Question

如何从以下

获取网址

[bpfb_link url ='http：//www.abcd.com/'title ='abcd Technologies :: 主页'image ='http：//www.abcd.com//images/itwall_b1.gif'] [/ bpfb_link]

以上链接i retrieved从php方法转换为xml。如何从formatted url获得above tag。我想用componentsseperatedBystring来获取字符串，但格式每次都会改变。我正在执行以下操作以删除不需要的字符串

NSRange rangeLink = [[[arrUpdates valueForKeyPath:@"content"]objectAtIndex:intIndexPath] rangeOfString:@"bpfb_link"];
            if (rangeLink.location != NSNotFound) 
            {

            NSArray *arrNewLink=[[[arrUpdates valueForKeyPath:@"content"]objectAtIndex:intIndexPath] componentsSeparatedByString:@"["];
                NSLog(@"arrNewLink  %@",[arrNewLink objectAtIndex:0]);
                //[[arrNewLink objectAtIndex:1] componentsSeparatedByString:@"]"];

                NSArray *arrEditedLinks=[[arrNewLink objectAtIndex:1]componentsSeparatedByString:@"]"];
                 NSLog(@"arrEditedLinks  %@",[arrEditedLinks objectAtIndex:1] );
}

请帮帮我

Answer 1

如果您的回答是不变的，那么您可以这样做

NSString *temp=@"[bpfb_link url='http://www.abcd.com/' title='abcd Technologies : : Home Page' image='http://www.abcd.com//images/itwall_b1.gif'] [/bpfb_link]";
    NSArray *arr=[[NSArray alloc]init];
    arr=[temp componentsSeparatedByString:@"'"];
    NSLog(@"arr %@",[arr objectAtIndex:1]);

out put is-- http://www.abcd.com/

如何从此字符串中获取URL

1 个答案: