I downloaded bc*.jar files(bcprov, bcpkix, bcmail, bcpg) and put them into my project. But Eclipse cannot parse org.bouncycastle.asn1.*. The documentation apparently lists asn1 related functions.
My code:
X500Principal subject = new X500Principal("C=NO");
ContentSigner signGen = new JcaContentSignerBuilder("SHA1withRSA").build(pk);
PKCS10CertificationRequestBuilder builder = new JcaPKCS10CertificationRequestBuilder(subject, pub);
PKCS10CertificationRequest csr = builder.build(signGen);
Error: The type org.bouncycastle.asn1.x500.X500Name cannot be resolved. It is indirectly referenced from required .class files
Is this a correct way to generate csr?
Appreciate!
The exception is throws because some of your classes has a dependency on org.bouncycastle.asn1.x500.X500Name, and this class is not in the classpath. In your case for example PKCS10CertificationRequestBuilder use internally org.bouncycastle.asn1.x500.X500Name so probably the problem is there.
The thing is that PKCS10CertificationRequestBuilder is on bcpkix.jar and org.bouncycastle.asn1.x500.X500Name is on bcprov.jar, so simply add bcprov.jar to your classpath.
Anyways if you want to generate a certificate signing request using java maybe it's easy using directly keytool which is a tool distributed with JVM, and it's located on $JAVA_HOME/bin/keytool. There are a lot of information on internet about how to generate a csr using keytool you can search or if you prefer to show one you can take a look here
Hope this helps,
Related
I'm new to Akeneo, and I discovered profile configuration for assets.
So I imported my YML in order to add asset transformations, and now, cli based, I can't find a command that allows me to generate the variation file for all assets. I saw the command to do that asset by asset and channel by channel, but I need to do that for all of them.
Do you know how I can manage to do that ? I already tried pim:asset:generate-missing-variation-files but that didn't change anything
There is no built-in command to do that, however you could develop a very simple command to achieve this.
You can use the pimee_product_asset.finder.asset service to call retrieveVariationsNotGenerated() in order to retrieve every variation that are not yet genreated, then finally use the pimee_product_asset.variation_file_generator to generate the variation with generate().
Not tested code, but this would be like that:
$finder = $this->get('pimee_product_asset.finder.asset');
$generator = $this->get('pimee_product_asset.variation_file_generator');
$variations = $finder->retrieveVariationsNotGenerated();
foreach ($variations as $variation) {
$generator->generate($variation);
}
I am trying to create an aws instance using jclouds 1.9.0 and then run a script on it (via ssh). I am following the example locate here but I am getting authentication failed errors when the client (java program) tries to connect at the instance. The AWS console show that instance is up and running.
The example tries to create a LoginCrendentials object
String user = System.getProperty("user.name");
String privateKey = Files.toString(new File(System.getProperty("user.home") + "/.ssh/id_rsa"), UTF_8);
return LoginCredentials.builder().user(user).privateKey(privateKey).build();
which is latter used from the ssh client
responses = compute.runScriptOnNodesMatching(
inGroup(groupName), // predicate used to select nodes
exec(command), // what you actually intend to run
overrideLoginCredentials(login) // use my local user & ssh key
.runAsRoot(false) // don't attempt to run as root (sudo)
.wrapInInitScript(false));
Some Login information are injected to the instance with following commands
Statement bootInstructions = AdminAccess.standard();
templateBuilder.options(runScript(bootInstructions));
Since I am on Windows machine the creation of LoginCrendentials 'fails' and thus I alter its code to
String user = "ec2-user";
String privateKey = "-----BEGIN RSA PRIVATE KEY-----.....-----END RSA PRIVATE KEY-----";
return LoginCredentials.builder().user(user).privateKey(privateKey).build();
I also to define the credentials while building the template as described in "EC2: In Depth" guide but with no luck.
An alternative is to build instance and inject the keypair as follows, but this implies that I need to have the ssh key stored in my AWS console, which is not currently the case and also breaks the functionality of running a script (via ssh) since I can not infer the NodeMetadata from a RunningInstance object.
RunInstancesOptions options = RunInstancesOptions.Builder.asType("t2.micro").withKeyName(keypair).withSecurityGroup(securityGroup).withUserData(script.getBytes());
Any suggestions??
Note: While I am currently testing this on aws, I want to keep the code as decoupled from the provider as possible.
Update 26/10/2015
Based on #Ignasi Barrera answer, I changed my implementation by adding .init(new MyAdminAccessConfiguration()) while creating the bootInstructions
Statement bootInstructions = AdminAccess.standard().init(new MyAdminAccessConfiguration());
templateBuilder.options(runScript(bootInstructions));
Where MyAdminAccessConfiguration is my own implementation of the AdminAccessConfiguration interface as #Ignasi Barrera described it.
I think the issue relies on the fact that the jclouds code runs on a Windows machine and jclouds makes some Unix assumptions by default.
There are two different things here: first, the AdminAccess.standard() is used to configure a user in the deployed node once it boots, and later the LoginCredentials object passed to the run script method is used to authenticate against the user that has been created with the previous statement.
The issue here is that the AdminAccess.standard() reads the "current user" information and assumes a Unix System. That user information is provided by this Default class, and in your case I'm pretty sure it will fallback to the catch block and return an auto-generated SSH key pair. That means, the AdminAccess.standard() is creating a user in the node with an auto-generated (random) SSH key, but the LoginCredentials you are building don't match those keys, thus the authentication failure.
