是否有更好的方法可以重写URL来获取 @2x图像资源

Question

我已经写了这一点代码来处理字符串形式的URL，以尝试获取可能的视网膜图像。意图是转弯

scheme://host/path/name.extension

进入

scheme://host/path/[e-mail]

URL是可以预测的，因此假设例如一个'。在最终文件中是安全的。但是我有iOS的感觉，我想知道这是否太多了，无法完成此操作；是否有更好的，也许是API提供的方法？

这只是启动请求的方法的开始。

- (NSData *)fetchResourceWithURLLocation:(NSString *)location
                             requestedBy:(id<DataRequestDelegate>)requester
                          withIdentifier:(id)requestID {
    NSData *resultData = nil;
    // some disassembly to get rid of the double '/' in scheme
    NSURL *locationURL = [NSURL URLWithString:location];
    NSString *host = [locationURL host];
    NSString *scheme = [locationURL scheme];
    NSString *dataPath = [locationURL relativePath];
    // if this is a retina display, rewrite the dataPath to use @2x
    if ([[UIScreen mainScreen] respondsToSelector:@selector(displayLinkWithTarget:selector:)] && ([UIScreen mainScreen].scale == 2.0)) {
        NSArray *pathComponents = [dataPath componentsSeparatedByString:@"/"];
        NSString *nameComponent = [pathComponents lastObject];
        NSArray *nameComponents = [nameComponent componentsSeparatedByString:@"."];
        NSAssert(([nameComponents count] == 2), @"nameComponent contains more than one "."");
        nameComponent = [NSString stringWithFormat:@"%@@2x.%@", [nameComponents objectAtIndex:0], [nameComponents lastObject]];
        dataPath = @"";
        for(NSString *pathComponent in pathComponents) {
            if(pathComponent != [pathComponents lastObject])
                dataPath = [dataPath stringByAppendingString:[NSString stringWithFormat:@"%@/", pathComponent]];
        }
        dataPath = [dataPath stringByAppendingString:nameComponent];
    }
    // reassembly to check existence
    NSString *processedLocation = [NSString stringWithFormat:@"%@://%@%@", scheme, host, dataPath];

Answer 1

您可以认为这是将字符串插入地址字符串中的巨大代码。您想做的是相当简单的，因此，这是一个示例方法，它只是将"@2x"弹出到最后一个"."之前。如果没有"."，它将返回零。代码注释中的进一步说明。

-(NSString *)retinaURLStringForString:(NSString *)nonRetinaAddress{
    // Find the range (location and length of ".")
    // Use options parameter to start from the back.
    NSRange extDotRange = [nonRetinaAddress rangeOfString:@"." options:NSBackwardsSearch];
    // You can check whether the "." is there or not like this:
    if (extDotRange.location == NSNotFound){
        // Handle trouble
        return nil;
    }
    // We can use NSString's stringByReplacingCharactersInRange:withString: method to insert the "@2x".
    // To do this we first calculate the range to 'replace'.
    // For location we use the location of the ".".
    // We use 0 for length since we do not want to replace anything.
    NSRange insertRange = NSMakeRange(extDotRange.location, 0);
    // Lastly simply use the stringByReplacingCharactersInRange:withString: method to insert "@2x" in the insert range.
    NSString *retinaAddress = [nonRetinaAddress stringByReplacingCharactersInRange:insertRange withString:@"@2x"];
    return retinaAddress;
}

进行测试时：

NSString *nonRetinaAddress = @"scheme://host/path/name.extension";
NSString *retinaAddress = [self retinaURLStringForString:nonRetinaAddress];
NSLog(@"%@",nonRetinaAddress);
NSLog(@"%@",retinaAddress);

它登录：

scheme://host/path/name.extension
scheme://host/path/[e-mail]

小侧注：

如果原始字符串是可变的字符串，则可以使用[mutableAddressString insertString:@"@2x" atIndex:extDotRange.location]而不是计算替换范围。