I want to save my file to S3 and every item will have it own path, such as base/item._id/filename. How does one accomplish this?
It seems the filename is auto generated and the option doesn't work.
Keystone's Types.S3File type has a filename option that you can set to a custom function in order to determine your own file name. Documentation Link
{
type: Types.S3File,
filename: function(item, filename) {
// prefix file name with object id
return item._id + '-' + filename;
}
}
item has all the properties of the current item, so you could set it to item.name if that value exists.
Related
I want to use my own S3 storage and display the image that was just uploaded. How can I get the filename that was uploaded to S3?
For example I can upload a jpg image to UploadCare.
This is the output I can get:
fileInfo.cdnUrl: "https://ucarecdn.com/6314bead-0404-4279-9462-fecc927935c9/"
fileInfo.name: "0 (3).jpg"
But if I check my S3 bucket this is the file name that was actually uploaded to S3: https://localdevauctionsite-us.s3.us-west-2.amazonaws.com/6314bead-0404-4279-9462-fecc927935c9/03.jpg
Here is the javascript I have so far:
var widget = uploadcare.Widget('[role=uploadcare-uploader]');
widget.onChange(group => {
group.files().forEach(file => {
file.done(fileInfo => {
// Try to list the file from aws s3
console.log('CDN url:', fileInfo.cdnUrl);
//https://uploadcare.com/docs/file_uploader_api/files_uploads/
console.log('File name: ', fileInfo.name);
});
});
});
Filenames are sanitized before copying to S3, so the output file name can contain the following characters only:
abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789_
The following function can be used to get a sanitized file name:
function sanitize(filename) {
var extension = '.' + filename.split('.').pop();
var name = filename.substring(0, filename.length - extension.length);
return name.replace(/[^A-Za-z0-9_]+/g, '') + extension;
}
The final S3 URL can be composed of the S3 base URL, a file UUID, which you can obtain from the fileInfo object, and a sanitized name of the uploaded file.
I have a folder with my text files that can read and write and work in eclipse. But, when I export to jar, it fails because the files are not found, meaning they are not exported and I don't know how to make eclipse do that. I'm sure the solution is out there, but I don't know exactly what I'm searching for. Do I make a relative directory and how? Or another source folder? What exactly do I need to do?
This shows I have a folder called conf where my files are stored but it is not there on export.
Scanner in = new Scanner(new FileReader("conf/Admins.txt"));
FileWriter out = new FileWriter("conf/CurrentUser.txt");
int id = 0;
String name = "";
String pass = "";
boolean found = false;
while(in.hasNext()) {
id = in.nextInt();
name = in.next();
pass = in.next();
if(id == userID) {
out.write(id + " " + name + " " + pass + "\n");
found = true;
break;
}
}
All I had to do was once exported, put my conf file in the same place as the exported jar file. I don't know if, theres a better way but this is a win for me.
I have this script to save my spreadsheet in the Google Drive Squads folder..
The name of the saved file is according to the value that is in cell H2 of the Gerais page, sometimes the name repeats and when saving, a new file is created instead of subscribing to the existing one, I would like to add in this code the option that if a file with the same name as this new one already exists, instead of having two files with the same name in the Google Drive Squads folder, the old one will disappear completely and only make the new file available
//Create PDF
SpreadsheetApp.flush();
var theurl = 'https://docs.google.com/a/mydomain.org/spreadsheets/d/' + // Best to place the line break after '+'
'XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX' + // SHEET ID
'/export?format=pdf' +
'&size=0' +
'&portrait=true' +
'&fitw=true' +
'&top_margin=0' +
'&bottom_margin=0' +
'&left_margin=0' +
'&right_margin=0' +
'&sheetnames=false&printtitle=false' +
'&pagenum=false' +
'&gridlines=false' +
'&fzr=FALSE' +
'&gid=' +
'XXXXXXXXXXXX'; //SHEET PAGE ID
var token = ScriptApp.getOAuthToken();
var docurl = UrlFetchApp.fetch(theurl, { headers: { 'Authorization': 'Bearer ' + token } });
var pdfBlob = docurl.getBlob();
// Get filename from sheet "Gerais", cell "H2"
var fileName = spreadsheet.getSheetByName("Gerais").getRange("H2").getValue();
// Create file from blob and name it
// The newFile is placed in the root folder by default
var newFile = DriveApp.createFile(pdfBlob).setName(fileName);
// if folder exists use next
if (DriveApp.getFoldersByName("Squads").hasNext()){
var folder = DriveApp.getFoldersByName("Squads").next();
// if folder does not exist
} else {
var folder = DriveApp.createFolder("Squads");// new folder created in the root folder by default
}
folder.addFile(newFile); // add new file to folder
DriveApp.removeFile(newFile); // remove file from root folder
I tried to create an interaction between IF and ELSE too to make the file name match the same as I did for the folder name but I was unsuccessful in trying
Problem
Overwriting existing File in a Folder by its name (presumably unique).
