Question

我正在使用以下代码......

openPhoneDialler(mobile) {
    let url = 'tel:' + mobile;
    Linking.canOpenURL(url).then(supported => {
      if (!supported) {
        throw 'Cant open phone dialler';
      } else {
        return Linking.openURL(url);
      }
    }).catch(err =>
      Alert.alert('Error','Error occurred',[{text: 'OK', onPress: () => {}}])
    );
  }

我在没有SIM卡功能的iPad上测试（仅限Wifi），canOpenURL得到解决，并且调用Linking.openURL（url），显示带有手机号码的警报和两个选项'取消'和'呼叫' 。当我按下“呼叫”时，警报会关闭，没有任何事情发生，因为它只是Wifi iPad。

但是当我按下“取消”时，应用程序会因以下错误屏幕崩溃

JSON value '{
    code = EUNSPECIFIED;
    domain = RCTErrorDomain;
    framesToPop = 1;
    nativeStackIOS =     (
        "0   workforce_mobile                    0x00000001000d4d00 RCTJSErrorFromCodeMessageAndNSError + 152",
        "1   workforce_mobile                    0x0000000100053e4c __41-[RCTModuleMethod processMethodSignature]_block_invoke_2.224 + 192",
        "2   workforce_mobile                    0x000000010015e2f0 -[RCTLinkingManager openURL:resolve:reject:] + 352",
        "3   CoreFoundation                      0x000000018164ce80 <redacted> + 144",
        "4   CoreFoundation                      0x00000001815422b4 <redacted> + 292",
        "5   CoreFoundation                      0x0000000181546e7c <redacted> + 60",
        "6   workforce_mobile                    0x0000000100055098 -[RCTModuleMethod invokeWithBridge:module:arguments:] + 1704",
        "7   workforce_mobile                    0x00000001000aea54 -[RCTBatchedBridge callNativeModule:method:params:] + 568",
        "8   workforce_mobile                    0x00000001000addf4 __33-[RCTBatchedBridge handleBuffer:]_block_invoke.477 + 1276",
        "9   libdispatch.dylib                   0x00000001804fe9e0 <redacted> + 24",
        "10  libdispatch.dylib                   0x00000001804fe9a0 <redacted> + 16",
        "11  libdispatch.dylib                   0x000000018050cad4 <redacted> + 928",
        "12  libdispatch.dylib                   0x00000001805022cc <redacted> + 884",
        "13  libdispatch.dylib                   0x000000018050ea50 <redacted> + 540",
        "14  libdispatch.dylib                   0x000000018050e7d0 <redacted> + 124",
        "15  libsystem_pthread.dylib             0x00000001807071d0 _pthread_wqthread + 1096",
        "16  libsystem_pthread.dylib             0x0000000180706d7c start_wqthread + 4"
    );
    userInfo = "<null>";
}' of type NSDictionary cannot be converted to NSString

注意：如果您在没有任何按下的情况下将警报打开大约10秒钟，也会发生同样的崩溃。

测试：iPad（仅限Wifi和物理设备）操作系统版本10.3.1

这是React-Native Linking Api中的错误还是上述代码中的错误。

Answer 1

Linking.openUrl返回一个promise对象。按取消按钮，您的承诺将被拒绝。只需添加一个捕获来吞下被拒绝的承诺。

Linking.openUrl().catch(() => null)

以下是有关该文档的文档：

https://facebook.github.io/react-native/docs/linking.html#openurl

将Api的openURL与iPad

1 个答案: