从状态栏崩溃中获取蜂窝信号

Question

以下是我的代码，我试图通过状态栏获得信号强度。

UIApplication *app = [UIApplication sharedApplication];
NSArray *subviews = [[[app valueForKey:@"statusBar"] valueForKey:@"foregroundView"] subviews];

崩溃发生在（iOS 11，xcode 9，iPhone x）： -

[uistatusbar_modern valueforundefinedkey：]：此类不适合键入键值编码

Answer 1

您可以使用此

NSArray *subviews = nil;
id statusBar = [[UIApplication sharedApplication] valueForKey:@"statusBar"];
if ([statusBar isKindOfClass:NSClassFromString(@"UIStatusBar_Modern")]) {
    subviews = [[[statusBar valueForKey:@"statusBar"] valueForKey:@"foregroundView"] subviews];
} else {
    subviews = [[statusBar valueForKey:@"foregroundView"] subviews];
}

Answer 2

使用以下：

- (NSString*)newtworkType {
    NSArray *subviews = nil;
    id statusBar = [[UIApplication sharedApplication] valueForKey:@"statusBar"];
    if ([statusBar isKindOfClass:NSClassFromString(@"UIStatusBar_Modern")]) {
        subviews = [[[statusBar valueForKey:@"statusBar"] valueForKey:@"foregroundView"] subviews];
    } else {
        subviews = [[statusBar valueForKey:@"foregroundView"] subviews];
    }
    NSNumber *dataNetworkItemView = nil;
    for (id subview in subviews) {
        if([subview isKindOfClass:[NSClassFromString(@"UIStatusBarDataNetworkItemView") class]]) {
            dataNetworkItemView = subview;
            break;
        }
    }
    switch ([[dataNetworkItemView valueForKey:@"dataNetworkType"]integerValue]) {
        case 0:
            return @"No wifi or cellular";
            break;
        case 1:
            return @"2G";
            break;
        case 2:
            return @"3G";
            break;
        case 3:
            return @"4G";
            break;
        case 4:
            return @"LTE";
            break;
        case 5:
            return @"WIFI";
            break;

        default:
            break;
    }
    return @"";
}