Solution
How about checking if a file with the name written into fileName variable exists and if so, removing it before adding the newly created one? Modification will look something like this (your script has be authorized with one the scopes required for the API):
//prepare new file and Squads folder;
var existing = folder.getFilesByName(fileName); //returns file iterator;
var hasFile = existing.hasNext(); //check if iterator isn't empty;
if(hasFile) {
var duplicate = existing.next(); //access file;
//delete file;
var durl = 'https://www.googleapis.com/drive/v3/files/'+duplicate.getId();
var dres = UrlFetchApp.fetch(durl,{
method: 'delete',
muteHttpExceptions: true,
headers: {'Authorization': 'Bearer '+token}
});
if(dres.getResponseCode()>=400) {
//handle errors;
}
}
//continue: add file, remove from root, etc;
Notes
The way you define query parameters is the common case of a pyramid of doom. Besides, each parameter is hardcoded, which makes the export inflexible. To avoid this, use objects to configure them (as query is nothing more than a set of key-value pairs):
var params = {
format: 'pdf',
size: 0,
portrait: true,
fitw: true,
top_margin: 0,
bottom_margin: 0,
left_margin: 0,
right_margin: 0,
sheetnames: false,
printtitle: false,
pagenum: false,
gridlines: false,
fzr: 'FALSE',
gid: 'XXXXXXXXXXXX'
};
//append params to the url;
var theurl = 'base?'+Object.keys(params).map(function(key){
return key+'='+params[key];
}).join('&');
Updates
Corrected UrlFetchApp.fetch() call as method should be a parameters object property and not an argument.
Reference
Drive API delete reference;
I am trying to write a DigitalMicrograph script which opens all images containing a specific string in the file name.
I know how I can open an image using OpenImage( filename ) and I have seen in the documentation that a command GetFilesInDirectory() exists, which seems to be what I need. However, I do not understand how I can use this command. Can somebody give me a code snippet demonstrating this, please?
The command GetFilesInDirectory() gives you a TagList of all files / subfolders in a given directory. This is shown in the following example:
String folder
TagGroup FileList
number fFiles = 1
number fFolders = 2
If ( !GetDirectoryDialog( "Select base folder", "", folder ) )
Exit(0)
FileList = GetFilesInDirectory( folder, fFiles + fFolders )
If ( FileList.TagGroupCountTags() > 0 )
FileList.TagGroupOpenBrowserWindow( "Files & Folders", 0 )
This script will show you the resulting TagGroup in a browser window like the one below. Each list entry is itself a TagGroup which contains a single tag "Name". This tag contains the file or folder name. You can use the command to either give you only files, only subfolders, or both.
Once you have the TagGroup of all entries, you process is like any other TagGroup in DigitalMicrograph. For example, you can browse the list to read out the strings and simple print them to the results window like this:
number nTags = FileList.TagGroupCountTags()
for ( number I = 0; I < nTags; i++ )
{
TagGroup entryTG
FileList.TagGroupGetIndexedTagAsTagGroup( i, entryTG )
if ( entryTG.TagGroupIsValid() )
{
string filestr
if ( entryTG.TagGroupGetTagAsString( "Name", filestr ) )
{
Result( "\n File:" + filestr )
}
}
}
I am trying to upload a file via a form and then save in in SQL as a blob.
