I have to test the code coverage of web services hosted on Tomcat during Unit Testing. I'll be doing manual testing means i'll pass different values of parameters in my http request and need to have a coverage report for each test case means for each request sent to the server. How can i do that with Emma or is there any other tool that I can use?
or
Please tell the steps to instrument WAR using Emma
create an ANT target in your project which does a compilation of the source code and instruments the class files and create war file.
Copy the emma-2.1.5320.jar jar to java/jre/lib/ext/ directory(whichever java your tomcat uses).
create a setenv.sh file in tomcat/bin/ directory which contains below lines:
!/bin/sh
export JAVA_OPTS="$JAVA_OPTS -XX:-UseSplitVerifier -Demma.coverage.out.merge=true"
restart tomcat server and check if the above properties are updated in tomcat process (ps -aef | grep tomcat)
Then copy this generated war file to the tomcat/webapps/ directory.
start your test cases and now you should see the below line in your catalina.out file
EMMA: collecting runtime coverage data
Related
I have read here that the log4j library can be "nested" within other files that are deployed with an application.
I can find files with 'log4j' in the filename but don't know how to find log4j in these "nested jars". Is there a way to do this from the command line?
update
This question has moved to SuperUser here.
Run the following command search for log4j jar files in an application:
dir /s /b <application_root>\*log4j*.jar
If any files are displayed, check the version number that is part of the file name. For example Tomcat\webapps\ROOT\WEB-INF\lib\log4j-core-2.15.0.jar is version 2.15.
If the version is between 2 and 2.14 (2.15 is not vulnerable), the application is vulnerable to CVE-2021-44228 and one of the following mitigations must be applied.
Actually, I'm trying to add new language to Streama Media Server. I downloaded source code, added new language file (as guided here). After that, I want to build a jar with that project, I build it with IntelliJ Idea 2021.1 (here is how I did). So, When I run that jar file (in Ubunt 20.04), it fails and gives this error: Screenshot
When you have made adjustments to the source code, it is likely that you will want to create a new .jar file and deploy it on your server. For this, you can use a simple command:
# for unix based systems
**./gradlew assemble**
# for windows
**./gradlew.bat assemble**
This will create 2 new .jar files under build/libs,
streama-{version}.jar
streama-{version}.jar.original
all you will need is the streama-{version}.jar.
This file is an executable, so you can just copy it into your deployment directory / your server and start it as usual.
I want to read a json file placed in resources/data/info.json in my webapp. Got to know that need to use Convertor.class.getResourceAsStream("\\WEB-INF\\classes\\data\\Address.json"); where Convertor is my util class for some reason am unable to read the file and i see the InputStream returning null and i see the below exception :
com.fasterxml.jackson.databind.exc.MismatchedInputException: No content to map due to end-of-input
at [Source: UNKNOWN; line: 1, column: 0]
at com.fasterxml.jackson.databind.exc.MismatchedInputException.from(MismatchedInputException.java:59)
at com.fasterxml.jackson.databind.ObjectMapper._initForReading(ObjectMapper.java:4133)
at com.fasterxml.jackson.databind.ObjectMapper._readMapAndClose(ObjectMapper.java:3988)
at com.fasterxml.jackson.databind.ObjectMapper.readValue(ObjectMapper.java:3058)
at com.wf.hrca.util.Convertor.unMarshal(Convertor.java:84)
resources is a folder of your source project on your development machine. Tomcat doesn't care about how your source project looks like. And by the way, when you'll deploy the application in production, there won't be any source project on the machine.
All Tomcat cares about is the war file you deploy. Inside this war file, the WEB-INF/classes directory, along with all the jar files under WEB-INF/lib, constitute the classpath of the application.
Convertor.class.getResourceAsStream(), as the javadoc explains (but you really need to read it to know) expects a /-separated path. If the path starts with a /, then the path starts at one of the roots of the classpath (i.e. / refers to WEB-INF/classes, and to the root of each jar file of WEB-INF/lib).
