]

You can use braces {2,3} for 2-3 numbers - but also <code>*+</code> isn't right.

<pre><code>/^(([0-9]{1,3}\s-\s)?[0-9]{1,3})\s/
</code></pre>

I nested the braces so you only want the first result from the regex.

it breaks up like this

<pre><code>([0-9]{1,3}\s-\s)?
</code></pre>

first, Is there a 1-3 digit number with a space-dash-space - OPTIONAL

then.. does it end in a 1-3 digit number followed by a space.

blocks|key|1192051|text|以您为例，您可以：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1192052|/%5E(\d%2B[-\s\d]*)\s/gm|code-block|syntax|javascript|1192053|演示|offset|length|1192054|解释：|1192055|/%5E(\d%2B[-\s\d]*)\s/gm
+%5E++++++++++++++++++++++start+of+line
++++%5E+++++++++++++++++++at+least+1+digit+and+as+many+digits+as+possible
+++++++%5E++++++++++++++++any+character+of+the+set+-,+space,+digit
+++++++++++++%5E++++++++++zero+or+more
++++++++++++++++%5E+++++++trailing+space
++++++++++++++++++++%5E+++multiline+for+the+%5E+start+of+line+assertion|1192056|entityMap|0|LINK|mutability|MUTABLE|url|https://regex101.com/r/7v14Fv/1/^0|0|0|0|2|0|0|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|X|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Y|8|@]|9|@$I|Z|J|10|1|11]]|A|$]]|$1|K|3|L|5|6|7|12|8|@]|9|@]|A|$]]|$1|M|3|N|5|D|7|13|8|@]|9|@]|A|$E|F]]|$1|O|3|-4|5|6|7|14|8|@]|9|@]|A|$]]]|P|$Q|$5|R|S|T|A|$U|V]]]]

For your example, you can do:

<pre><code>/^(\d+[-\s\d]*)\s/gm
</code></pre>

<a href="https://regex101.com/r/7v14Fv/1/" rel="nofollow noreferrer">Demo</a>

Explanation:

<pre><code>/^(\d+[-\s\d]*)\s/gm
 ^ start of line
 ^ at least 1 digit and as many digits as possible
 ^ any character of the set -, space, digit
 ^ zero or more
 ^ trailing space
 ^ multiline for the ^ start of line assertion
</code></pre>

blocks|key|1192083|text|从你的正则表达式开始：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1192084|%5E[0-9][0-9]\s*%2B(\s*-\s*[0-9][0-9]%2B)|offset|length|style|CODE|1192085|在第二个街区有一个额外的空格匹配器：|1192086|%5E[0-9][0-9]\s*%2B(-\s*[0-9][0-9]%2B)|1192087|我建议你把[0-9]换成\d|1192088|%5E[\d][\d]\s*%2B(-\s*[\d][\d]%2B)|1192089|使用%2B代替\d的O2副本，意味着至少有一个数字：|1192090|%5E[\d]%2B\s*%2B(-\s*[\d]%2B)|1192091|使最后一个块是可选的，因此它与99+Fake+Address匹配|1192092|%5E[\d]%2B\s*%2B(-\s*[\d]%2B)?|1192093|如果你知道只有一个空格，你可以用\s*代替\s?|1192094|%5E[\d]%2B\s?(-\s?[\d]%2B)?|1192095|它们都应该匹配:D|1192096|entityMap^0|0|0|Z|0|0|0|W|0|5|5|C|2|0|0|S|0|2|1|5|2|0|0|L|0|F|F|0|0|M|0|G|3|L|3|0|0|L|0|0^^$0|@$1|2|3|4|5|6|7|15|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|16|8|@$D|17|E|18|F|G]]|9|@]|A|$]]|$1|H|3|I|5|6|7|19|8|@]|9|@]|A|$]]|$1|J|3|K|5|6|7|1A|8|@$D|1B|E|1C|F|G]]|9|@]|A|$]]|$1|L|3|M|5|6|7|1D|8|@$D|1E|E|1F|F|G]|$D|1G|E|1H|F|G]]|9|@]|A|$]]|$1|N|3|O|5|6|7|1I|8|@$D|1J|E|1K|F|G]]|9|@]|A|$]]|$1|P|3|Q|5|6|7|1L|8|@$D|1M|E|1N|F|G]|$D|1O|E|1P|F|G]]|9|@]|A|$]]|$1|R|3|S|5|6|7|1Q|8|@$D|1R|E|1S|F|G]]|9|@]|A|$]]|$1|T|3|U|5|6|7|1T|8|@$D|1U|E|1V|F|G]]|9|@]|A|$]]|$1|V|3|W|5|6|7|1W|8|@$D|1X|E|1Y|F|G]]|9|@]|A|$]]|$1|X|3|Y|5|6|7|1Z|8|@$D|20|E|21|F|G]|$D|22|E|23|F|G]]|9|@]|A|$]]|$1|Z|3|10|5|6|7|24|8|@$D|25|E|26|F|G]]|9|@]|A|$]]|$1|11|3|12|5|6|7|27|8|@]|9|@]|A|$]]|$1|13|3|-4|5|6|7|28|8|@]|9|@]|A|$]]]|14|$]]

Starting from your regex:

<code>^[0-9][0-9]\s*+(\s*-\s*[0-9][0-9]+)</code>

You got an extra white space matcher in the second block: 

