am trying to learn dropbox php-api.
i have done upload, but unable to create a new folder(if not exist) in dropbox account
code snippet :
// Create new directory
$create_new_folder = $dropbox->createFolder('new_folder','dropbox');
// Create a temporary file and write some data to it
$tmp = tempnam('/tmp', 'dropbox');
$data = 'This file was uploaded using the Dropbox API!';
file_put_contents($tmp, $data);
// Upload the file with an alternative filename
$put = $dropbox->putFile($tmp, 'abc.txt');
// Unlink the temporary file
unlink($tmp);
// Dump the output
var_dump($put);
error:
Call to undefined method Dropbox\API::createFolder() in C:\wamp\www\BenTheDesigner-Dropbox-b49576c\examples\putFile.php on line 18
just use it
require_once('bootstrap.php');
// Create new directory
$create_new_folder = $dropbox->create('docs','dropbox');
now check in your dropbox :)
This error seems to be indicating that you are trying to call a function named "createFolder" that doesn't exist. Checking the library at:
https://github.com/BenTheDesigner/Dropbox/blob/master/Dropbox/API.php
It looks like the function is actually just named "create":
/**
* Creates a folder
* #param string New folder to create relative to root
* #return object stdClass
*/
public function create($path)
{
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 already have the path of the Excel file, but whenever I call save method, the program writes new content to UTC.xlsx file.
string pathString = "C:\Users\ADMIN-PC\Documents\SUMMER 2018\PRN192\CreateFile\SubFolder\UTC.xlsx";
ef.Save(pathString);
var path = #"C:\Users\ADMIN-PC\Documents\SUMMER 2018\PRN192\CreateFile\SubFolder\UTC.xlsx";
// Load the workbook:
var workbook = ExcelFile.Load(path);
// Manipulate the workbook …
// … and save it over the original
workbook.Save(path);
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.
I have built a Server that you can upload files to and download, using Eclipse, servlet and jsp, it's all very new to me. (more info).
Currently the upload system works with the file's name. I want to programmatically assign each file a random key. And with that key the user can download the file. That means saving the data in a config file or something like : test.txt(file) fdjrke432(filekey). And when the user inputs the filekey the servlet will pass the file for download.
I have tried using a random string generator and renameTo(), for this. But it doesn't work the first time, only when I upload the same file again does it work. And this system is flawed, the user will receive the file "fdjrke432" instead of test.txt, their content is the same but you can see the problem.
Any thoughts, suggestions or solutions for my problem?
Well Sebek, I'm glad you asked!! This is quite an interesting one, there is no MAGIC way to do this. The answer is indeed to rename the file you uploaded. But I suggest adding the random string before the name of the file; like : fdjrke432test.txt.
Try this:
filekey= RenameRandom();
File renamedUploadFile = new File(uploadFolder + File.separator+ filekey+ fileName);
item.write(renamedUploadFile);
//remember to give the user the filekey
with
public String RenameRandom()
{
final int LENGTH = 8;
StringBuffer sb = new StringBuffer();
for (int x = 0; x < LENGTH; x++)
{
sb.append((char)((int)(Math.random()*26)+97));
}
System.out.println(sb.toString());
return sb.toString();
}
To delete or download the file from the server you will need to locate it, the user will input the key, you just need to search the upload folder for a file that begins with that key:
filekey= request.getParameter("filekey");
File f = new File(getServletContext().getRealPath("") + File.separator+"data");
File[] matchingFiles = f.listFiles(new FilenameFilter() {
public boolean accept(File dir, String name) {
return name.startsWith(filekey);
}
});
String newfilename = matchingFiles[0].getName();
// now delete or download newfilename