无法使用NSRegularExpression(正则表达式模式)提取字符串-iPhone [英] Unable to extract string using NSRegularExpression (regex pattern) - iphone

问题描述

我想在[]括号之间提取字符串

I want to extract string between [ ] brackets

例如原始字符串:@这是一个测试[获取此字符串]."

for Example Original string: @"this is a test [to get this string]."

我想获取[]之间的提取字符串，即本例中的获取此字符串".

I want to get extract string between [] i.e "to get this string" in this example.

请在下面找到我的代码

NSMutableString *predicateString = [[NSMutableString alloc] initWithFormat:@"%@",@"this is a test [to get this string]."];

NSRegularExpression *regex = [NSRegularExpression regularExpressionWithPattern:@"\[[^\].]*\]"
                     options:NSRegularExpressionCaseInsensitive
                    error:&error];


 NSMutableArray *rangeArray = [[NSMutableArray alloc] init];
 __block NSUInteger count = 0;
 [regex enumerateMatchesInString:predicateString options:0 range:NSMakeRange(0, [predicateString length]) usingBlock:^(NSTextCheckingResult *match, NSMatchingFlags flags, BOOL *stop){
  NSRange matchRange = [match range];
  matchRange.length = matchRange.length+1;
  NSMutableDictionary *dict = [[NSMutableDictionary alloc] init];
  NSLog(@"matchRange.location: %d",matchRange.location);
  NSLog(@"matchRange.length: %d",matchRange.length);

  if (++count >= 100) *stop = YES;

 }]; 

如果我做错了事，请告诉我.

Please let me know if I am doing something wrong.

谢谢.

推荐答案

看起来表达式是错误的，您需要转义斜线(例如: @"\\ [[^ \\].] * \\]").

Looks like the expression is wrong, you need to escape the slashes (eg : @"\\[[^\\].]*\\]").

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