I have a installer class in my winform application by using which i am creating setup of my application.Now i want the values entered into setup textboxes to get retrieved into installer class and write into text file using custom action .Here is the parameter i am trying to send using custom action of Setup project..
/targetdir="[TARGETDIR]\"/Param1="[EDITA1]"/Param2="[EDITA2]"/Param3="[EDITA3]"
And here is the way i am trying to retrieve and write into newly created text file which is not happening..
public override void Install(System.Collections.IDictionary stateSaver)
{
base.Install(stateSaver);
string targetDirectory = Context.Parameters["targetdir"];
string param1 = Context.Parameters["Param1"];
string param2 = Context.Parameters["Param2"];
string param3 = Context.Parameters["Param3"];
try
{
File.Create("D:\\Yourfile.txt");
FileStream fs1 = new FileStream("D:\\Yourfile.txt", FileMode.OpenOrCreate, FileAccess.Write);
StreamWriter writer = new StreamWriter(fs1);
writer.Write("Hello", writer.NewLine);
writer.Write(param1, writer.NewLine);
writer.Write(param2, writer.NewLine);
writer.Write(param3, writer.NewLine);
writer.Close();
}
catch {
}
}
I am not able to get even Hello in my textfile.Please tell me where i am going wrong..
Thanks..
You're creating a file and then trying to access it , which generates an IOException, which you happily swallow with the empty catch.
You can simply comment out the following line File.Create("D:\\Yourfile.txt");, and then it'll work for you.
Do note that you're writting lines, and relaying on the writer.NewLine. I would probably use the following if this is what you want:
writer.WriteLine("hello");
Other than that, You can also just write to a string or an array of strings, and do :
var first_ten = read_all_lines.Take(10);
var out_path = "D:\\Yourfile.txt";
System.IO.File.WriteAllLines(out_path, first_ten);
Related
I know this is a bit of a "Day one" question, but I'm still having trouble understanding why my Stream writer is writing empty lines after each time it writes
namespace PostFinder
{
class HistorySaver
{
public static void Save(string item, string path)
{
StreamReader sre = new StreamReader(path);
string historyList = sre.ReadToEnd();
sre.Dispose();
StreamWriter sr = new StreamWriter(path);
sr.WriteLine(historyList+sr.NewLine+item);
sr.Dispose();
}
}
}
sr.WriteLine(historyList+sr.NewLine+item);
The .WriteLine() method puts an end-of-line character after the contents you pass to it. If you don't want that character, use .Write().
It looks like all you are wanting to do is append text to a file, so there is really no need to open a streamreader to read in the existing contents and then write them back out with your new content.
You can use the below to do all that you want in one step. If the input path file doesn't exist it will create a new one, and if it already exists it will just append your new item.
namespace PostFinder
{
class HistorySaver
{
public static void Save(string item, string path)
{
File.AppendAllText(path, item + Environment.NewLine);
}
}
}
I have a rich text editor that I have created in C#. One of the features I am now trying to add is templates. I do not want the user to have to use an OpenFileDialog to navigate to the template and open the file. I would like to specify the filepath myself so that the user only has to click one button in order to open the template.
Currently, I am trying to achieve this using the following code:
private void formalLetterToolStripMenuItem_Click(object sender, EventArgs e)
{
try
{
FileStream fileStream = new FileStream(#".\templates\tmp1.rtf", FileMode.Open);
String str;
str = fileStream.ToString();
string fileContents = File.ReadAllText(filepath);
fileContents = fileStream.ToString();
try
{
if (richTextBoxPrintCtrl1.Modified == true);
{
NewFile();
}
richTextBoxPrintCtrl1.Rtf = fileContents;
}
catch (Exception exception)
{
MessageBox.Show("There was an error opening the template. " + exception, "Error", MessageBoxButtons.OK, MessageBoxIcon.Error);
}
}
catch (Exception exception)
{
MessageBox.Show("There was an error opening the template. " + exception, "Error", MessageBoxButtons.OK, MessageBoxIcon.Error);
}
However, whenever I try to open the template, I get an exception that is as follows:
System.ArgumentsException: File format is not valid.
However, I have tried to open the file using my OpenFileDialog and that works fine. Could somebody assist me in getting this working correctly?
Your problem is that you're trying to convert the file to a string using str = fileStream.ToString(); however, this converts the filestream to a string which is not the same thing.
