正则表达式在ios中提取href url并丢弃锚标签的其余部分？ [英] Regular expression in ios to extract href url and discard rest of anchor tag?

问题描述

我想在目标C中编写一个url提取函数。输入文本可以是任何内容，可以包含也可以不包含html定位标记。

请考虑这一点：

  NSString * input1 = @This is cool site< a href =https://abc.com/coolstuff> ;尽情探索< / a>; 
 NSString * input2 = @This is cool site< a target =_ blankhref =https://abc.com/coolstuff>必须访问< / a>; 
 NSString * input3 = @This is cool site< a href =https://abc.com/coolstufftarget =_ blank>立即试用< / a>; 
  

我想修改字符串为这是很棒的网站https：// abc .com / coolstuff

忽略锚标记之间的所有文本，并且需要考虑其他属性，如锚标记中的_target

我可以做一些类似于

  static NSRegularExpression * regex = [NSRegularExpression regularExpressionWithPattern：@< ; a \ shref = \（。*？）\>。*？< / a>options：NSRegularExpressionCaseInsensitive error：nil] ;; 
 NSString * modifiedString = [regex stringByReplacingMatchesInString：inputString options：0 range：NSMakeRange（0，[inputString length]）withTemplate：@$ 1]; 
  

在input1下工作正常，但在其他情况下失败。

谢谢 解决方案

<

试试这个：

 < a [^>] + href = \（。*？）\\ \\[^>] *>。*？< / a> 
  

I want to write a url extracting function in objective C. The input text can be anything and may or may not contain html anchor tags.

Consider this:

NSString* input1 = @"This is cool site <a   href="https://abc.com/coolstuff"> Have fun exploring </a>";
NSString* input2 = @"This is cool site <a target="_blank" href="https://abc.com/coolstuff"> Must visit </a>";
NSString* input3 = @"This is cool site <a href="https://abc.com/coolstuff" target="_blank" > Try now </a>";

I want modified string as "This is cool site https://abc.com/coolstuff

Ignoring all text between anchor tag. And need to consider other attributes like _target in anchor tag

I can do something like

static NSRegularExpression *regex = [NSRegularExpression regularExpressionWithPattern:@"<a\shref=\"(.*?)\">.*?</a>" options:NSRegularExpressionCaseInsensitive error:nil];;
NSString* modifiedString = [regex stringByReplacingMatchesInString:inputString options:0 range:NSMakeRange(0, [inputString length]) withTemplate:@"$1"];

Works fine with input1 but fails in other cases.

Thanks

解决方案

Try this one:

<a[^>]+href=\"(.*?)\"[^>]*>.*?</a>

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