I am writing a winforms program in C# that uses the openfiledialog. I would like it to be able to take the file that the user selected and open it as text, regardless of the file type.
I tried it like this:
if (openFileDialog1.ShowDialog() == DialogResult.OK)
{
textBox1.Text = Process.Start("notepad.exe", openFileDialog1.ToString()).ToString();
}
However, that didn't work and I'm not sure if I"m even on the right track.
You should use this code:
First add this namespace :
using System.IO;
Then add this codes to your function:
OpenFileDialog openFileDialog = new OpenFileDialog();
openFileDialog.Filter = "txt files (*.txt)|*.txt|All files (*.*)|*.*";
if (openFileDialog.ShowDialog()== DialogResult.OK)
{
textBox1.Text = File.ReadAllText(openFileDialog.FileName);
}
To open the file using notepad, you need to pass the file name as second parameter of Start method. For example:
using (var ofd = new OpenFileDialog())
{
if(ofd.ShowDialog()== DialogResult.OK)
{
System.Diagnostics.Process.Start("notepad.exe", ofd.FileName);
}
}
Also if for any reason while knowing that not all file contents are text, you are going to read the file content yourself:
using (var ofd = new OpenFileDialog())
{
if(ofd.ShowDialog()== DialogResult.OK)
{
var txt = System.IO.File.ReadAllText(ofd.FileName);
}
}
What you are doing at the moment is starting a Process with the argument openFileDialog1.ToString(), calling ToString() on the process and setting this as text in the TextBox. If the path was valid, the result would probably be something like "System.Diagnostics.Process". But since you use openFileDialog1.ToString() as a path, your application probably crashes with a file not found error.
To get the selected file of an OpenFileDialog, use openFileDialog1.FileName. (See the docs here)
What I think you actually want to do, is read from the file and write its contents as text to the TextBox. To do this, you need a StreamReader, like so:
if (openFileDialog1.ShowDialog() == DialogResult.OK)
{
using(var reader = new StreamReader(openFileDialog1.FileName))
{
textBox1.Text = reader.ReadToEnd();
}
}
This way, you open the file with the StreamReader, read its contents and then assign them to the text box.
The using statement is there because a StreamReader needs to be disposed of after you're done with it so that the file is no longer in use and all resources are released. The using statement does this for you automatically.
Related
Using Visual Studio 2017 and Windows 10 I want to be able to open a file explorer and navigate to a file outside of the program. Once my file is collected I want to get the file path and the complete file name for the file explorer.
OpenFileDialog openFileDialog1 = new OpenFileDialog();
openFileDialog1.Filter = "All files (*.*)|*.*";
if (openFileDialog1.ShowDialog() == DialogResult.OK)
textBoxFolderpath.Text= openFileDialog1.FileName;
if (textBoxFolderpath.Text.Contains('\\'))
textBoxFolderpath.Text = textBoxFolderpath.Text.Substring(0, textBoxFolderpath.Text.LastIndexOf('\\'));
This is what I have tried so far. The textBoxFolderpath is being used to hold the values.
What am I missing or doing incorrectly?
When working with filenames and paths I'd highly recommend using built-in classes to handle this, like the Path class:
Path.GetDirectoryName(openFileDialog1.FileName)
This returns the path, without the filename. It's also cross-platform compatible.
Thanks Zer0 - below is what I ended up using.
private void btnCurrentFolder_Click(object sender, EventArgs e)
{
OpenFileDialog openDialog1 = new OpenFileDialog();
// Determine starting directory
if (chkSetToRoot.Checked)
{
openDialog1.InitialDirectory = #"K:\RESULTS";
}
openDialog1.Title = "Select A Test File";
openDialog1.Filter = "All Files (*.*)|*.*";
if (openDialog1.ShowDialog() == DialogResult.OK)
{
textBoxFolderpath.Text = Path.GetDirectoryName(openDialog1.FileName);
textBoxFileName.Text = Path.GetFileName(openDialog1.FileName);
}
}
I'm basically just trying to get a file path to save a file to but my SaveFileObject won't let me access the SelectedPath. I've checked the other forums and can't figure out why it won' tlet me, here's my code;
SaveFileDialog filePath = new SaveFileDialog();
DialogResult result = filePath.ShowDialog();
if (result == DialogResult.OK)
{
string folderPath = filePath.;
}
It'll let me select filePath.ShowDialog again and filePath.ToString etc... Where am I going wrong?
You actually want the file name from the FileName property from your SaveFileDialog. That will give you the full path and file name for the file your user wants to save.
SaveFileDialog saveDialog = new SaveFileDialog();
DialogResult result = saveDialog.ShowDialog();
if (result == DialogResult.OK)
{
String fileName = saveDialog.FileName;
//your code to save the file;
}
Although, since .ShowDialog() returns a DialogResult, you can use it directly in the if to spare one line of code (yup! I'm greedy)
I'm using the following code to open multiple XML files and read the contents of the files but it doesn't work.
