Question

我使用Secure.Android_Id获取设备的唯一ID。但我需要将其作为服务的GUID传递给它。有没有简单的方法来创建它？

我写的基本代码看起来像 this ：

String android_id = getAndroidId(); long leastSignificantBits = 0; long mostSignificantBits = 0; UUID uuid = null;



    if (android_id != null) {
        String asciiConvertedText = "";
        int len = android_id.length();
        for (int i = 0; i < len; i++) {
            asciiConvertedText += "" + ((short) android_id.charAt(i));
        }

        len = asciiConvertedText.length();
        if (len > 16) {
            leastSignificantBits = Long.parseLong(asciiConvertedText.substring(0, 15));

            if (asciiConvertedText.length() > 31) {
                mostSignificantBits = Long.parseLong(asciiConvertedText.substring(16, 31));
            } else {
                mostSignificantBits = Long.parseLong(asciiConvertedText.substring(32,
                        asciiConvertedText.length()));
            }

            uuid = new UUID(mostSignificantBits, leastSignificantBits);
        } else if (len > 0) {

            leastSignificantBits = Long.parseLong(asciiConvertedText);
            uuid = new UUID(mostSignificantBits, leastSignificantBits);
        } else {
            uuid = UUID.randomUUID();
        }
    }

    if (uuid != null) {
        deviceUUID = uuid.toString();
    }

if (android_id != null) { String asciiConvertedText = ""; int len = android_id.length(); for (int i = 0; i < len; i++) { asciiConvertedText += "" + ((short) android_id.charAt(i)); } len = asciiConvertedText.length(); if (len > 16) { leastSignificantBits = Long.parseLong(asciiConvertedText.substring(0, 15)); if (asciiConvertedText.length() > 31) { mostSignificantBits = Long.parseLong(asciiConvertedText.substring(16, 31)); } else { mostSignificantBits = Long.parseLong(asciiConvertedText.substring(32, asciiConvertedText.length())); } uuid = new UUID(mostSignificantBits, leastSignificantBits); } else if (len > 0) { leastSignificantBits = Long.parseLong(asciiConvertedText); uuid = new UUID(mostSignificantBits, leastSignificantBits); } else { uuid = UUID.randomUUID(); } } if (uuid != null) { deviceUUID = uuid.toString(); }

Answer 1

new UUID(Secure.Android_Id.hashCode(), Secure.Android_Id.hashCode()怎么样？

编辑：这是我的“唯一”ID代码：

public static String getDeviceId() {
    final TelephonyManager tm = (TelephonyManager)Globals.Line2App.getSystemService(Context.TELEPHONY_SERVICE);

    final String tmDevice, tmSerial, androidId;
    tmDevice = "" + tm.getDeviceId();
    tmSerial = "" + tm.getSimSerialNumber();
    androidId = "" + Secure.getString(Globals.Line2App.getContentResolver(), Secure.ANDROID_ID);

    UUID deviceUuid = new UUID(androidId.hashCode(), ((long)tmDevice.hashCode() << 32) | tmSerial.hashCode());
    return deviceUuid.toString();
}

我不确定上面是否对你有所帮助，但是我使用了多个值的组合（我想我在stackoverflow上从另一个线程获得了这个代码）。

Android_id要guid？

1 个答案: