This particularly relates to files created in Drag and Drop functionality, but applies to any file uploaded.
Is it possible to use a workflow or script to create a naming convention, such as {internalID}+{date}, so that all files uploaded are named automatically to refer to the record they are attached to?
I use this function and I just send as parameter the name of the file
function createfile( content, fileName) {
var fileObj = null,
fileObj = file.create( {
name: fileName+ '.json',
fileType: file.Type.PLAINTEXT,
contents: content,
description: 'TEST FILE .JSON',
encoding: file.Encoding.UTF8,
folder: 57,
isOnline: true
});
return fileObj.save();
}
Related
I found this script that will upload files from google forms response to a folder using the first response (name of applicant) to name the folder.
function onFormSubmit(e) {
const folderId = "###"; // ID of the destination folder
const form = FormApp.getActiveForm();
const formResponses = form.getResponses();
const itemResponses = formResponses[formResponses.length - 1].getItemResponses();
// Prepare the folder.
const destFolder = DriveApp.getFolderById(folderId);
const folderName = itemResponses[0].getResponse();
Logger.log(folderName)
const subFolder = destFolder.getFoldersByName(folderName);
const folder = subFolder.hasNext() ? subFolder.next() : destFolder.createFolder(folderName);
// Move files to the folder.
itemResponses[1].getResponse().forEach(id => DriveApp.getFileById(id).moveTo(folder));
}
I tried using it but it will only upload one file - the one correspnding to the specific question number in the line itemResponses[1].getResponse()
How Do i modify this script so it will upload all of the files in the response regardless of their position without creating multiple folders?
i.e. is it possible to set a range of questions that all the files will upload?
Also, i want to modify the script so that name of the folder will also consist of the response sequence number as a prefix to the name of responder, how is that possible?
I have file which name contains single quote like Test'Quote.html
in order to upload file to server, I'm using HttpClient and MultipartFormDataContent like:
using (var formContent = new MultipartFormDataContent($"{new string('-', 10)}-{DateTime.Now.Ticks}"))
{
foreach (var file in files)
{
var fileContent = new StreamContent(file.Stream);
fileContent.Headers.ContentType = MediaTypeHeaderValue.Parse(file.ContentType);
formContent.Add(fileContent, file.Name, file.Filename);
}
await client.PostAsync(address, formContent)
}
I check formContent and I see:
And I cannot upload file to server because it says that filename is invalid, seems that last filename is taken in consideration instead of first one.
How to disable or avoid encoding such name ?
I am trying to transfer ownership of all my .pdf files to another account with more space. I am testing the code with a single folder in my drive.
function transfer() {
var user = Session.getActiveUser();
var folder = DriveApp.getFolderById('123folder-id456789-VxdZjULVQkPAaJ');
var files = folder.getFilesByType(MimeType.PDF);
while (files.hasNext()) {
var file = files.next();
if (file.getOwner() == user) file.setOwner('example#gmail.com');
}
}
When I run the code, none of the files change ownership.
How about this modification?
Modification points:
In your script, it tries to compare the objects of Session.getActiveUser() and file.getOwner(). I think that this is the reason of your issue.
So how about this modification? Please think of this as just one of several answers.
Modified script:
function transfer() {
var user = Session.getActiveUser().getEmail(); // Modified
var folder = DriveApp.getFolderById('123folder-id456789-VxdZjULVQkPAaJ');
var files = folder.getFilesByType(MimeType.PDF);
while (files.hasNext()) {
var file = files.next();
if (file.getOwner().getEmail() == user) file.setOwner('example#gmail.com'); // Modified
}
}
In this modification, the emails are compared.
References:
getActiveUser()
getOwner()
Class User
If this didn't resolve your issue, I apologize.
Currently drive can't transefere the ownership of file that are not build-in, like pdf, zip, etc. So you must download them and reupload from the other account. I wrote a colab to do that without consume my bandwith. It can recursively transfere an entire folder with both build-in file types and other file types.
I'm attempting to save a screenshot using a generic method in protractor. Two features, it creates the folder if it does not exist and it saves the file (with certain conditions).
export function WriteScreenShot(data: string, filename: string) {
let datetime = moment().format('YYYYMMDD-hhmmss');
filename = `../../../test-reports/${filename}.${datetime}.png`;
let path =filename.substring(0, filename.lastIndexOf('/'));
if (!fs.existsSync(path)) {
fs.mkdirSync(path);
}
let stream = fs.createWriteStream(filename);
stream.write(new Buffer(data, 'base64'));
stream.end();
}
This can be used by calling browser.takeScreenshot().then(png => WriteScreenShot(png, 'login/login-page')); Using this example call, a file will be created, I assumed, in the path relative where my WriteScreenShot method's file resides. But that does not appear to be the case.
For example, when I run my spec test in the spec's folder, the image gets saved in the correct place. But if I run it at the project root, an error is capture. Obviously, this has to do with my relative path reference. How do I capture the project's root directory and build from that so that I can run the test from any directory?
This is a classical directory access error. Let me just explain what is happening to your code -
let path =filename.substring(0, filename.lastIndexOf('/'));
The above line outputs to ../../../test-reports
fs.existsSync checks whether thispath exists or not -
case 1 :(postive flow) Your spec folder is in the same current working directory in which you are trying to create reports folder. When you run your test, the path exists, it generates the test-reports directory & screenshots and your code works fine.
case 2:(negative flow) When you try to run it from the root directory which is the current working directory now, fs.existsSync tries to check the path & the reports folder inside it. If it doesn't exist , fs.mkdirSync tries to create your directories but it would fail as it cannot create multiple directories.
You should be using native path module of nodejs to extract the path instead of using file substring and the mkdirp external module for creating multiple directories.
import * as path from 'path';
let {mkdirp} = require('mkdirp'); // npm i -D mkdirp
export function WriteScreenShot(data: string, filename: string) {
let datetime = moment().format('YYYYMMDD-hhmmss');
filename = `../../../test-reports/${filename}.${datetime}.png`;
let filePath = path.dirname(filename); // output: '../../..' (relative path)
// or
let filePath = path.resolve(__dirname); // output: 'your_root_dir_path' (absolute path)
// or
let filePath = path.resolve('.'); // output: 'your_root_dir_path' (absolute path)
if (!fs.existsSync(filePath )) {
mkdirp.sync(filePath); // creates multiple folders if they don't exist
}
let stream = fs.createWriteStream(filename);
stream.write(new Buffer(data, 'base64'));
stream.end();
}
If you are curious to know the difference btw mkdir & mkdir-p please read this SO thread.
I have an application that uploads a file. I need to pass this file into another program but for that I need the file name only. Is there any simple code for that, using only Java or a servlet procedure?
while (files.hasMoreElements())
{
name = (String)files.nextElement();
type = multipartRequest.getContentType(name);
filename = multipartRequest.getFilesystemName(name);
originalFilename = multipartRequest.getOriginalFileName(name);
//extract the file extension - this can be use to reject a
//undesired file extension
extension1 = filename.substring
(filename.length() - 4, filename.length());
extension2 = originalFilename.substring
(originalFilename.length() - 4, originalFilename.length());
//return a File object for the specified uploaded file
File currentFile = multipartRequest.getFile(name);
//InputStream inputStream = new BufferedInputStream
(new FileInputStream(currentFile));
if(currentFile == null) {
out.println("There is no file selected!");
return;
}
There's a method in apache commons-io to get the file's extension. There's also guava Files class, with its getFileExtension method.