java android.provider.Settings.Secure.ANDROID_ID 返回“android_id”
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/26244512/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
android.provider.Settings.Secure.ANDROID_ID returns "android_id"
提问by Fabii
Is there reason android.provider.Settings.Secure.ANDROID_ID is return the constant "android_id" instead of a 64-bit number as a hex string ?
android.provider.Settings.Secure.ANDROID_ID 是否有理由返回常量“android_id”而不是 64 位数字作为十六进制字符串？
Description : android.provider.Settings.Secure.ANDROID_ID
描述：android.provider.Settings.Secure.ANDROID_ID
- A 64-bit number (as a hex string) that is randomly generated when the user first sets up the device and should remain constant for the lifetime of the user's device. The value may change if a factory reset is performed on the device.
- 用户首次设置设备时随机生成的 64 位数字（作为十六进制字符串），并应在用户设备的整个生命周期内保持不变。如果在设备上执行恢复出厂设置，该值可能会更改。
I am using a Samsung Galaxy S4 w /
我正在使用三星 Galaxy S4 w /
<uses-sdk android:minSdkVersion="13" android:targetSdkVersion="19" />
Cheers
干杯
回答by Mike Laren
The android.provider.Settings.Secure.ANDROID_IDis a constant that can be used in android.provider.Settings.Secure.getString(ContentResolver resolver, String name). By default it is set to 'android_id' since that's the name of the property containing the actual Android ID.
的android.provider.Settings.Secure.ANDROID_ID是，可以用在一个恒定android.provider.Settings.Secure.getString(ContentResolver resolver, String name)。默认情况下，它设置为“android_id”，因为这是包含实际 Android ID 的属性的名称。
Use this code to get the actual id:
使用此代码获取实际 ID：
String androidId = Secure.getString(getContext().getContentResolver(), Secure.ANDROID_ID);
回答by RenniePet
This is just to elaborate on the answer by @MikeLaren, which is correct. But there are a couple of gotchas to be aware of, as documented in the following code, which is what I'm currently using:
这只是详细说明@MikeLaren 的答案，这是正确的。但是有几个问题需要注意，如以下代码中所述，这是我目前正在使用的：
// Get the unique (supposedly) ID for this Android device/user combination
long androidId = convertHexToLong(Settings.Secure.getString(
_applicationContext.getContentResolver(), Settings.Secure.ANDROID_ID));
....
....
// Method to convert a 16-character hex string into a Java long. This only took me about an hour,
// due to a known bug in Java that it took them 13 years to fix, and still isn't fixed in the
// version of Java used for Android development.
// http://bugs.java.com/bugdatabase/view_bug.do?bug_id=4215269
// http://stackoverflow.com/questions/1410168/how-to-parse-negative-long-in-hex-in-java
// On top of that it turns out that on a HTC One the Settings.Secure.ANDROID_ID string may be 15
// digits instead of 16!
private long convertHexToLong(String hexString) {
hexString = "0000000000000000" + hexString;
int i = hexString.length();
hexString = hexString.substring(i - 16, i);
try {
return Long.parseLong(hexString.substring(0, 8), 16) << 32 |
Long.parseLong(hexString.substring(8, 16), 16);
} catch (Exception e) {
return 0L;
}
}
回答by CrandellWS
android.provider.Settings.Secure.ANDROID_ID is to big so use this answer from: https://stackoverflow.com/a/10151694/1815624
android.provider.Settings.Secure.ANDROID_ID 太大，所以请使用以下答案：https: //stackoverflow.com/a/10151694/1815624
new BigInteger(string, 16).longValue()
For any value of someLong:
对于 someLong 的任何值：
new BigInteger(Long.toHexString(someLong), 16).longValue() == someLong
In other words, this will return the long you sent into Long.toHexString()for any long value, including negative numbers. It will also accept strings that are bigger than a long and silently return the lower 64 bits of the string as a long. You can just check the string length <= 16 (after trimming whitespace) if you need to be sure the input fits in a long.
换句话说，这将返回您发送Long.toHexString()的任何 long 值的 long，包括负数。它还将接受大于 long 的字符串，并默默地将字符串的低 64 位返回为 long。如果您需要确保输入适合很长，您可以只检查字符串长度 <= 16（修剪空格后）。
