I am trying to do file upload using quartz job scheduler. but, it is not working.
Is it possible to upload file to the server using job or windows sevice.
I have googled but not understand much. please provide some link for reference.
It is possible to upload file using windows service i guess. We build a similar kind of windows service for inserting to database recursively.[1]: https://www.c-sharpcorner.com/article/create-windows-services-in-c-sharp/
protected override void OnStart(string[] args)
{
WriteToFile("Service is started at " + DateTime.Now);
Fileupload();
timer.Elapsed += new ElapsedEventHandler(OnElapsedTime);
timer.Interval = 1000*60*60*10; //number in milisecinds
//timer.Interval = 1000 * 60;
timer.Enabled = true;
}
private void OnElapsedTime(object source, ElapsedEventArgs e)
{
Fileupload();
WriteToFile("Service is recall at " + DateTime.Now);
}
//you can use the above 2 functions in Serivce Cs file and you define your Custom Fileupload Function for uploading file. Hope this works for you
In others words reaching out to the command line and running another command while capturing the standard output.
The reading that've been doing so far seems to indicate that this is a clear violation of the sandbox model and therefore not possible.
You can easily do this is in Android:
//This is just an example don't get hanged up on the actual command.
Process process = Runtime.getRuntime().exec("cat somefile.txt");
BufferedReader bufferedReader = new BufferedReader(new InputStreamReader(process.getInputStream()));
String line;
while ((line = bufferedReader.readLine()) != null) {
logCatTraces.append(line);
logCatTraces.append("\r\n");
}
I have a running Electron app and is working great so far. For context, I need to run/open a external file which is a Go-lang binary that will do some background tasks.
Basically it will act as a backend and exposing an API that the Electron app will consume.
So far this is what i get into:
I tried to open the file with the "node way" using child_process but i have fail opening the a sample txt file probably due to path issues.
The Electron API expose a open-file event but it lacks of documentation/example and i don't know if it could be useful.
That's it.
How i open an external file in Electron ?
There are a couple api's you may want to study up on and see which helps you.
fs
The fs module allows you to open files for reading and writing directly.
var fs = require('fs');
fs.readFile(p, 'utf8', function (err, data) {
if (err) return console.log(err);
// data is the contents of the text file we just read
});
path
The path module allows you to build and parse paths in a platform agnostic way.
var path = require('path');
var p = path.join(__dirname, '..', 'game.config');
shell
The shell api is an electron only api that you can use to shell execute a file at a given path, which will use the OS default application to open the file.
const {shell} = require('electron');
// Open a local file in the default app
shell.openItem('c:\\example.txt');
// Open a URL in the default way
shell.openExternal('https://github.com');
child_process
Assuming that your golang binary is an executable then you would use child_process.spawn to call it and communicate with it. This is a node api.
var path = require('path');
var spawn = require('child_process').spawn;
var child = spawn(path.join(__dirname, '..', 'mygoap.exe'), ['game.config', '--debug']);
// attach events, etc.
addon
If your golang binary isn't an executable then you will need to make a native addon wrapper.
Maybe you are looking for this ?
dialog.showOpenDialog refer to: https://www.electronjs.org/docs/api/dialog
If using electron#13.1.0, you can do like this:
const { dialog } = require('electron')
console.log(dialog.showOpenDialog({ properties: ['openFile', 'multiSelections'] }))
dialog.showOpenDialog(function(file_paths){
console.info(file_paths) // => this gives the absolute path of selected files.
})
when the above code is triggered, you can see an "open file dialog" like this (diffrent view style for win/mac/linux)
Electron allows the use of nodejs packages.
In other words, import node packages as if you were in node, e.g.:
var fs = require('fs');
To run the golang binary, you can make use of the child_process module. The documentation is thorough.
Edit: You have to solve the path differences. The open-file event is a client-side event, triggered by the window. Not what you want here.
I was also totally struggling with this issue, and almost seven years later the documentation is quite not clear what's the case with Linux.
So, on Linux it falls under Windows treatment in this regard, which means you have to look into process.argv global in the main processor, the first value in the array is the path that fired the app. The second argument, if one exist, is holding the path that requested the app to be opened. For example, here is the output for my test case:
Array(2)
0: "/opt/Blueprint/b-test"
1: "/home/husayngonzalez/2022-01-20.md"
length: 2
So, when you're creating a new window, you check for the length of process.argv and then if it was more than 1, i.e. = 2 it means you have a path that requested to be opened with your app.
Assuming you got your application packaged with the ability to process those files, and also you set the operating system to request your application to open those.
I know this doesn't exactly meet your specification, but it does cleanly separate your golang binary and Electron application.
