file-sharing

how to programmatically download a public Google Drive file?

混江龙づ霸主 提交于 2019-12-06 17:37:35
问题 I have a backend server that has to download a file on Google Drive that has been shared to public by the owner. Our clients are emailing us the link and we get the link from the email -- no UI is involved. The program needs to get or download the file and process it offline . The google samples seem to be runnig using the UI where as this is a automated backend process.Also, this application is to running on my server and not on Google App Engine . Do I still need to register my app if all I

Allow iPhone application to sync with iTunes

℡╲_俬逩灬. 提交于 2019-12-06 09:15:45
问题 In the Apps tab of iTunes, there is a 'File Sharing' section with a list of apps that can synchronize their documents with your computer. What do you need to do to tell iTunes to sync your apps Documents folder? 回答1: Add the UIFileSharingEnabled key to your Info.plist. 回答2: For iOS 7/8 and Xcode 6: Go to properties of your Target in Xcode. Click the Info tab. Click the + on the last line under "Custom iOS Properties." Type "UIFileSharingEnabled" and press Enter. It should change to

Dropbox API - Using Dropbox as a server

末鹿安然 提交于 2019-12-05 21:50:12
I was wanting to use a file sharing server to keep certain files up-to-date and constant across multiple instances of my application across multiple computers - like (for example) writing a multiplayer game, which stores all the player's positions in a text file, and uses something like Dropbox to keep the text file constant across all the applications, and each application instance can change the file with that application's player's position, and then the rest of the applications can update accordingly. This is only an example, and is not what I intend to do using this technology. What I

how to programmatically download a public Google Drive file?

女生的网名这么多〃 提交于 2019-12-04 23:11:18
I have a backend server that has to download a file on Google Drive that has been shared to public by the owner. Our clients are emailing us the link and we get the link from the email -- no UI is involved. The program needs to get or download the file and process it offline . The google samples seem to be runnig using the UI where as this is a automated backend process.Also, this application is to running on my server and not on Google App Engine . Do I still need to register my app if all I am doing is acessing files shared to public by the owner. For reference , please note that it is

How to share files between 2 local ios apps without url scheme or external server?

空扰寡人 提交于 2019-12-03 19:42:29
问题 I have 2 local app "A" and "B". I want that the application "A" call the application "B" and "B" send a response to "A" with a file. Url scheme is a good solution to call Application "B" but it's a bad solution to send files to "A". How do I send a file from one application to another without url scheme and without external server ? 回答1: Other than URL option, you have an best option to use iOS keychain. See this article. There are couple of other options mentioned here. Another option is to

File sharing not working as expected

我与影子孤独终老i 提交于 2019-12-03 17:13:10
问题 I have a file sharing issue where my process is trying to read a log file whilst it is currently still open by NLog. In diagnosing the issue, I found something surprising. The following fails: using (var fileStream1 = new FileStream("test.file", FileMode.Append, FileAccess.Write, FileShare.Read)) using (var fileStream2 = new FileStream("test.file", FileMode.Open, FileAccess.Read, FileShare.Read)) { } The second FileStream constructor call fails with: System.IO.IOException was unhandled

File sharing not working as expected

Deadly 提交于 2019-12-03 05:31:54
I have a file sharing issue where my process is trying to read a log file whilst it is currently still open by NLog. In diagnosing the issue, I found something surprising. The following fails: using (var fileStream1 = new FileStream("test.file", FileMode.Append, FileAccess.Write, FileShare.Read)) using (var fileStream2 = new FileStream("test.file", FileMode.Open, FileAccess.Read, FileShare.Read)) { } The second FileStream constructor call fails with: System.IO.IOException was unhandled Message=The process cannot access the file 'c:\...\test.file' because it is being used by another process.

Identify client in peer to peer filesharing using Java and Corba. And build socket socket between 2 Clients

*爱你&永不变心* 提交于 2019-12-02 13:53:53
问题 I am developing a peer to peer file sharing application usig Java and CORBA. I am new to CORBA. So far I have successfully created a client program that registers a file in the SQL Database via CORBA. The table stores the following information : filename Share(Basically, it stores the option of sharing or not sharing the file in the network) HostName PortNumber I am stuck at this point: How do I find the hostname and portnumber of a client when registering the files to SQL?. Because I believe

File transfer over the air with iPad

萝らか妹 提交于 2019-12-02 13:03:27
问题 I have an iPad app that outputs a PDF file that needs to be uploaded to a remote server. What are my options here? Is there a sFTP or FTP implementation for iOS that would help me achieve this? Can it be done using other protocol? Thanks! 回答1: FTP Client Wrapper in Objective-C for iPhone ConnectionKit (might not be iOS compatible) 回答2: How about HTTP? http://www.cocoadev.com/index.pl?HTTPFileUpload EDIT: SCP using libssh2? http://www.libssh2.org/examples/scp_write.html 回答3: This on costs

share a photo in instagram

我是研究僧i 提交于 2019-12-02 07:37:30
问题 i'm developping an app and i'm trying to share a picture in Instagram . i'm using the code below : Intent i = new Intent(Intent.ACTION_SEND); i.setType("image/jpeg"); i.putExtra(Intent.EXTRA_STREAM, Uri.parse("file:///"+savedPhotoPath));//savedPhotoPath is the path of my picture stored somewhere in the sdcard startActivity(Intent.createChooser(i, "Share Image")); the problem is that the instagramm is launched with the home activity,( i want to have the activity of share that picture to be