Since the AdminAccess entity is immutable, the better and cleaner approach to fix this is to create your own implementation of the AdminAccessConfiguration interface. You can just copy the entire Default class and change the Unix specific bits to accommodate the SSH setup in your Windows machine. Once you have the implementation class, you can inject it by creating a Guice module and passing it to the list of modules provided when creating the jclouds context. Something like:
// Create the custom module to inject your implementation
Module windowsAdminAccess = new AbstractModule() {
#Override protected void configure() {
bind(AdminAccessConfiguration.class).to(YourCustomWindowsImpl.class).in(Scopes.SINGLETON);
}
};
// Provide the module in the module list when creating the context
ComputeServiceContext context = ContextBuilder.newBuilder("aws-ec2")
.credentials("api-key", "api-secret")
.modules(ImmutableSet.<Module> of(windowsAdminAccess, new SshjSshClientModule()))
.buildView(ComputeServiceContext.class);
I have a standalone java application which will interact with my web application running on node. I am trying to add new rules dynamically through web UI. So far I am unable to figure it out, how to create and add rules. Any suggestions for the right direction would be helpful
This is basically a duplicate of https://stackoverflow.com/questions/25036973 so the following is basically a duplicate of my answer to that question...
It's probably best to just look at the Drools examples source code. For instance the KieFileSystem example shows how to create a rule from a String and launch a session which includes it.
The essentials are that you create a KieServices, which contains a virtual file system. You then add rules to that file system. A little bit like the following:
KieServices ks = KieServices.Factory.get();
KieRepository kr = ks.getRepository();
KieFileSystem kfs = ks.newKieFileSystem();
kfs.write("src/main/resources/my/rules/therule.drl", "The source code of the rule");
KieBuilder kb = ks.newKieBuilder(kfs);
kb.buildAll();
you can add multiple Compiled rule DRL files like
knowledgebuilder.add(new ByteArrayResource(compiledDRL.getBytes()),ResourceType.DRL);
Get all the knowledgePackages and fire the all rules
knowledgeBase kbase = knowledgeBaseFactory.newKnowledgeBase();
kbase.addknowledgePackages(knowledgeBuilder.getKnowledgePackages());
knowledgeSession ksession = kbase.newStatefullKnowledgeSession();
ksession.insert(inputObject);
ksession.fireAllRules();
ksession.dispose();
I need to embed some resource in a pure compiled dll written in php using phalanger.
These are txt files tha I set in visual studio as "Embedded Resource".
My problem is that I cannot use the Assembly class to get the resource using GetManifestResourceStream.
I tried code like this:
use System\Reflection\Assembly
$asm = Assembly::GetExecutingAssembly(); //this gives me mscorlib instead of my dll
$str = $asm->GetManifestResourceStream("name");
My question is: how do I get access to embedded resources in phalanger?
Many thanks
I'm not sure, why Assembly::GetExecutingAssembly() returns an incorrect value. Anyway to workaround the $asm value, use following code:
$MyType = CLRTypeOf MyProgram;
$asm = $MyType->Assembly;
Then you can access embedded resources as you posted
$asm->GetManifestResourceStream("TextFile1.txt");
or you can include standard resource file (.resx) into your project, and use \System\Resources\ResourceManager
$this->manager = new \System\Resources\ResourceManager("",$asm);
$this->manager->GetObject("String1",null);
Just note, currently there can be just one .resx within Phalanger project
This question is old, but the part of the Phalanger code (Php.Core.Emit.AddResourceFile() method) responsible for this hasn't changed since this was asked. I faced the same problem and solved it in (almost) non-hacky way. You have to provide alternative name (/res:/path/to/filename,alternative-name) for this to work though.
$asm = clr_typeof('self')->Assembly;
$resourceStream = $asm->GetManifestResourceStream("filename");
$reader = new \System\Resources\ResourceReader($resourceStream);
$type = $data = null;
$reader->GetResourceData("alternative-name", $type, $data);
// and still there are 4 excess bytes
// representing the length of the resource
$data = \substr($data, 4);
$stream = new IO\MemoryStream($data);
// after this $stream is usable as you would expect
Straightforward GetManifestResourceStream() (as suggested by Jakub) does not work because Phalanger does not use System.Reflection.Emit.ModuleBuilder.DefineManifestResource() (like I think it should when supplied with unrecognized file format). It uses ModuleBuilder.DefineResource() which returns ResourceWriter instead, that only really suited for .resources files. And this is what dictates the requirement to use ResourceReader when you need to read your resource.
Note: This answer applies to Phalanger master branch at the time of writing and prior versions since circa 2011. Noted because it looks like a bug (especially the need to use both original and alternative names).
I am trying to run tests by adding a version of tornado downloaded from github.com in the sys.path.
[tests]
recipe = zc.recipe.testrunner
extra-paths = ${buildout:directory}/parts/tornado/
defaults = ['--auto-color', '--auto-progress', '-v']
But when I run bin/tests I get the following error :
ImportError: No module named tornado
Am I not understanding how to use extra-paths ?
Martin
Have you tried looking into generated bin/tests script if it contains your path? It will tell definitely if your buildout.cfg is correct or not. Maybe problem is elsewhere. Because it seem that your code is ok.
If you happen to regularly include various branches from git/mercurial or elsewhere to buildout, you might be interested in mr.developer. mr.developer can download and add package to develop =. You wont need to set extra-path in every section.