The way I have done it is to expose the golang binary as a web service. Like this
package main
import (
"fmt"
"net/http"
)
func handler(w http.ResponseWriter, r *http.Request) {
//TODO: put your call here instead of the Fprintf
fmt.Fprintf(w, "HI there from Go Web Svc. %s", r.URL.Path[1:])
}
func main() {
http.HandleFunc("/api/someMethod", handler)
http.ListenAndServe(":8080", nil)
}
Then from Electron just make ajax calls to the web service with a javascript function. Like this (you could use jQuery, but I find this pure js works fine)
function get(url, responseType) {
return new Promise(function(resolve, reject) {
var request = new XMLHttpRequest();
request.open('GET', url);
request.responseType = responseType;
request.onload = function() {
if (request.status == 200) {
resolve(request.response);
} else {
reject(Error(request.statusText));
}
};
request.onerror = function() {
reject(Error("Network Error"));
};
request.send();
});
With that method you could do something like
get('localhost/api/somemethod', 'text')
.then(function(x){
console.log(x);
}
This is my stored procedure
Using connection = New SqlConnection
' Dim connectionStringName = Me.DataWorkspace.XtraReportsServiceData.Details.Name
connection.ConnectionString = "Data Source=WIN-BTU6KNHNEGR\SQLEXPRESS;Initial Catalog=SapyLiveSystem;User ID=sa;Password=tytyty#21;Connection Timeout=3000000;"
Dim procedure = "UpdateColorMatching"
Using command = New SqlCommand(procedure, connection)
command.CommandType = CommandType.StoredProcedure
connection.Open()
command.ExecuteNonQuery()
End Using
End Using
'result = webservicesql.CallStoredProcuder
Log.log("Called after Stored Procedure")
It runs 100% on my development pc and if I call the stored procedure from a web service it runs 100% if I install it on the server and try and call that procedure from a webservice through light switch it wont run. If I call the webservice it runs 100% fine. If I run from lightswitch on development pc it works 100% fine buy not from server any help
I have a simple application that watches a folder for any changes as following:
private void Form1_Load(object sender, EventArgs e)
> {
> FileSystemWatcher w = new FileSystemWatcher();
> w.Path = #"C:\temp";
> w.Changed += new FileSystemEventHandler(OnChanged);
> w.Created += new FileSystemEventHandler(OnChanged);
> w.Deleted += new FileSystemEventHandler(OnChanged);
> w.Renamed += new RenamedEventHandler(OnChanged);
> // Begin watching.
> w.EnableRaisingEvents = true;
}
// Define the event handlers.
private static void OnChanged(object source, FileSystemEventArgs e)
{
// Specify what is done when a file is changed, created, or deleted.
MessageBox.Show("File: " + e.FullPath + " " + e.ChangeType + Path.GetFileName(e.FullPath));
}
I added the same as service from the command prompt as
sc create <service name> binPath= <path of the exe file>
This added the exe in the services and also made the entries in Registry. But when I tried to start the service as
sc start <service name>
it showed up the "Interactive Service Detection" message.
I want to avoid this message from popping up and start the service.
I also need this to be done in c# but if anyone has any idea about doing it in cmd I can add it as a batch file and execute the same.
EDIT I
As #Seva suggested I created a service that calls the exe that I wish. I wrote the following code to start the exe on start of the service:
protected override void OnStart(string[] args)
{
base.OnStart(args);
BackgroundWorker bw = new BackgroundWorker();
bw.DoWork += new DoWorkEventHandler(bw_DoWork);
bw.RunWorkerAsync();
}
private void bw_DoWork(object sender, DoWorkEventArgs e)
{
p.StartInfo.CreateNoWindow = false;
p.StartInfo.WindowStyle = ProcessWindowStyle.Normal;
p.StartInfo.WorkingDirectory = #"<my exe path>";
p.StartInfo.FileName = "<myexe.exe>";
p.StartInfo.Arguments = #"<my exe path>";
p.StartInfo.UseShellExecute = false;
p.StartInfo.RedirectStandardOutput = true;
p.Start();
p.WaitForExit();
base.Stop();
}
I installed the service successfully but is not starting the exe on starting.
EDIT II
The exe started. The service's property had to be configured to allow service interaction with desktop, but then again the "Interactive service detection" message is coming up.
You will have to rearchitecture your windows service into two parts -- a GUI-less service process and a separate UI app that runs on user desktop. There are many ways service can communicate with UI app. These SO questions will get you started:
GUI and windows service communication
Communication between windows service and desktop app
There is no other way around. BTW, your existing approach is already broken -- for Non-admin users and for remote desktop sessions -- they won't see UI from a service even if they want to.