通过比较PackageInfo的包名和签名,我在android应用程序中实现了不兼容应用程序的验证。我通过以下方式获得有关已安装软件包的信息:
val packages = context.ctx.packageManager.getInstalledPackages(PackageManager.GET_META_DATA or PackageManager.GET_SIGNATURES)
之后,只需比较包名和签名的字符串。我通过以下函数将PackageInfo.signatures转换为十六进制:
fun bytesToHex(bytes: ByteArray): String {
val hexArray = charArrayOf('0', '1', '2', '3', '4', '5', '6', '7', '8', '9', 'A', 'B', 'C', 'D', 'E', 'F')
val hexChars = CharArray(bytes.size * 2)
var v: Int
for (j in bytes.indices) {
v = bytes[j].toInt() and 0xFF
hexChars[j * 2] = hexArray[v.ushr(4)]
hexChars[j * 2 + 1] = hexArray[v and 0x0F]
}
return String(hexChars)
}而且效果很好。
现在,我想为这个功能实现几个单元测试,并且遇到了困难:如何模拟PackageInfo.signatures?如何将SHA1指纹字符串转换为android.content.pm.Signature?
发布于 2018-11-27 17:51:11
我不建议模拟框架类,因为这将是脆弱的。相反,只需创建自己的接口来提供签名,并在依赖于框架的实际应用程序中使用一个实现。例如:
interface SignatureProvider {
val signatures: List<String>
}
class AndroidSignatureProvider(private val packageManager: PackageManager): SignatureProvider {
private val flags = PackageManager.GET_META_DATA or PackageManager.GET_SIGNATURES
override val signatures: List<String>
get() = packageManager.getInstalledPackages(flags)
.flatMap { it.packageInfo.signatures.toList() }
.map { it.toCharsString() }
}然后,在您的测试中,您只需模拟一个SignatureProvider,返回要对其进行测试的不同签名字符串:
private lateinit var signatureProvider: SignatureProvider
@Before fun setup() {
signatureProvider = mock(SignatureProvider::class.java)
doReturn(listOf("sigA", "sigB")).`when`(signatureProvider)`.signatures
}发布于 2019-06-26 09:13:33
我相信你可以通过这样做来实现这一点:
testPackage() {
final ShadowPackageManager packageManager = shadowOf(RuntimeEnvironment.application.getPackageManager());
packageManager.installPackage(getPackageInfo(com.yourapp));
final PackageInfo packageInfo = packageManager.getInternalMutablePackageInfo(com.yourapp);
packageInfo.signatures = new Signature[1];
packageInfo.signatures[0] = new Signature("your_signature");
}
private static PackageInfo getPackageInfo(String packageName) {
final PackageInfoBuilder builder = PackageInfoBuilder.newBuilder();
builder.setPackageName(packageName);
return builder.build();
}预期的格式是由Signature#toChars()生成的。
现在,我同意@kcoppock,因为使用签名和它们不同的表示方式操作很麻烦,而且容易出错。
https://stackoverflow.com/questions/53501818
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