Objective c 在NSURL中进行更改

我尝试在NSURL中插入一些内容。我从jSon字符串获取URL：

imageUrl = [NSURL URLWithString:[[jsonPost objectAtIndex:countPage] objectForKey:@"excerpt"]];

imageURL现在包含以下内容：

http://www.site.com/images/uploads/2011/04/2010-12-14-image.gif

我试图实现的是将URL更改为：

http://www.site.com/images/uploads/2011/04/2010-12-14-image_r.gif

因此，将在扩展名之前添加_r。

方法1：（如果您知道文件扩展名为

.gif

，则可以正常工作）

编辑： 更可靠的解决方案：

NSRange range;
range.location = 0;//starting from the first character
range.length = string.length - 4;//excluding the last 4 characters
//of course you have to make sure the .extension part is 4 characters long(at least fixed length), not like .torrent or .rmvb
NSString *newString = [urlString substringWithRange:range];//this should give you the url part without the file extension

//then append the newString with something and make your url with this string

方法1：（如果您知道文件扩展名为

.gif

，则工作正常）

编辑： 更可靠的解决方案：

NSRange range;
range.location = 0;//starting from the first character
range.length = string.length - 4;//excluding the last 4 characters
//of course you have to make sure the .extension part is 4 characters long(at least fixed length), not like .torrent or .rmvb
NSString *newString = [urlString substringWithRange:range];//this should give you the url part without the file extension

//then append the newString with something and make your url with this string

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谢谢你，这真的为我指明了正确的方向！我使用以下代码修复了我的问题（也解决了图像扩展问题）：

NSString*url=[[jsonPost objectAtIndex:countPage]objectForKey:@“摘录”]
NSArray*数组=[url组件由字符串分隔：@]；NSString*retinaUrl=[NSString stringWithFormat:@“%@.%@.%@.%@.%@.%@，[array objectAtIndex:0]、[array objectAtIndex:1]、[array objectAtIndex:2]、[array objectAtIndex:3]
imageUrl=[NSURL-URLWithString:retinaUrl]谢谢这真的为我指明了正确的方向！我使用以下代码修复了我的问题（也解决了图像扩展问题）：
NSString*url=[[jsonPost objectAtIndex:countPage]objectForKey:@“摘录”]
NSArray*数组=[url组件由字符串分隔：@]；NSString*retinaUrl=[NSString stringWithFormat:@“%@.%@.%@.%@.%@.%@，[array objectAtIndex:0]、[array objectAtIndex:1]、[array objectAtIndex:2]、[array objectAtIndex:3]
imageUrl=[NSURL-URLWithString:retinaUrl]