I've created a new file to write some information:
import org.apache.jmeter.services.FileServer;
out = new FileOutputStream("myfilePathName", true);
String outAggr="";
out << outAggr;
out.close();
I need to overwrite it or clear it before to add new information (out << outAggr).
I don't want to create another file each time.
I've already tried to:
out < outAggr;
out < "";
but I got an exception
Can you help me with this?
If you want to overwrite any existing contents, you can just do:
new File("myfilePathName").text = ''
To override file you need to call newWriter, for example:
String outAggr="";
writer = new File(myfilePathName).newWriter("UTF-8", true)
writer.write(outAggr)
writer.close()
newWriter() methods have been added. It's now possible to specify the encoding used to write files, and optionnaly to specify wether we're in append mode.
Related
I need to read a text file with readLines() and I've already found this question, but the code in the answers always uses some variation of javaClass; it seems to work only inside a class, while I'm using just a simple Kotlin file with no declared classes. Writing it like this is correct syntax-wise but it looks really ugly and it always returns null, so it must be wrong:
val lines = object {}.javaClass.getResource("file.txt")?.toURI()?.toPath()?.readLines()
Of course I could just specify the raw path like this, but I wonder if there's a better way:
val lines = File("src/main/resources/file.txt").readLines()
Thanks to this answer for providing the correct way to read the file. Currently, reading files from resources without using javaClass or similar constructs doesn't seem to be possible.
// use this if you're inside a class
val lines = this::class.java.getResourceAsStream("file.txt")?.bufferedReader()?.readLines()
// use this otherwise
val lines = object {}.javaClass.getResourceAsStream("file.txt")?.bufferedReader()?.readLines()
According to other similar questions I've found, the second way might also work within a lambda but I haven't tested it. Notice the need for the ?. operator and the lines?.let {} syntax needed from this point onward, because getResourceAsStream() returns null if no resource is found with the given name.
Kotlin doesn't have its own means of getting a resource, so you have to use Java's method Class.getResource. You should not assume that the resource is a file (i.e. don't use toPath) as it could well be an entry in a jar, and not a file on the file system. To read a resource, it is easier to get the resource as an InputStream and then read lines from it:
val lines = this::class.java.getResourceAsStream("file.txt").bufferedReader().readLines()
I'm not sure if my response attempts to answer your exact question, but perhaps you could do something like this:
I'm guessing in the final use case, the file names would be dynamic - Not statically declared. In which case, if you have access to or know the path to the folder, you could do something like this:
// Create an extension function on the String class to retrieve a list of
// files available within a folder. Though I have not added a check here
// to validate this, a condition can be added to assert if the extension
// called is executed on a folder or not
fun String.getFilesInFolder(): Array<out File>? = with(File(this)) { return listFiles() }
// Call the extension function on the String folder path wherever required
fun retrieveFiles(): Array<out File>? = [PATH TO FOLDER].getFilesInFolder()
Once you have a reference to the List<out File> object, you could do something like this:
// Create an extension function to read
fun File.retrieveContent() = readLines()
// You can can further expand this use case to conditionally return
// readLines() or entire file data using a buffered reader or convert file
// content to a Data class through GSON/whatever.
// You can use Generic Constraints
// Refer this article for possibilities
// https://kotlinlang.org/docs/generics.html#generic-constraints
// Then simply call this extension function after retrieving files in the folder.
listOfFiles?.forEach { singleFile -> println(singleFile.retrieveContent()) }
In order to have the same url that work for both Jar or in local, the url (or path) needs to be a relative path from the repository root.
..meaning, the location of your file or folder from your src folder.
could be "/main/resources/your-folder/" or "/client/notes/somefile.md"
The url must be a relative path from the repository root.
it must be "src/main/resources/your-folder/" or "src/client/notes/somefile.md"
Now you get the drill, and luckily for Intellij Idea users, you can get the correct path with a right-click on the folder or file -> copy Path/Reference.. -> Path From Repository Root (this is it)
Last, paste it and do your thing.
def file= new File(path + 'fileName.csv');
List.unique().each { element ->
file<< element << newLine
}
Now the script is appending the file
But i want to clear all the data and over write the fileName.csv if it already exists
If the file doesn't exists, create the file and write data to file.
The easiest is to delete the file prior to writing to it via i.e. Files.deleteIfExists() function Add the next line to your script which will remove the file if it's present already:
java.nio.file.Files.deleteIfExists(file.toPath())
Full code just in case:
def file = new File(path + 'fileName.csv');
java.nio.file.Files.deleteIfExists(new File().toPath())
List.unique().each { element ->
file << element << newLine
}
See The Groovy Templates Cheat Sheet for JMeter for more hints
The following code works fine when returning a file on a mac since it automatically appends the
file extension to the name of the file.
On windows however i have to type in the extension of the file as part of the file name in order for it to return with that extension....even though the extension is selected in the 'save type as' pulldown menu.
is there a way to automatically append the extension to the name when returning a file from the filechooser on windows?
FileChooser.ExtensionFilter extFilter = new FileChooser.ExtensionFilter(fileExtension.toUpperCase()+" files(*."+fileExtension+")", "*."+fileExtension);
fileChooser.getExtensionFilters().add(extFilter);
//Show save file dialog
final File file = fileChooser.showSaveDialog(MyStage.this);
I ran into the same issue. My solution was to make a new File, and append the file extension as a string in the File constructor.
