I'm trying to create link for the file upload. I'm calling file upload function from .cfc file. The file should be uploaded in a different folder. Here is what I get if after this code is executed in component.cfc page:
<cfset thisPath = ExpandPath( "./" ) />
C:\\wwwroot\\myapp\\components\\
I need to go level back in myapp and open bug folder. Then in a bug folder I need to direct the path either to folder1/ documents or folder2/documents. That will depend on the form field, here is example:
<cfset folder = trim(form.type) EQ 1 "folder1" : "folder2">
The path should either point to:
C:\\wwwroot\\myapp\\bug\\folder1\\documents\\
or
C:\\wwwroot\\myapp\\bug\\folder2\\documents\\
I'm looking for a solution that will work even if I roll this code to a different server with the same directory structure. Is there a good way to achieve this in ColdFusion?
Related
I have images and I want to load them to my appplication.
my_app/lib/... - here my source files
my_app/assets/images/... - here my images
I need to get list of files in assets/images/ and after that show some of them (according to other logic)
I'm trying this
final dir = Directory("assets/images/");
print(dir.existsSync()); // <---- it also print: false
var files = dir.listSync().toList();
files.forEach((e) => list.add(MyImageItem(e.path)));
The problem is: I recieve exception
FileSystemException: Directory listing failed, path = 'assets/images/'
(OS Error: No such file or directory, errno = 2)
I've tried different ways: assets/images/, images/, images and so on
My pubspec.yaml
flutter:
assets:
- assets/images/
- assets/
When I create Image directly all is fine
new Image(image: AssetImage("assets/images/cat.png"))
I knew that previously (month ago) each resource has to be declared in pubspec.yaml directly, but now assets/images/ is ok.
I can load file by direct path. Why I can't access a directory? How to get list of files in directory to get them from my code?
When you add files to your assets, it means that you already know their paths.
Store all the images paths in a list and access them whenever you need.
For example if you have 3 images in
your_app/assets/images/image1.jpg
your_app/assets/images/image2.jpg
your_app/assets/images/image3.jpg
create a list and store all of them like:
List<String> imagePaths = ['assets/images/image1.jpg', 'assets/images/image2.jpg', 'assets/images/image3.jpg'];
and for example if you want to access your first image, use
AssetImage(imagePaths[0])
I keep a json file inside assets which records the file tree of assets folder.
When file list is needed, I just read from the json file.
Such json file can be easily generated by code.
However, this does not solve your problem directly when u have 5-10k images.
Only one json file might be too large to read.
Images should be grouped and recorded in separated json files.
The following code runs without a hitch:
On the other hand, I get an access-denied error with this:
The destination is in my personal folder and I have full control. The directory is not read-only. Anyway, in either of those cases, the first code sample should not run either! I appreciate the help ...
In the second sample, you have two problems:
There are back slashes instead of forward slashes, so some of them may get interpreted as escape sequences.
You completely ignore the first parameter of write and specify what I assume is a folder as destination. You can't open a file stream on a folder, no wonder you get access denied.
This should work:
let write filename (ms:MemoryStream) =
let path = System.IO.Path.Combine( "C:/Users/<whatever>/signal_processor", filename )
use fs = new FileStream( path, FileMode.Create )
ms.WriteTo(fs)
We are trying to upload a document from our local machine to server. We know that .NET browse control can be used for this but our requirement is to just have a button and call that as "Save Document". On click we know the path and also the document name, all we need to do is search the document in local machine (inside temp folder) and if the document is available then pick it and save it to our APP data folder on the server.
it's easy to achieve this using .NET borwse control but not sure how to do that same using normal button?
To find a file in a folder:
FileInfo file = Directory.GetFiles(#"c:\folder")
.FirstOrDefault(f => f.name = whatever);
To copy file:
File.Copy(file.FullName, targetPath, true);
I want to read file from "res" folder on blackberry. The file that i used is a file javascript.
I used this code InputStream in = classs.getResourceAsStream("file.js");. But i get "could not find this path" and I use also
String srcFile = "/res/ressourcesWeb/file.js";
FileConnection srcConn = (FileConnection) Connector.open(srcFile, Connector.READ);
InputStream in = srcConn.openInputStream();
but i got an exception.
Can any one help me to read the file and give me the right path that should I use?
Your res folder has to be inside src folder to be accessed from your code.
src folder is the root folder of your project package. And all folders outside of src folder are invisible for the code at runtime.
Check this post for more details: Blackberry runtime error: FRIDG: could not find img/logo.png
There's file location principle described.
You actually do not need to put your resources under the src folder for them to be accessible from your code.
That is one way to do it, but I don't think it's the best way. Files under the src folder should really be source code, not images, or other resources. For JavaScript resources, it's debatable whether those should be under src or not. Most projects I've seen have used the src folder for only Java source code.
In any case, if you would like to keep your file (or other resources, like images) outside the src folder, you can do so. The BlackBerry plugin for Eclipse actually sets it up like this by default, when you create a new project. There is a res folder at the top level, next to (not under) src.
If you have
src\
src\com\mycompany\myapp\
res\
res\resourcesWeb\
res\resourcesWeb\file.js
Then, you can open the file like this:
String jsPath = "/resourcesWeb/file.js";
InputStream input = getClass().getResourceAsStream(jsPath);
byte [] content = IOUtilities.streamToBytes(input);
String contentAsString = new String(content);
P.S. You also can probably do this:
String jsPath = "/file.js";
InputStream input = getClass().getResourceAsStream(jsPath);
and not specify the path to the resource. Obviously, this will only work if there are no naming conflicts in your resource folders (e.g. you don't have /res/resourcesWeb/file.js and also /res/otherPath/file.js)
While trying to upload the Image in struts2 the image is not uploading and showing following msg Unable to find 'struts.multipart.saveDir' property setting. Defaulting to javax.servlet.context.tempdir
You are getting this message due to build din feature of S2.you need to tell S2 file upload interceptor where you want to place the uploaded file using struts.multipart.saveDir property either in the struts.property file or as a constant in struts.xml file like
Struts property file
struts.multipart.saveDir=location of your choice
struts.xml file
<constant name="struts.multipart.saveDir" value="your location" />
If this property is not set it defaults to javax.servlet.context.tempdir.Sine S2 put the upload file in the temp directory in case you have not specified the saveDir but once the process complete, in order for the clean up process it will remove that temp file, so make sure either you have set the saveDir parameter or you should put logic in your action class to pick file from there and move it to your choice location.
For details about S2 file upload read the official doc
handling-file-uploads
This is manual configuration, Once the file uploaded you have to explicitly handle to save the file(due to security reasons server delete the file immediately.)
import org.apache.commons.io.FileUtils;
String fullFileName = "d:/temp/"+uploadFileName;
File destFile = new File(fullFileName);
FileUtils.copyFile(upload, destFile);
Source : FileUpload
You can configure it in struts.properties file
struts.multipart.saveDir - The directory where the uploaded files will be placed. If this property is not set it defaults to javax.servlet.context.tempdir.
put the uploaded files in /tmp. My application will move them to their final destination
struts.multipart.saveDir=/tmp