How to do File creation and manipulation in functional style? - scala

I need to write a program where I run a set of instructions and create a file in a directory. Once the file is created, when the same code block is run again, it should not run the same set of instructions since it has already been executed before, here the file is used as a guard.
var Directory: String = "Dir1"
var dir: File = new File("Directory");
dir.mkdir();
var FileName: String = Directory + File.separator + "samplefile" + ".log"
val FileObj: File = new File(FileName)
if(!FileObj.exists())
// blahblah
else
{
// set of instructions to create the file
}
When the programs runs initially, the file won't be present, so it should run the set of instructions in else and also create the file, and after the first run, the second run it should exit since the file exists.
The problem is that I do not understand new File, and when the file is created? Should I use file.CreateNewFile? Also, how to write this in functional style using case?

It's important to understand that a java.io.File is not a physical file on the file system, but a representation of a pathname -- per the javadoc: "An abstract representation of file and directory pathnames". So new File(...) has nothing to do with creating an actual file - you are just defining a pathname, which may or may not correspond to an existing file.
To create an empty file, you can use:
val file = new File("filepath/filename")
file.createNewFile();
If running on JRE 7 or higher, you can use the new java.nio.file API:
val path = Paths.get("filepath/filename")
Files.createFile(path)
If you're not happy with the default IO APIs, you an consider a number of alternative. Scala-specific ones that I know of are:
scala-io
rapture.io
Or you can use libraries from the Java world, such as Google Guava or Apache Commons IO.
Edit: One thing I did not consider initially: I understood "creating a file" as "creating an empty file"; but if you intend to write something immediately in the file, you generally don't need to create an empty file first.

Related

How to copy whole directories containing subdirectories to /boot (i.e bootfs) in Yocto while inheriting core-image class?

I have a directory which again contains subdirectories, which are built has part of other recipe and moved to DEPLOY_DIR_IMAGE using deploy bb class. So now I want to copy it to main image boot partition.
If it was a single file then appending required filename to IMAGE_EFI_BOOT_FILES variable, then yocto copies it to /boot. But same doesn't work for directories containing subdirectories please provide style to include even the subdirectories. Thank you
PS: I have tried appending IMAGE_EFI_BOOT_FILES += "parent_dir/*" didnt work.
It is obvious that IMAGE_EFI_BOOT_FILES is acting like the well known IMAGE_BOOT_FILES and other variables that are responsible for having the files necessary to be shipped in the boot partition. And that needs files and not directories.
So, if you do not need to specify all the files by hand, but instead you want to pass the directory, I suggest you use a python method to collect the files for you and append them to the variable.
See the following example I developed and tested:
def get_files(d, dir):
import os
dir_path = dir
if not os.path.exists(os.path.dirname(dir)):
dir_path = d.getVar(dir)
return ' '.join(f for f in os.listdir(d.getVar(dir)) if os.path.isfile(f))
IMAGE_EFI_BOOT_FILES += "${#get_files(d, 'DEPLOY_DIR_IMAGE')}"
The method will test if the argument is a real path then it will directly check for files, if not it will assume that it is a bitbake variable and it will get its content, so if DEPLOY_DIR_IMAGE is, for example, /home/user/dir, passing DEPLOY_DIR_IMAGE or /home/usr/dir will give the same result.
IMPORTANT
It is obvious also that IMAGE_EFI_BOOT_FILES is used in a .conf file such as local.conf or a custom machine configuration file. So adding that python function in .conf file will not work. I suggest creating a class for it and inherit it globally in your .conf file:
meta-custom/classes/utils.bbclass
local.conf:
INHERIT += "utils"
IMAGE_EFI_BOOT_FILES += "${#get_files(d, 'DEPLOY_DIR_IMAGE')}"
Try this and let me know in the comments.
EDIT
I have just realized that bitbake already imports os within python expressions expansions, so you can do it in one line without any need for a separate python function:
PATH = "/path/to/directory/" or
PATH = "A variable containing the path"
IMAGE_EFI_BOOT_FILES += "${#' '.join('%s' % f for f in os.listdir('${PATH}') if os.path.isfile(f))}"
Note: I am looking for Yocto built-in which can achieve solution for above mentioned , would like to share other way to resolve the functionality for community's benefit.
Add following in bb file if you are using one or refer to talel-belhadjsalem answer to use utils.bbclass.
def add_directory_bootfs(d, dirname, bootfs_dir):
file_list = list()
boot_files_list = list()
deploy_dir_image = d.getVar('DEPLOY_DIR_IMAGE')
for (dirpath, dirnames, filenames) in os.walk(os.path.join(deploy_dir_image, dirname)):
file_list += [os.path.join(dirpath, file) for file in filenames]
for file in file_list:
file_rel_path = os.path.relpath(file, os.path.join(deploy_dir_image, dirname))
boot_file_entry = os.path.join(dirname, file_rel_path) + ';' + os.path.join(bootfs_dir, dirname, file_rel_path)
boot_files_list.append(boot_file_entry)
return ' '.join(boot_files_list)
IMAGE_EFI_BOOT_FILES += "${#add_directory_bootfs(d, 'relative_path_to_dir_in_deploy_dir_image', 'EFI/BOOT')}"

