是否有原因android.provider.Settings.Secure.ANDROID_ID返回常量“android_id”而不是64位数作为十六进制字符串?
描述:android.provider.Settings.Secure.ANDROID_ID
我正在使用三星Galaxy S4 w /
<uses-sdk android:minSdkVersion="13" android:targetSdkVersion="19" />
干杯
答案 0 :(得分:19)
android.provider.Settings.Secure.ANDROID_ID
是一个可以在android.provider.Settings.Secure.getString(ContentResolver resolver, String name)
中使用的常量。默认情况下,它设置为&#39; android_id&#39;因为那是包含实际Android ID的属性的名称。
使用此代码获取实际ID:
String androidId = Secure.getString(getContext().getContentResolver(), Secure.ANDROID_ID);
答案 1 :(得分:3)
这只是为了详细说明@MikeLaren的答案,这是正确的。但是有一些问题需要注意,如下面的代码所示,这就是我目前使用的:
// 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;
}
}
答案 2 :(得分:0)
android.provider.Settings.Secure.ANDROID_ID很大,所以请使用以下答案: https://stackoverflow.com/a/10151694/1815624
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对于someLong的任何值:
new BigInteger(string, 16).longValue()
换句话说,这将返回您发送到new BigInteger(Long.toHexString(someLong), 16).longValue() == someLong
的任何长值,包括负数。它还将接受大于long的字符串,并以字符串的长度静默返回字符串的低64位。你可以检查字符串长度&lt; = 16(在修剪空格之后),如果你需要确保输入适合长。