我有一个像这样的字符串:
"This is test string http://www.google.com and it is working."
我只想http://www.google.com
从上面的字符串中获取链接()。我怎么才能得到它?
它应该像这样工作:
NSString *test = @"This is test string http://www.google.com and it is working.";
NSString *string = [test stringByAppendingString:@" "];
NSError *error = NULL;
NSRegularExpression *regex = [NSRegularExpression regularExpressionWithPattern:@"https?://[^ ]* "
options:0
error:&error];
NSArray *matches = [regex matchesInString:string
options:0
range:NSMakeRange(0, [string length])];
for (NSTextCheckingResult *match in matches) {
NSRange matchRange = [match range];
NSString *url = [string substringWithRange:matchRange];
NSLog(@"Found URL: %@", url);
}
您可以在此处找到有关使用 NSRegularExpression 的更多信息:
查看 NSString 类文档 - 那里有很多方法可以找到某些子字符串格式的位置、在分隔符上拆分字符串以及提取子字符串等。
例如,在上面的示例中,如果您想提取字符串中的任何嵌入式 url,您可以首先使用以下方法拆分字符串:
NSArray *substrings = [myString componentsSeparatedByString:@" "];
然后在结果数组中,遍历它并找出它是否有一个'http'字符串:
for (int i=0;i<[substrings length];i++) {
NSString aStr = [substrings objectAtIndex:i];
if ([aStr rangeOfString:@"http"].location != NSNotFound) {
NSLog(@"I found a http url:%@", aStr);
}
}
这将像一个 CHARM 一样工作:
NSString *totalString = @"This is test string http://www.google.com and it is working.";
NSLog(@"%@", totalString);
NSRange urlStart = [totalString rangeOfString: @"http"];
NSRange urlEnd = [totalString rangeOfString: @".com"];
NSRange resultedMatch = NSMakeRange(urlStart.location, urlEnd.location - urlStart.location + urlEnd.length);
NSString *linkString = [totalString substringWithRange:resultedMatch];
NSLog (@"%@", linkString);