C# - OpenFileDialog too slow only first time on pressing OK - c#

I have a button which opens an OpenFileDialog. When I compile the application, run it for the first time, press the button, select file(s) and then press the accept dialog button, it waits for about a minute before adding the selected file(s) into my list box.
If I close the application, restart it and do the same thing as above, everything works fast and normal. From then on it always works fast. It's only the very first time I run it after the compilation when it's too slow.
The code extract is below. What could be wrong with the dialog? Why does it run slowly for the first time? Thank you.
void ButtonAddClick(object sender, EventArgs e)
{
this.openFileDialog.FileName = String.Empty;
this.openFileDialog.InitialDirectory = this.openPath;
if (this.openFileDialog.ShowDialog() == DialogResult.OK)
{
foreach (string file in this.openFileDialog.FileNames)
{
if (!File.Exists(file))
{
this.ShowStatus("Error occured selecting file " + Path.GetFileName(file));
}
else if (!this.listBoxFiles.Items.Contains(file))
{
this.listBoxFiles.Items.Insert(0, file);
}
else{
this.ShowStatus("File " + Path.GetFileName(file) + " already selected");
}
}
}
if (this.listBoxFiles.Items.Count > 0)
{
this.openPath = Path.GetDirectoryName(this.listBoxFiles.Items[0].ToString());
this.listBoxFiles.Enabled = true;
this.buttonClear.Enabled = true;
this.buttonFolder.Enabled = true;
}
}

If you are facing this problem of slowness at first initialisation, my suggestion is listed below to rectify this.
For VS debugging, in Visual Studio IDE, just go to Tools>Options>Debugging. Find settings page named [Symbols]. Click it to load the page content. On the right side panel if “Microsoft Symbol Servers” checkbox is checked, just uncheck it and press “Ok” to save the settings. Now on running compiled exe, if the problem still persists then I suggest to perform some cleanup operation on your pc. Also make sure that all timely VS updates are incorporated in your VS. Let me know if this helps if not we can figure out something else.

Try calling Directory.GetFiles(folderPath); prior to showing the OpenFileDialog window. That might trigger the same caching that's occurring after you do OpenFileDialog the first time with your current method.

Related

c# Form Project won't save Setting changes of Console Project

Background
Console Application:
I have a console application that retrieves data from a spreadsheet using the google sheets API. This application is automated by running it every 5 minutes with windows scheduler.
Form Application:
In the same solution I have created a windows form project that can be run manually, outside of the automation process to tweak any settings without disturbing the 5 minute process. (i.e. If we want to change spreadsheet ID to fetch data form a different spreadsheet, or if i want to change the output folder of where the data is being fetched)
My Goal
I'm trying to develop a form project that will edit the "settings.settings" file of another project in the same solution. Below is a screenshot of how i have my solution laied out:
What I've done so far
I've already added a reference from my sheetstocsv project in my SettingsUI Project, and i've successfully created a form that accesses sheetstocsv's Settings and edits them when a save button is clicked. Shown below is the function that's supposed to save the new settings from the form.
private void Save_Button_Click(object sender, EventArgs e)
{
sheetstocsv.Properties.Settings.Default.outputdir = OutputDirectory_TextBox.Text;
sheetstocsv.Properties.Settings.Default.spreadsheetID = SpreadsheetID_Textbox.Text;
sheetstocsv.Properties.Settings.Default.Entity = Entity_Texbox.Text;
sheetstocsv.Properties.Settings.Default.headernum = (int)Columns_NumericUpDown.Value;
string headers = "";
for (int i = 0; i < (int)Columns_NumericUpDown.Value; i++)
{
headers = headers + Columns_DataGridView.Rows[0].Cells[i].Value.ToString() + ",";
}
sheetstocsv.Properties.Settings.Default.headers = headers;
sheetstocsv.Properties.Settings.Default.configured = true;
sheetstocsv.Properties.Settings.Default.Save();
MessageBox.Show("Saving Complete!", "Settings",
MessageBoxButtons.OK, MessageBoxIcon.Asterisk);
}
How I'm trying to do it
Below is a code snippet of how i'm only letting the sheetstocsv project continue if the configuration file has configured it first
//Check for if configured
if (Properties.Settings.Default.configured == false)
{
Console.WriteLine("Program has not been configured yet! Please Run SettingsUI first to start this program.");
Console.ReadLine();
Environment.Exit(0);
}
My Problem
Whenever I run my settingsUI, save, and then run my sheetstocsv project the changes that were supposed to be saved are not. and it will not allow the program move foward.
Edit
Below is the full form .cs that shows how i'm editing the properties of the other project
using sheetstocsv;
namespace SettingsUI
{
public partial class Form1 : Form
{
public Form1()
{
InitializeComponent();
Columns_DataGridView.RowCount = 1;
Columns_DataGridView.ColumnCount =(int)Columns_NumericUpDown.Value;
}
// Event handlers for other UI elemets here.......
private void Save_Button_Click(object sender, EventArgs e)
{
sheetstocsv.Properties.Settings.Default.outputdir = OutputDirectory_TextBox.Text;
sheetstocsv.Properties.Settings.Default.spreadsheetID = SpreadsheetID_Textbox.Text;
sheetstocsv.Properties.Settings.Default.Entity = Entity_Texbox.Text;
sheetstocsv.Properties.Settings.Default.headernum = (int)Columns_NumericUpDown.Value;
string headers = "";
for (int i = 0; i < (int)Columns_NumericUpDown.Value; i++)
{
headers = headers + Columns_DataGridView.Rows[0].Cells[i].Value.ToString() + ",";
}
sheetstocsv.Properties.Settings.Default.headers = headers;
sheetstocsv.Properties.Settings.Default.configured = true;
sheetstocsv.Properties.Settings.Default.Save();
MessageBox.Show("Saving Complete!", "Settings", MessageBoxButtons.OK, MessageBoxIcon.Asterisk);
}
}
}
I don't quite have enough info from your question (see comment from Rufus), but I am guessing that your console app launches your WinForm? If this is the case, what is likely happening is this:
1) Console app starts, and loads the config file into memory.
2) Your Winform launches and changes the config file.
3) Console app overwrites the changes to the config file on exit since it read it before the changes were made.
You could just move the winform into your console project, then just do something like this:
Properties.Settings.Default.outputdir = OutputDirectory_TextBox.Text;
...
Adding More Info (Oct 1):
I must be honest, doing what you're doing is a little strange, and I would really rethink referencing an executable to change its default settings. Just because VS lets you do this, doesn't necessarily mean you should. But if you must, there are a couple other things you can try:
1) You can simply parse the xml settings file and change node values of the settings you want to change. System.Xml namespace will make very short work of this. The post below has good examples:
Change xml node value
2) Try wrapping the code that is currently in your button click event handler with some method inside the console project. This way, you're calling a method that changes the settings, rather than trying to change the settings from an external assembly.
Note: neither of those methods will work if your console app is running while you're making changes to its config file. On shut down, the console app will overwrite the settings file with the settings it loaded into memory when it started up. If the console app isn't running, 1) will definitely work.

