如何获取Android设备唯一ID

原文 标签 IT工具网 java android

问题

每一个android设备都有唯一ID吗?如果有?怎么用java最简单取得呢?

回答1(最佳)

如何取得android唯一码?

好处: - 1.不需要特定权限. - 2.在99.5% Android装置(包括root过的)上,即API => 9,保证唯一性. - 3.重装app之后仍能取得相同唯一值.

伪代码:

if API => 9/10: (99.5% of devices)

return unique ID containing serial id (rooted devices may be different)

else

return unique ID of build information (may overlap data - API < 9)

代码:


/**
 * Return pseudo unique ID
 * @return ID
 */public static String getUniquePsuedoID() {
    // If all else fails, if the user does have lower than API 9 (lower
    // than Gingerbread), has reset their device or 'Secure.ANDROID_ID'
    // returns 'null', then simply the ID returned will be solely based
    // off their Android device information. This is where the collisions
    // can happen.
    // Thanks http://www.pocketmagic.net/?p=1662!
    // Try not to use DISPLAY, HOST or ID - these items could change.
    // If there are collisions, there will be overlapping data
    String m_szDevIDShort = "35" + (Build.BOARD.length() % 10) + (Build.BRAND.length() % 10) + (Build.CPU_ABI.length() % 10) + (Build.DEVICE.length() % 10) + (Build.MANUFACTURER.length() % 10) + (Build.MODEL.length() % 10) + (Build.PRODUCT.length() % 10);

    // Thanks to @Roman SL!
    // http://stackoverflow.com/a/4789483/950427
    // Only devices with API >= 9 have android.os.Build.SERIAL
    // http://developer.android.com/reference/android/os/Build.html#SERIAL
    // If a user upgrades software or roots their device, there will be a duplicate entry
    String serial = null;
    try {
        serial = android.os.Build.class.getField("SERIAL").get(null).toString();

        // Go ahead and return the serial for api => 9
        return new UUID(m_szDevIDShort.hashCode(), serial.hashCode()).toString();
    } catch (Exception exception) {
        // String needs to be initialized
        serial = "serial"; // some value
    }

    // Thanks @Joe!
    // http://stackoverflow.com/a/2853253/950427
    // Finally, combine the values we have found by using the UUID class to create a unique identifier
    return new UUID(m_szDevIDShort.hashCode(), serial.hashCode()).toString();}

回答2

好处: - 1.不需要特定权限. - 2.在100% Android装置(包括root过的)上,保证唯一性.

坏处 - 1.重装app之后不能取得相同唯一值.

private static String uniqueID = null;
private static final String PREF_UNIQUE_ID = "PREF_UNIQUE_ID";

public synchronized static String id(Context context) {
    if (uniqueID == null) {
        SharedPreferences sharedPrefs = context.getSharedPreferences(
                PREF_UNIQUE_ID, Context.MODE_PRIVATE);
        uniqueID = sharedPrefs.getString(PREF_UNIQUE_ID, null);
        if (uniqueID == null) {
            uniqueID = UUID.randomUUID().toString();
            Editor editor = sharedPrefs.edit();
            editor.putString(PREF_UNIQUE_ID, uniqueID);
            editor.commit();
        }
    }
    return uniqueID;
}

回答3(需要有电话卡)

好处: 1.重装app之后仍能取得相同唯一值.

代码:

    final TelephonyManager tm = (TelephonyManager) getBaseContext().getSystemService(Context.TELEPHONY_SERVICE);
    final String tmDevice, tmSerial, androidId;
    tmDevice = "" + tm.getDeviceId();
    tmSerial = "" + tm.getSimSerialNumber();
    androidId = "" + android.provider.Settings.Secure.getString(getContentResolver(), android.provider.Settings.Secure.ANDROID_ID);
    UUID deviceUuid = new UUID(androidId.hashCode(), ((long)tmDevice.hashCode() << 32) | tmSerial.hashCode());
    String deviceId = deviceUuid.toString();

谨记:要取得以下权限

<uses-permission android:name="android.permission.READ_PHONE_STATE" />

stackoverflow链接: http://stackoverflow.com/questions/2785485/is-there-a-unique-android-device-id

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