So, find where the json file is located inside the war file. If it's under /WEB-INF/classes/data/info.json, then you should use
Convertor.class.getResourceAsStream("/data/info.json");
I have a war file,. myapp.war (it happens to be a grails app, but this is not material)
I upload this to an s3 bucket, say myapp in us-west-2
I set up an OpsWorks using the S3 repository type:
Repository Type: S3
Repository URL: https://myapp.s3-us-west-2.amazonaws.com/myapp.war
Access key ID: A key with read permission on the above bucket
Secret access key: the secret for this key
Deploy to an instance in Java layer (Tomcat 7)
All lights are green, deployments succeeded
But the app isn't actually deployed
Shelling in to the instance and looking in /usr/share/tomcat7/webapps I find a directory called 'myapp'. Inside this directory is a file called 'archive'. 'archive' appears to be a war file, but it is not named 'archive.war', and it is in a subdirectory of webapps, so tomcat isn't going to deploy it anyway.
Now, the OpsWorks docs say the archive should be a 'zip' file. But:
zipping up myapp.war into a zip archive 'myapp.war.zip' and changing the path to this file results in 'myapp' containing 'myapp.war'. No deployment, since tomcat isn't looking for war files in 'webapps/myapp'
Changing the name of 'myapp.war' to 'myapp.zip' and changing the repository path results in 'myapp' containing the single file 'archive' again.
So. Can anyone describe how to properly provide a war file to OpsWorks from S3?
It appears that the problem has to do with how the zip archive is made.
Jars, war, and the like created with the java 'jar' tool do not work. Zip archives created with a zip tool, and then renamed to have a '.war' extension do.
This is explained here: https://forums.aws.amazon.com/thread.jspa?messageID=559582򈧞
Quoting that post's answer:
Our current extract script doesn't correctly identify WAR files. If
you unpack the WAR file and use zip to pack it, it should work until
we update our script.
So the procedure that works is to:
Explode the war made by your development environment (In the case of grails, the war build cleans up the staging directory for the war, so you don't have an exploded war directory laying around to zip up yourself, you have to unzip it first.)
Zip the contents of the directory created by exploding the war using a zip tool (or, if your build tool leaves the exploded war directory there, then just zip it directly)
Optionally, rename the new zip archive to have a '.war' extension.
Resume the procedure from the original question, step 3 -- that is, upload the war to the s3 bucket and
Specify the S3 path to the war file as the repository in the OpsWorks setup.
EDIT:
After answering this, I discovered that Grails can produce an exploded war directory after all.
// BuildConfig.groovy
...
grails.project.war.exploded.dir = "path/to/exploded/war-directory"
grails.war.exploded=true
...
That directory can be zipped or jarred or whatever you want by your builder/deployer.
From this wiki page you see that a WAR file is just a special JAR file. And if you check out what a JAR is here then you see it is just zipped up compiled java code.
This SuperUser question also touches on the .WAR vs .zip business. Basically, a WAR is just a special ZIP. So when you upload a WAR, you are uploading a ZIP.
Make sure it's a WAR file in the S3 bucket.
Provide the entire link to the S3 WAR file. To get this, right-click the WAR file in S3 and select Properties and then copy the link.
I have created a Setup file i.e. setup.msi , this file contains a website installer
I have a License.xml file that I want to give to my client with the installer (setup.msi).
Before running the setup.msi , client has to make sure that the License.xml file should be in the same directory where setup.msi resist.
I want my setup.msi file to copy the License.xml file to the Destination directory (where website will be installed , this path will be prompted to user for customization)
I am using MSIFactory for creating setup.msi. I am not able to do this. I searched over net but did not get any accurate answer.
I am not familiar with MSIfactory, but in installshield what I would do is, call a batch script to do the copy and pass the target directory as a parameter to it. or use the 'XCopyFile' function to do the copying.