我们有一个在嵌入式平台上运行的遗留应用程序,我们使用Java 6作为JVM。我们可以从需要TLS1.2支持的应用程序访问https。我们使用的JVM没有提供这种功能。如何实现TLS1.2对应用程序的支持?
发布于 2020-11-26 07:59:30
利用弹跳城堡库可以实现对TLS1.2的支持。
以下是详细的解决方案
Maven依赖
<dependency>
<groupId>org.bouncycastle</groupId>
<artifactId>bcprov-jdk15to18</artifactId>
<version>1.64</version>
</dependency>
<dependency>
<groupId>org.bouncycastle</groupId>
<artifactId>bctls-jdk15to18</artifactId>
<version>1.64</version>
</dependency>或者,您可以更新JRE/lib/security/java.security
security.provider.1=org.bouncycastle.jce.provider.BouncyCastleProvider
security.provider.2=org.bouncycastle.jsse.provider.BouncyCastleJsseProvider
security.provider.3=sun.security.provider.Sun
security.provider.4=sun.security.rsa.SunRsaSign
security.provider.5=com.sun.net.ssl.internal.ssl.Provider
security.provider.6=com.sun.crypto.provider.SunJCE
security.provider.7=sun.security.jgss.SunProvider
security.provider.8=com.sun.security.sasl.ProviderBC库需要位于顶部(1&2)
测试
HttpsURLConnection urlConnection = null;
try {
URL url = new URL("https://www.nist.gov/");
urlConnection = (HttpsURLConnection) url.openConnection();
String data = IOUtils.toString(urlConnection.getInputStream(), "UTF-8");
System.out.println(data);
} catch (IOException ex) {
ex.printStackTrace();
try {
if (urlConnection != null) {
code = ((HttpURLConnection) urlConnection).getResponseCode();
message = ((HttpURLConnection) urlConnection).getResponseMessage();
} else {
message = ex.toString();
}
} catch (IOException ex2) {
message = ex2.toString();
}
System.out.println("Response : " + message);
} finally {
if (urlConnection != null) {
urlConnection.disconnect();
}
}https://stackoverflow.com/questions/65018035
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