Old feature file path is used even after updating a new path

I am new to cucumber and I am automating a scenario. Initially I kept my features files in the path C:\Users\test\eclipse-workspace\Automation\src\test\resources\featureFile. Then I moved the feature files to a different path (C:\Users\test\eclipse-workspace\Automation\src\test\com\test]automation\features). I have updated the same in CucumberOptions as shown below.
#CucumberOptions(features = {
"src/test/java/com/test/automation/features/CO_Self_Service_Home_Page_Personalizations.feature" }, glue = {
"src/test/java/com/oracle/peoplesoft/HCM/StepDefinitions" })
But when I try to run the feature, I am getting the below exception stating the feature file is not found. Here the path shown in the exception is the old path. I am not sure from where it is fetched as I have updated the new path in Cucumber options. Can you please help me understand the cause of this issue.
Exception in thread "main" java.lang.IllegalArgumentException: Not a
file or directory:
C:\Users\test\eclipse-workspace\Automation\src\test\resources\featureFile\Self_Service_Home_Page_Personalizations.feature
at
cucumber.runtime.io.FileResourceIterator$FileIterator.(FileResourceIterator.java:54)
at
cucumber.runtime.io.FileResourceIterator.(FileResourceIterator.java:20)
at
cucumber.runtime.io.FileResourceIterable.iterator(FileResourceIterable.java:19)
at
cucumber.runtime.model.CucumberFeature.loadFromFeaturePath(CucumberFeature.java:103)
at
cucumber.runtime.model.CucumberFeature.load(CucumberFeature.java:54)
at
cucumber.runtime.model.CucumberFeature.load(CucumberFeature.java:34)
at
cucumber.runtime.RuntimeOptions.cucumberFeatures(RuntimeOptions.java:235)
at cucumber.runtime.Runtime.run(Runtime.java:110) at
cucumber.api.cli.Main.run(Main.java:36) at
cucumber.api.cli.Main.main(Main.java:18)
There are a couple of points you need to take care as follows :
As per Best Practices cerate the directory features which will contain the featurefile(s) strictly through your IDE only (not through other softwares Notepad or Textpad or SubLime3) as per the image below (New -> File) :
Create the featurefile i.e. CO_Self_Service_Home_Page_Personalizations.feature within features directory strictly through your IDE only.
Keep your Project Structure simple by placing the directory containing the featurefile(s) just under Project Workspace. For Featurefiles Cucumber works with directory names. So create the features directory just under your project space Automation (same hierarchy as src). So the location of the Self_Service_Home_Page_Personalizations.feature will be :
C:\Users\test\eclipse-workspace\Automation\features\Self_Service_Home_Page_Personalizations.feature
Again, as in your Class file containing #CucumberOptions you have mentioned glue = {"StepDefinitions" } ensure that the Class file containing #CucumberOptions must be in the similar hierarchy as the figure below :
So your CucumberOptions will be as follows :
#CucumberOptions(features = {"features" }, glue = {"StepDefinitions" })
Execute your Test
Note : Do not move/copy feature file(s)/directory(ies). Delete the unwanted and create a new one through your IDE only.

How to access static resources in jar (that correspond to src/main/resources folder)?