If you want users to be able to select and overwrite an existing file, make sure and add a check to make sure the initial save file does not contain the particular extension already before appending or else you will get something like "test.xls.xls".
FileChooser fc = new FileChooser();
FileChooser.ExtensionFilter extFilter = new FileChooser.ExtensionFilter("XLS File (*.xls)", "*.xls");
fc.getExtensionFilters().add(extFilter);
File save = fc.showSaveDialog(stage);
save = new File(save.getAbsolutePath()+".xls");
FileOutputStream fileOut = new FileOutputStream(save);
I need to read a file from the file system and load the entire contents into a string in a groovy controller, what's the easiest way to do that?
String fileContents = new File('/path/to/file').text
If you need to specify the character encoding, use the following instead:
String fileContents = new File('/path/to/file').getText('UTF-8')
The shortest way is indeed just
String fileContents = new File('/path/to/file').text
but in this case you have no control on how the bytes in the file are interpreted as characters. AFAIK groovy tries to guess the encoding here by looking at the file content.
If you want a specific character encoding you can specify a charset name with
String fileContents = new File('/path/to/file').getText('UTF-8')
See API docs on File.getText(String) for further reference.
A slight variation...
new File('/path/to/file').eachLine { line ->
println line
}
In my case new File() doesn't work, it causes a FileNotFoundException when run in a Jenkins pipeline job. The following code solved this, and is even easier in my opinion:
def fileContents = readFile "path/to/file"
I still don't understand this difference completely, but maybe it'll help anyone else with the same trouble. Possibly the exception was caused because new File() creates a file on the system which executes the groovy code, which was a different system than the one that contains the file I wanted to read.
the easiest way would be
new File(filename).getText()
which means you could just do:
new File(filename).text
Here you can Find some other way to do the same.
Read file.
File file1 = new File("C:\Build\myfolder\myTestfile.txt");
def String yourData = file1.readLines();
Read Full file.
File file1 = new File("C:\Build\myfolder\myfile.txt");
def String yourData= file1.getText();
Read file Line Bye Line.
File file1 = new File("C:\Build\myfolder\myTestfile.txt");
for (def i=0;i<=30;i++) // specify how many line need to read eg.. 30
{
log.info file1.readLines().get(i)
}
Create a new file.
new File("C:\Temp\FileName.txt").createNewFile();
I'm currently trying to attach image files to a model directly from a zip file (i.e. without first saving them on a disk). It seems like there should be a clearer way of converting a ZipEntry to a Tempfile or File that can be stored in memory to be passed to another method or object that knows what to do with it.
Here's my code:
def extract (file = nil)
Zip::ZipFile.open(file) { |zip_file|
zip_file.each { |image|
photo = self.photos.build
# photo.image = image # this doesn't work
# photo.image = File.open image # also doesn't work
# photo.image = File.new image.filename
photo.save
}
}
end
But the problem is that photo.image is an attachment (via paperclip) to the model, and assigning something as an attachment requires that something to be a File object. However, I cannot for the life of me figure out how to convert a ZipEntry to a File. The only way I've seen of opening or creating a File is to use a string to its path - meaning I have to extract the file to a location. Really, that just seems silly. Why can't I just extract the ZipEntry file to the output stream and convert it to a File there?
So the ultimate question: Can I extract a ZipEntry from a Zip file and turn it directly into a File object (or attach it directly as a Paperclip object)? Or am I stuck actually storing it on the hard drive before I can attach it, even though that version will be deleted in the end?
UPDATE
Thanks to blueberry fields, I think I'm a little closer to my solution. Here's the line of code that I added, and it gives me the Tempfile/File that I need:
photo.image = zip_file.get_output_stream image
However, my Photo object won't accept the file that's getting passed, since it's not an image/jpeg. In fact, checking the content_type of the file shows application/x-empty. I think this may be because getting the output stream seems to append a timestamp to the end of the file, so that it ends up looking like imagename.jpg20110203-20203-hukq0n. Edit: Also, the tempfile that it creates doesn't contain any data and is of size 0. So it's looking like this might not be the answer.
So, next question: does anyone know how to get this to give me an image/jpeg file?
UPDATE:
I've been playing around with this some more. It seems output stream is not the way to go, but rather an input stream (which is which has always kind of confused me). Using get_input_stream on the ZipEntry, I get the binary data in the file. I think now I just need to figure out how to get this into a Paperclip attachment (as a File object). I've tried pushing the ZipInputStream directly to the attachment, but of course, that doesn't work. I really find it hard to believe that no one has tried to cast an extracted ZipEntry as a File. Is there some reason that this would be considered bad programming practice? It seems to me like skipping the disk write for a temp file would be perfectly acceptable and supported in something like Zip archive management.
Anyway, the question still stands:
Is there a way of converting an Input Stream to a File object (or Tempfile)? Preferably without having to write to a disk.
Try this
Zip::ZipFile.open(params[:avatar].path) do |zipfile|
zipfile.each do |entry|
filename = entry.name
basename = File.basename(filename)
tempfile = Tempfile.new(basename)
tempfile.binmode
tempfile.write entry.get_input_stream.read
user = User.new
user.avatar = {
:tempfile => tempfile,
:filename => filename
}
user.save
end
end
Check out the get_input_stream and get_output_stream messages on ZipFile.