I want to get the file of which I am deleting on Filepond. However when I use onremovefile={(file) => this.handleRemove(file)}, the file returns to null. What am I doing wrong?
Found the solution! The first parameter is a possible error response and the second one is the file item. Thus it's actually onremovefile={(errRes, file) => this.handleRemove(errRes, file)}
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.
Working in React Native. I'm trying to declare an array and then push things to said array, but I'm getting the error TypeError: Attempted to assign to readonly property
CONTEXT:
The app prints via a thermal printer.
The print method receives an array of commands
Example:
print([{appendText: "blah"}, {
appendCutPaper: StarPRNT.CutPaperAction.PartialCutWithFeed,
}]
The print method is asynchronous and if you attempt to call the method again before the last call has finished, it errors.
Because of #2, we created a queue system that accepts a job (array of commands) and then works through the jobs synchronously.
In a React component, I'm attempting to create a job by declaring an empty array named printJob
and then pushing various commands to it. In this case, we take a snapshot of a View and then push the commands returned by the printImage method to the printJob array.
onClick={() => {
const printJob = []
viewShot.current
.capture()
.then((uri) => {
printJob.push(...printImage(uri))
})
.catch((err) => alert(err))
newPrintJob(printJob)
}
printImage returns the array of commands to print an image and cut the paper:
const CUT_PAPER = {
appendCutPaper: StarPRNT.CutPaperAction.PartialCutWithFeed,
}
export function printImage(uri) {
return [{ appendBitmap: uri }, CUT_PAPER]
}
So the goal is to generate the array of commands and pass that to the queue as a job. Now, I could just do newPrintJob(printImage(uri)) in the above case, which works completely fine. However, there is a particular setting the user can configure where it will need to print multiple images, one per ticket (in other words, multiple printImages). I want to consider all of that one job, hence the need to create the printJob array.
THE PROBLEM:
I'm getting an error TypeError: Attempted to assign to readonly property which seems to be triggered by printJob.push(...printImage(uri)). If I comment that line out, the error doesn't get thrown.
I don't understand why this would happen because you can call push on an array, even if it's declared as a constant. I also tried declaring it with var and let and still received the same error.
I hope I've provided enough context here. LMK if I need to add more.
Additional info:
"react": "16.13.1"
"react-native": "~0.63.3"
Turns out the issue was not pushing to the array. The issue was was trying to add the job to the queue:
newPrintJob(printJob)
...outside of the async's callback. Solution was to move the newPrintJob line into the .then block.
I am fetching an array of objects from an RX/JS call from an http backend. It returns an object which I am then trying to work with. I am making changes to this object using a for loop (in this example I am trying the .forEach because I have tried a number of different things and none of them seem to work.
When I run the code, I get a very weird problem. If I return the values of the properties, I get the new values (i.e. correctionQueued returns as true, etc.) but in the very next line, when I return the object, those same values are the same as the original (correctionQueued === false, etc.) HOWEVER, correctionStatus (which does not exist on the original object from http) sets just fine.
I don't understand how
array[index].correctionQueued can return true, but
array[index] returns an object with correctionQueued as false.
After the loop, the original array (checklistCopy) is identical to the object before the forEach loop, except the new property (correctionStatus) is now set, but all properties that I changed that were part of the original object remain as they were.
I have tried using a for of, for in, and .forEach. I have used the index to alter the original array, always the same result. Preexisting properties do not change, new properties are added. I have even tried working on a copy of the object in case there is something special about the object returned from rxjs, but to no avail.
checklistCopy.forEach((checklistItem, index, array) => {
if (checklistItem.crCode.isirName === correctionSetItem) {
array[index].correctionQueued = true;
array[index].correctionValue = mostRecentCorrection.correctionValue;
array[index].correctionStatus = mostRecentCorrection.status;
console.log(array[index].correctionQueued, array[index].correctionValue, array[index].correctionStatus);
console.log(array[index]);
}
}
);
I don't get an error, but I get..
Original object is:
correctionQueued: false;
correctionValue: JAAMES;
--
console.log(array[index].correctionQueued, array[index].correctionValue, array[index].correctionStatus);
true JAMES SENT
but when I print the whole object:
console.log(array[index]);
correctionQueued: false;
correctionValue: JAAMES;
correctionStatus: "SENT'; <-- This is set correctly but does not exist on original object.
console.log(array[index]) (at least in Chrome) just adds the object reference to the console. The values do not resolve until you expand it, so your console log statement is not actually capturing the values at that moment in time.
Change your console statement to: console.log(JSON.stringify(array[index])) and you should discover that the values are correct at the time the log statement runs.
The behavior you are seeing suggests that something is coming along later and changing the object properties back to the original value. Unless you show a more complete example, we can't help you find the culprit. But hopefully this answers the question about why your logs show what they show.
Your output doesn't make sense to me either but cleaning up your code may help you. Try this:
checklistCopy.forEach(checklistItem => {
checklistItem.correctionQueued = checklistItem.crCode.isirName === correctionSetItem;
if (checklistItem.correctionQueued) {
checklistItem.correctionValue = mostRecentCorrection.correctionValue;
checklistItem.correctionStatus = mostRecentCorrection.status;
console.log('checklistItem', checklistItem)
}
}
);
I am building a code that lets the user open some files.
reference = warndlg('Choose the files for analysis.');
uiwait(reference);
filenames2 = uigetfile('./*.txt','MultiSelect', 'on');
if ~iscell(filenames2)
filenames2 = {filenames2}; % force it to be a cell array of strings
end
numberOfFiles = numel(filenames2);
data = importdata(filenames2{i},delimiterIn,headerlinesIn);
When I run the code, the prompts show up, I press OK, and then nothing happens. The code just stops, telling me :
Error using importdata (line 137)
Unable to open file.
Error in FreqVSChampB_no_spec (line 119)
data=importdata(filenames2{1},delimiterIn,headerlinesIn);
I just don't have the opportunity to select a file. The cellarray stays empty as showed in the following image.
MATLAB can't find the file that you have selected. Your variable filenames2 contains only the name of the file, not its full path. If you don't provide the full path to importdata, it will search for whatever file name you provide on the MATLAB path, and if it can't find it it will error as you see.
Try something like this - I'm just doing it with single selection for ease of description, but you can do something similar with multiple selection.
[fileName, pathName] = uigetfile('*.txt');
fullNameWithPath = fullfile(pathName, fileName);
importdata(fullNameWithPath)
fullfile is useful, as it inserts the correct character between pathName and fileName (\ on Windows, / on Unix).
You can try to add
pause(0.1);
just after uiwait(reference);
For me it works. In fact I've noticed the active windows changes when we use uiwait and uigetfile.
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.