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如何判断可可触摸设备是否可以打电话?[复制品]

ipod-touch url-scheme cocoa-touch objective-c ios

Will Harris · 技术社区 · 17 年前

可能重复:
iOS - Detecting whether or not device support phone calls?

我正在编写一个iPhone应用程序,它提供了一个呼叫电话号码的按钮。我用下面的代码用 tel: 通常的网址:

NSURL* contactTelURL = [NSURL
                        URLWithString:[NSString
                                       stringWithFormat:@"tel:%@",
                                       contactTel]];
[[UIApplication sharedApplication] openURL:contactTelURL];

它在真正的iPhone上运行良好,但我在模拟器中得到了一个“不支持的URL”警报。这大概也会发生在iPod touch上,尽管我还没有测试过。在不打电话的设备上运行时,最好删除按钮。

有没有一种方法可以通过编程检测可可触摸设备是否可以打电话?

4 回复 | 直到 8 年前

-1

Vadim Kotov First Zero 8 年前

你可以查询 [[UIDevice currentDevice] model] ,然后检查它是否是iPhone。

Community Mohan Dere 9 年前

来自诺亚威瑟斯彭 Make a call from my iPhone application

模拟器不支持很多iOS的URL方案,包括手机、地图、YouTube和短信应用程序的URL方案。对于iPod touch和iPad等没有电话功能的设备也是如此;在通过-openurl:使用任何URL方案之前,您应该使用-canopenurl:检查该方案的支持情况,根据当前设备是否支持您使用的URL方案,该方案将返回yes或no。

所以查询 [[UIApplication sharedApplication] canOpenURL:[NSURL URLWithString:@"tel://"]] 以确定设备是否可以拨打电话。

Kriem 17 年前

从 iphonedevelopment.blogspot.com

#import <sys/utsname.h>

enum {
    MODEL_IPHONE_SIMULATOR,
    MODEL_IPOD_TOUCH,
    MODEL_IPHONE,
    MODEL_IPHONE_3G
};

@interface DeviceDetection : NSObject

+ (uint) detectDevice;
+ (NSString *) returnDeviceName:(BOOL)ignoreSimulator;

@end


@implementation DeviceDetection

+ (uint) detectDevice {
    NSString *model= [[UIDevice currentDevice] model];

    // Some iPod Touch return "iPod Touch", others just "iPod"

    NSString *iPodTouch = @"iPod Touch";
    NSString *iPodTouchLowerCase = @"iPod touch";
    NSString *iPodTouchShort = @"iPod";

    NSString *iPhoneSimulator = @"iPhone Simulator";

    uint detected;

    if ([model compare:iPhoneSimulator] == NSOrderedSame) {
        // iPhone simulator
        detected = MODEL_IPHONE_SIMULATOR;
    } else if ([model compare:iPodTouch] == NSOrderedSame) {
        // iPod Touch
        detected = MODEL_IPOD_TOUCH;
    } else if ([model compare:iPodTouchLowerCase] == NSOrderedSame) {
        // iPod Touch
        detected = MODEL_IPOD_TOUCH;
    } else if ([model compare:iPodTouchShort] == NSOrderedSame) {
        // iPod Touch
        detected = MODEL_IPOD_TOUCH;
    } else {
        // Could be an iPhone V1 or iPhone 3G (model should be "iPhone")
        struct utsname u;

        // u.machine could be "i386" for the simulator, "iPod1,1" on iPod Touch, "iPhone1,1" on iPhone V1 & "iPhone1,2" on iPhone3G

        uname(&u);

        if (!strcmp(u.machine, "iPhone1,1")) {
            detected = MODEL_IPHONE;
        } else {
            detected = MODEL_IPHONE_3G;
        }
    }
    return detected;
}


+ (NSString *) returnDeviceName:(BOOL)ignoreSimulator {
    NSString *returnValue = @"Unknown";

    switch ([DeviceDetection detectDevice]) {
        case MODEL_IPHONE_SIMULATOR:
            if (ignoreSimulator) {
                returnValue = @"iPhone 3G";
            } else {
                returnValue = @"iPhone Simulator";
            }
            break;
        case MODEL_IPOD_TOUCH:
            returnValue = @"iPod Touch";
            break;
        case MODEL_IPHONE:
            returnValue = @"iPhone";
            break;
        case MODEL_IPHONE_3G:
            returnValue = @"iPhone 3G";
            break;
        default:
            break;
    }        
    return returnValue;
}

@end

Josh Peak 15 年前

这里有一个简单的代码片段,我用它来检查设备型号是否是一部手机而不是一个模拟器,以确保它可以打电话。

if ([[[UIDevice currentDevice] model] rangeOfString:@"Phone"].location != NSNotFound &&
    [[[UIDevice currentDevice] model] rangeOfString:@"Simulator"].location == NSNotFound ) {
            [[UIApplication sharedApplication] openURL:[NSURL URLWithString:[NSString stringWithFormat:@"tel:%@", number]  ] ];
}