I have a Spark Streaming application built with Maven (as jar) and deployed with the spark-submit script. The application project layout follows the standard directory layout:
myApp
src
main
scala
com.mycompany.package
MyApp.scala
DoSomething.scala
...
resources
aPerlScript.pl
...
test
scala
com.mycompany.package
MyAppTest.scala
...
target
...
pom.xml
In the DoSomething.scala object I have a method (let's call it doSomething()) that tries to execute a Perl script -- aPerlScript.pl (from the resources folder) -- using scala.sys.process.Process and passing two arguments to the script (the first one is the absolute path to a binary file used as input, the second one is the path/name of the produced output file). I call then DoSomething.doSomething().
The issue is that I was not able to access the script, not with absolute paths, relative paths, getClass.getClassLoader.getResource, getClass.getResource, I have specified the resources folder in my pom.xml. None of my attempts succeeded. I don't know how to find the stuff I put in src/main/resources.
I will appreciate any help.
SIDE NOTES:
I use an external Process instead of a Spark pipe because, at this step of my workflow, I must handle binary files as input and output.
I'm using Spark-streaming 1.1.0, Scala 2.10.4 and Java 7. I build the jar with "Maven install" from within Eclipse (Kepler)
When I use the getClass.getClassLoader.getResource "standard" method to access resources I find that the actual classpath is the spark-submit script's one.
There are a few solutions. The simplest is to use Scala's process infrastructure:
import scala.sys.process._
object RunScript {
val arg = "some argument"
val stream = RunScript.getClass.getClassLoader.getResourceAsStream("aPerlScript.pl")
val ret: Int = (s"/usr/bin/perl - $arg" #< stream).!
}
In this case, ret is the return code for the process and any output from the process is directed to stdout.
A second (longer) solution is to copy the file aPerlScript.pl from the jar file to some temporary location and execute it from there. This code snippet should have most of what you need.
object RunScript {
// Set up copy destination from the Java temporary directory. This is /tmp on Linux
val destDir = System.getProperty("java.io.tmpdir") + "/"
// Get a stream to the script in the resources dir
val source = Channels.newChannel(RunScript.getClass.getClassLoader.getResourceAsStream("aPerlScript.pl"))
val fileOut = new File(destDir, "aPerlScript.pl")
val dest = new FileOutputStream(fileOut)
// Copy file to temporary directory
dest.getChannel.transferFrom(source, 0, Long.MaxValue)
source.close()
dest.close()
}
// Schedule the file for deletion for when the JVM quits
sys.addShutdownHook {
new File(destDir, "aPerlScript.pl").delete
}
// Now you can execute the script.
This approach allows you to bundle native libraries in JAR files. Copying them out allows the libraries to be loaded at runtime for whatever JNI mischief you have planned.

Can't load a file in Play (always not found)

I can't load a file in Play:
val filePath1 = "/views/layouts/mylayout.scala.html"
val filePath2 = "views/layouts/mylayout.scala.html"
Play.getExistingFile(filePath1)
Play.getExistingFile(filePath2)
Play.resourceAsStream(filePath1)
Play.resourceAsStream(filePath2)
None of these works, they all return None.
You are essentially trying to read a source file at runtime. Which is not something you should usually do. If you want to read a file at runtime then I'd recommend putting it somewhere that will end up in the classpath and then use Play.resourceAsStream to read the file. The files in the conf directory and non-compiled files in the app dir should end up in the classpath.

Create new file in user directory (with InstalledFileLocator on NB Platform)

How to create new file in a user directory on NetBeans Platform application? I used:
System.getProperty("netbeans.user", "user.home") + "/myfile");
But the NB IDE 7.1.1 told me that it is depreceated and I should use InstalledFile Locator instead. Ok, I tried this:
File file = InstalledFileLocator.getDefault().locate("myfile", null, false);
It works fine, if the file already exists. I cannot see any way, how to create new with the InstalledFileLocator. But the javadoc say, this method allows to get folder. So I tried this:
File file = InstalledFileLocator.getDefault().locate("myfile", null, false);
if (file == null) {
file = new File(InstalledFileLocator.getDefault().locate("", null, false), "myfile");
}
Again without success, the method locate now fails that it can't find anything (the "/" is forbidden and does not work too).
So my question is, how to corectly load in my NetBeans Platform application an existing file in the user directory (it is for writing also, so it should not be in the program directory) and if it does not exist, create it?
You could use Places.getUserDirectory().
File file = InstalledFileLocator.getDefault().locate("myfile", null, false);
if (file == null)
{
file = new File(Places.getUserDirectory() + File. separator + "myfile");
}
From the netbeans platform docs InstalledFileLocator should not be used to find resources on the system filesystem. To find data in the system filesystem, use the Filesystems API. Ex:
FileObject fo = FileUtil.getConfigFile(myfile);
if (fo == null) {
fo = FileUtil.getConfigRoot().createData(myFile,ext);
}
Probably the easiest thing you can do is to include a simple empty file (say "here.txt") in your module that will be installed in the user directory automatically. You can see an example of this here (see the section "Lessons learned: bundling files with your NetBeans modules").
Basically you include the file in the "release/modules/ext/here.txt" directory of your module.
When the module is installed the platform will install the 'here.txt' file included in your module in the user directory automatically for you, so you don't have to worry about this.
Once your module is installed an running you want to locate the file like this:
File hereTXT = InstalledFileLocator.getDefault()
.locate("modules/ext/here.txt",
"a.b.c",
false);
(Where "a.b.c" is your module identifier.)
And then from that 'hereTXT' file you can get the directory with 'hereTXT.getParent()', and you're all set.