OpenFD.Filter = "XML Files (*.xml)|*.xml";
OpenFD.Multiselect = true;
if (OpenFD.ShowDialog() == DialogResult.OK)
{
foreach (string file in OpenFD.FileNames)
{
MessageBox.Show(file);
System.IO.Stream fileStream = OpenFD.OpenFile();
System.IO.StreamReader streamReader = new System.IO.StreamReader(fileStream);
using (streamReader)
{
MessageBox.Show(streamReader.ReadToEnd());
}
fileStream.Close();
}
}
For testing purposes, I created two xml files.
file1.xml (its content is "string1")
file2.xml (its content is "string2")
When I open the dialog and select the two files, I get four messages.
file1.xml
string1
file2.xml
string1
Even though the OpenFileDialog reads the file names correctly, I can't get to read the second file. It only reads the first file. So I'm guessing the problem is related to StreamReader, not to OpenFileDialog. What am I doing wrong?
You're using OpenFD.OpenFile() in each iteration, which:
Opens the file selected by the user, [...] specified by the FileName property.
Which in turn:
can only be the name of one selected file.
Use the file variable from your loop instead, and the StreamReader constructor that accepts a string:
using (var streamReader = new System.IO.StreamReader(file))
{
MessageBox.Show(streamReader.ReadToEnd());
}
This line is opening the file from the OpenFileDialog:
System.IO.Stream fileStream = OpenFD.OpenFile();
But there's no specification for which file. You need a way to distinguish which file you're opening. I would get rid of that line all together and just use the string file you have in the loop.
System.IO.StreamReader streamReader = new System.IO.StreamReader(file);
I'm using System.Windows.Forms.OpenFileDialog in my WPF application to select images. When user select an image, I'm displaying file name of selected file in a textbox as below.
OpenFileDialog fileDialog = new OpenFileDialog();
fileDialog.Title = "Select image";
fileDialog.InitialDirectory = "";
fileDialog.Filter = "Image Files (*.gif,*.jpg,*.jpeg,*.bmp,*.png)|*.gif;*.jpg;*.jpeg;*.bmp;*.png";
fileDialog.FilterIndex = 1;
fileDialog.RestoreDirectory = true;
if (fileDialog.ShowDialog() == System.Windows.Forms.DialogResult.OK)
{
txtImagePath.Text = fileDialog.FileName;
}
I have a button as Save in my application. When user click on this button, I need to rename this file to another name and copy it to another directory in my hard drive.
How can I achieve this?
Using File.Copy and methods in the Path class to extract the relevant part of your file
string newDir = #"D:\temp";
string curFile = Path.GetFileName(txtImagePath.Text);
string newPathToFile = Path.Combine(newDir, curFile);
File.Copy(txtImagePath.Text, newPathToFile);
Now the rename operation on the current dir using File.Move
string curDir = Path.GetDirectoryName(textImagePath.Text);
File.Move(txtImagePath.Text, Path.Combine(curDir, "NewNameForFile.txt"));
This code could be improved introducing some error handling
If you want to copy directly the old file in the new dir using the new name then you could write simply
string newPathToFile = #"D:\temp\NewNameForFile.txt";
File.Copy(txtImagePath.Text, newPathToFile);
and then do the rename on the current dir.
How about using:
System.IO.File.Copy
Alternatively, you can start a batch file, using processInfo
You can use File.Copy. Here are the details: http://msdn.microsoft.com/en-us/library/c6cfw35a.aspx. It takes the source file and copies it to a destination, possibly under a new name.
Is there a way to store the content of a file as string or as a dictionary instead of just its file path/name?
Below is the method that I am currently using for getting the file path from a Windows Form. Is there a way to adapt it or should I start from scratch? I am loading .ini files which is only text. LINQ seems to be one route but I am not familiar with it.
public void ShowSettingsGui()
{
System.Windows.Forms.OpenFileDialog ofd = new System.Windows.Forms.OpenFileDialog();
ofd.Multiselect = false;
ofd.Filter = "Data Sources (*.ini)|*.ini*|All Files|*.*";
if (ofd.ShowDialog() == DialogResult.OK)
{
string[] filePath = ofd.FileNames;
}
m_settings = Path.GetDirectoryName(ofd.FileName);
}
LINQ is indeed a nice way to do it: We simply convert the paths to a dictionary (where they become the keys). The values are determined by calling File.ReadAllText on every file path.
var dialog = new OpenFileDialog() { Multiselect = true,
Filter = "Data Sources (*.ini)|*.ini*|All Files|*.*" };
if (dialog.ShowDialog() != DialogResult.OK) return;
var paths = dialog.FileNames;
var fileContents = paths.ToDictionary(filePath => filePath, File.ReadAllText);
To help you understand what's going one here, take a look at the (roughly equivalent) non-LINQ version. Here, we explicitly iterate over the FileNames and add them as keys to our dictionary while again calling File.ReadAllText on every one of them.
// same as first snippet without the last line
foreach (var filePath in paths)
{
fileContents.Add(filePath, File.ReadAllText(filePath));
}
Set a breakpoint to the last line of each snippet, run them and take a look at the contents of the dictionary to determine the result.
EDIT: It wasn't clear in the question, but it seems you're only interested in a single file name. That means you don't need LINQ at all (m_settings needs to be a string).
var dialog = new OpenFileDialog{Filter = "Data Sources (*.ini)|*.ini*|All Files|*.*"};
if (dialog.ShowDialog() != DialogResult.OK) return;
m_settings = File.ReadAllText(dialog.FileName);
if you can add description of what are you trying to accomplish it would help.
just the same, I would use to store/read settings by using the settings class
here is a link to how to use it:
write user settings
I used in the past xml to parse a settings file, i find it much easier than reading ini in a sequential manner.
Hope it helps