使用正则表达式查找＆lt;＆gt;

Question

我有一个字符串

"Absolutely<U+653C><U+3E64><U+613C><U+3E30><U+623C><U+3E64><U+653C><U+3E64><U+623C><U+3E31><U+383C><U+3E64> Friendship goals exceeded here!! Sydney, Melbourne, Connecticut &amp; South Carolina<U+653C><U+3E64><U+613C><U+3E30><U+623C><U+3E64><U+653C><U+3E64><U+623C><U+3E31><U+383C><U+3E61>\r\n"

我想只分开＆lt;＆gt;部分来自字符串。我试过了<.*>，但它又回来了

<U+653C><U+3E64><U+613C><U+3E30><U+623C><U+3E64><U+653C><U+3E64><U+623C><U+3E31><U+383C><U+3E64> Friendship goals exceeded here!! Sydney, Melbourne, Connecticut &amp; South Carolina<U+653C><U+3E64><U+613C><U+3E30><U+623C><U+3E64><U+653C><U+3E64><U+623C><U+3E31><U+383C><U+3E61>

我不想要他们之间的话。我希望输出为，

 ["<U+653C><U+3E64><U+613C><U+3E30><U+623C><U+3E64><U+653C><U+3E64><U+623C><U+3E31><U+383C><U+3E64>", "<U+653C><U+3E64><U+613C><U+3E30><U+623C><U+3E64><U+653C><U+3E64><U+623C><U+3E31><U+383C><U+3E61>"]

任何帮助。我被困在python中。

Answer 1

您需要negative lookahead。此模式匹配，直到找到第一个UIWebView后面没有>：

<

Answer 2

您可以使用<.*?>(?!<)代替<.*>。

这就是你可以做的事情

s = "Absolutely<U+653C><U+3E64><U+613C><U+3E30><U+623C><U+3E64><U+653C><U+3E64><U+623C><U+3E31><U+383C><U+3E64> Friendship goals exceeded here!! Sydney, Melbourne, Connecticut &amp; South Carolina<U+653C><U+3E64><U+613C><U+3E30><U+623C><U+3E64><U+653C><U+3E64><U+623C><U+3E31><U+383C><U+3E61>\r\n"

result = re.findall('<.*?>(?!<)',s)