Im developing small admin panel with react-hook-forms form. In creates form I add to my subbmiting object array with files and img file. It is look like {arrFiles:[File, File, File], img:[File]}. For sending files to server I use base64. So its OK. But Im working with edit form right now. And I need to take default files values from server: file name, path in server and file date. How I can take this information? I thought I have to forming on server object for the each files like this {fileName:"name", path:"url/name.jpg", body:"base64...***"}. What do you think about that practies. Is this ok?
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can someone help me with my issue? I just started learning Express.
I have an images folder structure: enter image description here.
Now I store and receive them (at front) like this enter image description here.
I would like to store them at (express server) and make a get request at 'front' : get('https:// ...... /gallery')
get a list of folders in the array format [beruta16, borov , ..... etc], then iterate the array and make a new request for each element.
I think I can to make a list with folder names in JSON format like this app.use(URL,(req,res)=>res.json(array)) and also create a list of items(1,2,3 ... jpeg) for another request and send its. But I don't know how (handled create that madness).
Another way : fs.readdir(), but I am not sure if this is the right way. And for adding a new file I must do that again.
Mb's best way for this that store all URLlinks in MongoDB, but again make handle DB
Sorry, I can't find similar information for this
I am trying to export an excel sheet/file. Inside the file are product data and I am trying to insert the products' image along with other product data for each product. However everytime I run my export code I get an error which is something like, File http://path_to_file/image.jpg not found!. I am using CakePHP by the way. My code for exporting the excel sheet goes like this:
$objDrawing = new PHPExcel_Worksheet_Drawing();
$objDrawing->setName('Water_Level');
$objDrawing->setDescription('Water_Level');
$objDrawing->setPath(''.PATH_TO_IMAGES.'image.jpg');
$objDrawing->setHeight(20);
$objDrawing->setCoordinates('A1');
$objDrawing->setWorksheet($excel->getActiveSheet());
Is there something wrong with my code? Thanks.
I'm pretty sure the argument for PHPExcel_Worksheet_Drawing::setPath() is ment to be a local filesystem path, not a URL.
I want to save image to media library in windows phone 7. I`m using this example http://msdn.microsoft.com/en-us/library/ff769549(v=VS.92).aspx . It works fine, the only problem that i have is that after image modification i call save procedure with the same file name, exactly like in example
MediaLibrary library = new MediaLibrary();
Picture pic = library.SavePicture("SavedPicture.jpg", myFileStream);
myFileStream.Close();
but modification is saved to another file, even thought i use the same file name when i call SavePicture (and i want to override the image file). What am i doing wrong?
Reading between the lines a little you are seeing a new picture appear in the phone saved pictures collection where you were expecting an existing one to be replaced?
You should note that the code you have referenced creates duplicate pictures. One is stored in the phones saved pictures collection and another is saved in isolated storage for the application.
Its not possible for an application to mutate an existing picture in the saved pictures collection even if that application is the original creator of the picture. When saved, a new picture is created in the saved pictures collection.
On the other hand the existing content of the file in the isolated storage is replaced with the new content.
You can't.
It's only possible to read and add images in/to the MediaLibrary.
It is not possible to edit or delete images.
This is by design.
What function in Drupal gets file attachment path?
Edit: the attachments provided by Upload module in system section. But since you mentioned there may be another way around it. Maybe I could achieve may goals using FileField module so if you could tell me how to get a direct link of a file uploaded by FileField module that may also be very helpful.
Edit 2:
Ok I will start from my main and final objective:
It is to get an image linking to a file which visibility I could be managed using CCK module.
I am doing this step by step. So now I have written a code which generates the needed image and I have added that image to one of content types teaser. I found a way to warp my image in a tag and I had dug up simple attachments url. So my next step is to advance from simple upload attachment to the one added by FileFields and from odd looking HTML pace in all PHP document into beautifully managed CCK field.
How to fetch file path of a file uploaded FileFields?
Some plain and at the same time rich tutorials for making custom fields for CCK module.
Assuming you know how to get the $node object for a node containing a file attachment, this is super easy:
Upload (core)
$file = upload_load($node);
echo $file[1]->filepath;
Where 1 is the index of the file. Upload can let you upload more than one file, so file 2 would have an index of 2 and so on. If there's only one file, it'll always be 1.
More info: upload_load() API reference.
FileField
echo $node->field_fieldname[0]['filepath'];
Where field_filename is the short name you've used for the field, and 0 is the index of the field. CCK fields let you have multiple values in one field, so 0 would be the first, 1 would be the second, etc. If you only have one value, it'll always be 0.
Note: if you want to get the rendered output of the FileField using its formatters, you can just use $field_fieldname in your node template, or if you want to store the rendered output, you can use:
echo content_format('field_fieldname', $node->field_fieldname[0]);
More info: content_format() API reference.
Upload is a sad little module, and has been completely replaced with a core implementation of FileField in Drupal 7. If you have the option to use FileField, you should.
Don't direct links of file attachments appear below the uploaded file for the core Upload module?
Since you are trying to use images, you should use http://drupal.org/project/imagefield CCK module in order to add images to specified content type. After that using the Display fields tab in Content type configuration you can specify how image would be presented (as a link, image, image with link to node...) both in teaser and body view.
I want to receive multi file post from image uploader.(i use this)
Most examples show how to receive one image from post.
I tried many ways but never got the results.
For example
self.request.POST['Filename']
gives only first filename.
What to do when there are multiple files/images in post?
The reason for this is to resize before upload images, that are too big for google app engine
to upload.
EDIT:
self.request.POST.multi.__dict__
shows
{'_items':
[('Filename', 'camila1.jpg'),
('Filedata[]', FieldStorage('Filedata[]', 'camila1.jpg')),
('Upload', 'Submit Query\r\n--negpwjpcenudkacqrxpleuuubfqqftwm----negpwjpcenudkacqrxpleuuubfqqftwm\r\nContent-Disposition: form-data; name="Filename"\r\n\r\nbornToBeWild1.jpg'),
('Filedata[]', FieldStorage('Filedata[]', 'bornToBeWild1.jpg')),
('Upload', 'Submit Query')]}
Your flash uploader is designed to work with PHP and sends multiple Filedata[] fields (php interprets this as an array for easy access)
So you need to iterate and get them all:
def post(self):
for file_data in self.request.POST.getall('Filedata[]'):
logging.info(file_data.filename)
data should be file_data.value
Are you using the Django libraries available to you? If so, check this out.
Call self.request.POST.getall('Filename') to get a list of FieldStorage objects; each one contains one file. You can access the file data with .value, the name with .name, and the mimetype with .type.
I have no idea how that multi uploader works, I have made one in the past however and I just added a number on the end of input field name, then hide it. Then add a new file input field to add another file. The reason for this is that they don't let you play around with input file fields to much because you could make it upload files they didn't want you uploading.
Using my conventions, in your example the 2 files in your example would be "Filename0" and "Filename1". You could also use firebug to see what it renaming the input file fields to.
Edit: I had a look, and it's using flash. So i have no idea how it works.