How can I display a FolderBrowserDialog in LINQPad?

I want to browse for a folder in a LINQPad script so I tried using FolderBrowserDialog. It didn't work.
Here is a snippet showing what I'm doing?
string path = "";
var browser = new FolderBrowserDialog { ShowNewFolderButton = false };
if (browser.ShowDialog() == DialogResult.OK)
{
path = browser.SelectedPath;
}
This runs and hangs on the line with ShowDialog() with the yellow execution pointer against that line, but the folder browser dialog isn't visible.
I know that there is an overload for ShowDialog() that takes an IWin32Window owner argument and thought that might be the solution, but haven't yet figured out how to get an IWin32Window for the main LINQPad window. I hoped the Util class might provide a way but unless I'm missing it, it doesn't seem to.
Anyone have advice on getting around this problem?
Not solution, but an alternative, the FilePicker controle (part of new LinqPad Input Controls).
you can write:
new FilePicker().Dump().TextInput += (x, e) => ((FilePicker)x).Text.Dump();
or:
var picker = new FilePicker();
picker.TextInput += (x, e) => {
var fileName = picker.Text;
//action with the file...
};
full example:
void Main()
{
new FilePicker().Dump().TextInput += (x, e) => procces(((FilePicker)x).Text);
}
void procces(string file)
{
file.Dump("chosen file...");
//...
}
Setting the Run each query in its own process option to true is the cause of the problem. Setting that option back to the default false allows the code described above to run as expected.
However, making this change disables the built-in debugging. Furthermore the behaviour is still slightly problematic.
On first running the script the dialog is displayed and the script runs to completion after Ok or Cancel is selected. However, on running the script a second time it hangs as described in the question. After cancelling the execution and running it again the dialog displays but on the time after that it hangs again, and so on.
It was pointed out that setting the Always use fresh application domains option may resolve this and it does, allowing the dialog to display on every execution of the script.
I just came across this problem with LINQPad 5. I needed a folder picker similar to the file picker. Your solution worked without me having to modify LINQPad5 settings. The problem was the dialog was staying in the background. So here's how I got your snippet to work with that dialog in focus. Instead of using the FolderBrowseDialog.ShowDialog() I used the overload that passes in a windows form. I created a new form with description and window position then passed that to ShowDialog. That allowed me to set the description and window positioning.
string path = "";
using ( var browser = new System.Windows.Forms.FolderBrowserDialog { ShowNewFolderButton = false })
{
browser.Description = "Select Folder For Foo Processing";
var form = new System.Windows.Forms.Form(){TopMost = true, TopLevel = true};
var result = browser.ShowDialog(form);
if (result == System.Windows.Forms.DialogResult.OK)
{
path = browser.SelectedPath;
}
}
path.Dump();
I tried to initialize the form in ShowDialog with the settings, but had problems so I opted to declare it before show dialog. Hope this helps anyone with this problem.