<code>^[0-9][0-9]\s*+(-\s*[0-9][0-9]+)</code>

I would suggest you replace <code>[0-9]</code> with <code>\d</code>

<code>^[\d][\d]\s*+(-\s*[\d][\d]+)</code>

Use a <code>+</code> instead o 2 copies of <code>\d</code> meaning at least one number:

<code>^[\d]+\s*+(-\s*[\d]+)</code>

Make the last block optional, so it matches <code>99 Fake Address</code>:

<code>^[\d]+\s*+(-\s*[\d]+)?</code>

If you know there's only going to be 1 white space, you could replace <code>\s*</code> with <code>\s?</code>:

<code>^[\d]+\s?(-\s?[\d]+)?</code>

That should match all of them :D

blocks|key|1215769|text|另一种方式可能是|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1215770|In+[83]:+s+=+'955+-+959+Fake+Street'

In+[84]:+s1+=+'95-99+Fake+Street'

In+[85]:+s2+=+'95+-+99+Fake+Street'

In+[86]:+s3+=+'99+Fake+Street'

In+[87]:+d+=+re.search(r'%5E[0-9]%2B[+]*(-[+]*[0-9]%2B){0,1}',+s3)

In+[88]:+d.group()
Out[88]:+'99+'

In+[89]:+d+=+re.search(r'%5E[0-9]%2B[+]*(-[+]*[0-9]%2B){0,1}',+s2)

In+[90]:+d.group()
Out[90]:+'95+-+99'

In+[91]:+d+=+re.search(r'%5E[0-9]%2B[+]*(-[+]*[0-9]%2B){0,1}',+s1)

In+[92]:+d.group()
Out[92]:+'95-99'

In+[93]:+d+=+re.search(r'%5E[0-9]%2B[+]*(-[+]*[0-9]%2B){0,1}',+s)

In+[94]:+d.group()
Out[94]:+'955+-+959'|code-block|syntax|javascript|1215771|字符集0-9+cab由\d表示，如下所示|offset|length|style|CODE|1215772|d+=+re.search(r'%5E[\d]%2B[+]*(-[+]*[\d]%2B){0,1}',+s)|1215773|在这里，在所有示例中，我们在字符串的开头搜索一个序列，序列至少一个数字，后面有零个或多个空格，最多一个序列后面跟着一个-符号，后面跟着零或多个空格和至少一个或多个数字。|1215774|entityMap^0|0|0|3|3|B|2|0|0|1N|1|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|U|8|@$I|V|J|W|K|L]|$I|X|J|Y|K|L]]|9|@]|A|$]]|$1|M|3|N|5|D|7|Z|8|@]|9|@]|A|$E|F]]|$1|O|3|P|5|6|7|10|8|@$I|11|J|12|K|L]]|9|@]|A|$]]|$1|Q|3|-4|5|6|7|13|8|@]|9|@]|A|$]]]|R|$]]

Another way could be 

<pre><code>In [83]: s = '955 - 959 Fake Street'

In [84]: s1 = '95-99 Fake Street'

In [85]: s2 = '95 - 99 Fake Street'

In [86]: s3 = '99 Fake Street'

In [87]: d = re.search(r'^[0-9]+[ ]*(-[ ]*[0-9]+){0,1}', s3)

In [88]: d.group()
Out[88]: '99 '

In [89]: d = re.search(r'^[0-9]+[ ]*(-[ ]*[0-9]+){0,1}', s2)

In [90]: d.group()
Out[90]: '95 - 99'

In [91]: d = re.search(r'^[0-9]+[ ]*(-[ ]*[0-9]+){0,1}', s1)

In [92]: d.group()
Out[92]: '95-99'

In [93]: d = re.search(r'^[0-9]+[ ]*(-[ ]*[0-9]+){0,1}', s)

In [94]: d.group()
Out[94]: '955 - 959'
</code></pre>

the character set <code>0-9</code> cab be represented by <code>\d</code> like this

<pre><code>d = re.search(r'^[\d]+[ ]*(-[ ]*[\d]+){0,1}', s)
</code></pre>

Here, in all the examples, we are searching at the beginning of the string, for a sequence of at least one digit followed by zero or more spaces and optionally followed by at most one sequence of only one <code>-</code> symbol followed by zero or more spaces and at least one or more digits.

I have a list of addresses, currently quite unclean. They take the format:

<pre><code>955 - 959 Fake Street
95-99 Fake Street
4-9 M4 Ln
95 - 99 Fake Street
99 Fake Street
</code></pre>

What I would like to do is split up the street name and street number. I need a regex expression that is true for

<pre><code>955 - 959
95-99
4-9
95 - 99
99
</code></pre>

I currently have this:

<pre><code>^[0-9][0-9]\s*+(\s*-\s*[0-9][0-9]+)
</code></pre>

which works for the two digit addresses but does not work for the three or one digit addresses.

Thanks

Identify an address number in a 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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我有一个地址清单，目前相当不干净。它们的格式如下：955 - 959 Fake Street95-99 Fake Street4-9 M4 Ln95 - 99 Fake Street99 Fake Street我想做的是把街道名称和街道号码分开。我需要一个正则表达式，这个表达式对于955 - 95995-994-995 - 9999我现在有这样的情况：^[0-9][0-9]\s*+(\s*-\s*

问标识字符串中的地址号
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回答 5

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问标识字符串中的地址号EN

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