I already have my form working fine, my database is fully able to take the blob and I have a controller that take the file, saves it in a local directory:
[AcceptVerbs(HttpVerbs.Post)]
public ActionResult FileUpload(int id, HttpPostedFileBase uploadFile)
{
//allowed types
string typesNonFormatted = "text/plain,application/msword,application/pdf,image/jpeg,image/png,image/gif";
string[] types = typesNonFormatted.Split(',');
//
//Starting security check
//checking file size
if (uploadFile.ContentLength == 0 && uploadFile.ContentLength > 10000000)
ViewData["StatusMsg"] = "Could not upload: File too big (max size 10mb) or error while transfering the file.";
//checking file type
else if(types.Contains(uploadFile.ContentType) == false)
ViewData["StatusMsg"] = "Could not upload: Illigal file type!<br/> Allowed types: images, Ms Word documents, PDF, plain text files.";
//Passed all security checks
else
{
string filePath = Path.Combine(HttpContext.Server.MapPath("../Uploads"),
Path.GetFileName(uploadFile.FileName)); //generating path
uploadFile.SaveAs(filePath); //saving file to final destination
ViewData["StatusMsg"] = "Uploaded: " + uploadFile.FileName + " (" + Convert.ToDecimal(uploadFile.ContentLength) / 1000 + " kb)";
//saving file to database
//
//MISSING
}
return View("FileUpload", null);
}
Now all I am missing is putting the file in the database. I could not find anything on the subject... I found some way to do it in a regular website but nothing in MVC2.
Any kind of help would be welcome!
Thank you.
This could help: http://byatool.com/mvc/asp-net-mvc-upload-image-to-database-and-show-image-dynamically-using-a-view/
Since you have HttpPostedFileBase in your controllers method, all you need to do is:
int length = uploadFile.ContentLength;
byte[] tempImage = new byte[length];
myDBObject.ContentType = uploadFile.ContentType;
uploadFile.InputStream.Read(tempImage, 0, length);
myDBObject.ActualImage = tempImage ;
HttpPostedFileBase has a InputStream property
Hope this helps.
Alright thanks to kheit, I finaly got it working. Here's the final solution, it might help someone out there.
This script method takes all the file from a directory and upload them to the database:
//upload all file from a directory to the database as blob
public void UploadFilesToDB(long UniqueId)
{
//directory path
string fileUnformatedPath = "../Uploads/" + UniqueId; //setting final path with unique id
//getting all files in directory ( if any)
string[] FileList = System.IO.Directory.GetFiles(HttpContext.Server.MapPath(fileUnformatedPath));
//for each file in direcotry
foreach (var file in FileList)
{
//extracting file from directory
System.IO.FileStream CurFile = System.IO.File.Open(file, System.IO.FileMode.Open);
long fileLenght = CurFile.Length;
//converting file to a byte array (byte[])
byte[] tempFile = new byte[fileLenght];
CurFile.Read(tempFile, 0, Convert.ToInt32(fileLenght));
//creating new attachment
IW_Attachment CurAttachment = new IW_Attachment();
CurAttachment.attachment_blob = tempFile; //setting actual file
string[] filedirlist = CurFile.Name.Split('\\');//setting file name
CurAttachment.attachment_name = filedirlist.ElementAt(filedirlist.Count() - 1);//setting file name
//uploadind attachment to database
SubmissionRepository.CreateAttachment(CurAttachment);
//deleting current file fromd directory
CurFile.Flush();
System.IO.File.Delete(file);
CurFile.Close();
}
//deleting directory , it should be empty by now
System.IO.Directory.Delete(HttpContext.Server.MapPath(fileUnformatedPath));
}
(By the way IW_Attachment is the name of one of my database table)