Selenium WebDriver to test an ActiveX Control

I have been working to automate some tests for my companies website using Selenium's IWebDriver, and have run into some ActiveX controls (a dialog to select and upload a file) that I cannot seem to automate. I haven't been able to find any specific information on this on the internet.
I am, however, able to actually load the dialog box by triggering the "open" element within the page (the user will have to manually click the file destination and the open button), but it fails the test (this code doesn't make sense to me as to why it opens the dialog box, I originally had the SendKeys and Click in reverse order).
private void UploadFile()
{
foreach (var element in driver.FindElements(By.TagName("button")))
{
string open = element.Text;
if (open == "Open")
{
element.SendKeys(#"My\Relative\Path");
element.Click();
}
}
}
I've tried to execute JavaScript within my code to open the file, but my attempts have failed each time and my coworkers have told me that it wouldn't work anyways since ActiveX controls the file upload.
Any thoughts?
Thanks!
AutoIt was the solution for this. I was able to execute a script within my C# project.
private void UploadFile()
{
foreach (var element in driver.FindElements(By.TagName("button")))
{
string open = element.Text;
if (open == "Open")
{
element.SendKeys(#"C:\My\Relative\Path\");
element.Click();
string executable = #"C:\My\Relative\Path\fileUploadScript2.exe";
System.Diagnostics.Process.Start(executable);
}
}
}
Thanks to #SiKing for the push in the right direction.

Application hanging on DialogResult.OK

Im having some trouble my program hanging when selecting a file in a file dialog. This is the code that is showing the file browser dialog:
private void isForgeIncluded_btn_Click(object sender, EventArgs e)
{
this.isForgeIncluded.Text = FolderFileDialog("file", isForgeIncluded.Text, "Forge installer file (*.jar)|*.jar");
}
public string FolderFileDialog(string type, string current, string fileTypes = "All files (*.*)|*.*|All files (*.*)|*.*", string StartFolder = "C:\\")
string ReturnString = current;
if (current != "")
{
StartFolder = Path.GetFullPath(current);
}
if (type == "file")
{
OpenFileDialog minecraftFile = new OpenFileDialog();
minecraftFile.Title = "Select file";
minecraftFile.InitialDirectory = StartFolder;
minecraftFile.RestoreDirectory = true;
minecraftFile.Filter = fileTypes;
if (minecraftFile.ShowDialog() == DialogResult.OK)
{
ReturnString = minecraftFile.FileName;
return ReturnString;
}
minecraftFile = null;
}
return ReturnString;
}
I've narrowed the problem down to between the "if" statement and the "Return string = minecraftFile.FileName;" .. When using the debugger the program takes up to a five second break between those two lines. After its break it comes back and returns the value as if nothing was wrong. But IntelliTrace sometimes comes up with a "FileNotFound Exception" even though the program never shows me any error messages, and returns the correct value to the textbox as it should.
The wierd part is that it is not always this happens. Its random and can happen even though i select the same file as last time. The files i select are local files too on the system drive.
What could be wrong? Does the code look as it should?
Two basic problems can cause trouble with OpenFileDialog. First is misbehaving shell extension installed on your machine, they'll get loaded into your process when you open the dialog. A basic check is that you don't see trouble with Notepad's File + Open command and that you see nothing special in the Visual Studio Output window while the dialog is active. Enabling unmanaged debugging gives you more info.
The second cause is having an issue with the thread state of your UI thread. If must be initialized as a single-threaded apartment to be able to show this dialog properly. Go back to your Main() method, normally in Program.cs, and ensure it has the [STAThread] attribute.
If that doesn't pan out then we need more info about the code that's misbehaving. You need to enable unmanaged debugging, Project + Properties, Debug tab. And enable the Microsoft symbol server, Tools + Options, Debugging, Symbols. And disable Just My Code debugging, Tools + Options, Debugging, General. Hit Debug + Break All right after you get it to misbehave and post the content of the Call Stack window.

Launching JAR with Process.Start: javaw.exe runs but "nothing happens"?

Given the code. when the button is clicked nothing happens , i get no debug message etc in visual studio. However if i were to double click the .jar file in its folder i am able to run it. Anyone have any idea why?
Looking at Task manager when the button is clicked. javaw.exe is created but nothing happens.
private void btnKinderPuzzle_Click(object sender, RoutedEventArgs e)
{
// Check if this program is opened
if (IsProcessOpen("MTPuzzle"))
{
MessageBox.Show("KinderPuzzle is already running", "Kinder Package", MessageBoxButton.OK, MessageBoxImage.Information);
}
else
{
Process.Start(Directory.GetCurrentDirectory() + "\\Puzzle\\PuzzleGame\\MTPuzzle.jar");
}
}
Process.Start("java.exe",
Path.Combine("-jar " + Directory.GetCurrentDirectory(),
"Puzzle\\PuzzleGame\\MTPuzzle.jar"));
The path may be not correct. You might use instead
Process.Start(Path.Combine(Directory.GetCurrentDirectory(), "Puzzle\\PuzzleGame\\MTPuzzle.jar"));
If stills, then I think the problem in the setting of Java. To solve this potential problem, create a file run.cmd near your jar file and write this code into:
java -jar "MTPuzzle.jar"
Then, use Process.Start to start the file run.cmd

Categories