Instead just do string fileContents = File.ReadAllText(filepath); to get all of the files contents into a string. You only need to use a FileStream/StreamReader if you're going to do some type of processing on the file.
Also, your use of the FileStream is a little off. I think what you really want is a StreamReader with something like this (example from msdn);
using (StreamReader sr = new StreamReader("TestFile.txt"))
{
string line;
// Read and display lines from the file until the end of
// the file is reached.
while ((line = sr.ReadLine()) != null)
{
Console.WriteLine(line);
}
}
A FileStream cannot be used to read a file. It must be passed to a StreamReader in order to actually read the file and in this case there is no point in doing that because there is an overload of the constructor which takes a filepath. It's only useful if you don't know what kind of stream the reader is going to be reading.
Where you have;
FileStream fileStream = new FileStream(#".\templates\tmp1.rtf", FileMode.Open);
String str;
str = fileStream.ToString();
string fileContents = File.ReadAllText(filepath);
fileContents = fileStream.ToString();
You actually just want thins line; string fileContents = File.ReadAllText(filepath); , nothing else. There is no need for a FileStream when you're just reading all the text into a string.
You are making very heavy weather of loading RTF. Your code to read a file into a string will never work, as #evanmcdonnal explained. Did your file dialog based code that succeeded really do it like that? Remember that a file dialog is just UI that generates a file name in a string. If your code with a file dialog works, then it will work when the file dialog is replaced with a hard coded string.
I suspect that some part of your problem is that you are using a relative path. Perhaps the working directory is not what you expect it to be. You should specify the full path to the file.
In any case, to load RTF simply call the LoadFile method of the control. But I strongly recommend passing the full path to the file.
richTextBoxPrintCtrl1.LoadFile(fullPathToRtfFile);
I am making simple tool for manipulating images in a database. I want to show the output result in a txt file and because the outcome may be different each time, I want the file to be rewritten with the fresh data every time the data is executed.
Also I want (if possible) to use some default location where the txt file will be created even though I have an App.Config file and that's also an option.
The problem I am having is with this code:
string Resultfile =
System.Environment.GetFolderPath(Environment.SpecialFolder.MyDocuments) +
"\\PictureStatus.txt";
FileStream strm = new FileStream(Resultfile , FileMode.Create);
TextWriter tw = new StreamWriter(strm);
This populates the PictureStatus.txt only once and then I get the same text over and over again. I noticed that if I use some random destination the file is updated. Not sure if it's just random behavior or have something to do with using MyDocuments, but what I need is a way to be sure that I'll rewrite the file with the new data each time, and if possible, use some default destination that will work for other people.
You can try something like this
public partial Form2 : Form
{
public string path = Environment.CurrentDirectory + "/" + "Name.txt";
public Form2()
{
InitializeComponent();
if (!File.Exists(path))
{
File.Create(path);
}
}
private void button2_Click(object sender, EventArgs e)
{
using (StreamWriter sw = new StreamWriter(path, true))
{
sw.WriteLine("This text will be writen in the txt file", true);
sw.Close();
}
}
}
I have add to the button, when I pressed it will be written in the next line every time. If you remove "true" from code, it will be overwritten every time.
I have a textbox where a user can input their email, what I want to do is make it so that when they click a submit button. That email will be saved into a text file ( on my server ) called emails.txt
I managed to get this working using System.IO and then using the File.WriteAll method. However I want to make it so that it will add the email to the list ( on a new line ) rather then just overwriting whats already in there.
I've seen people mention using Append, but I can't quite grasp how to get it working.
This is my current code (that overwrites instead of appending).
public partial class _Default : Page
{
private string path = null;
protected void Page_Load(object sender, EventArgs e)
{
path = Server.MapPath("~/emails.txt");
}
protected void emailButton_Click(object sender, EventArgs e)
{
File.WriteAllText(path, emailTextBox.Text.Trim());
confirmEmailLabel.Text = "Thank you for subscribing";
}
}
You can use StreamWriter to get working with text file. The WriteLine method in true mode append your email in new line each time....
using (StreamWriter writer = new StreamWriter("email.txt", true)) //// true to append data to the file
{
writer.WriteLine("your_data");
}
From the official MSDN documentation:
using (StreamWriter w = File.AppendText("log.txt"))
{
MyWriteFunction("Test1", w);
MyWriteFunction("Test2", w);
}
Use StreamWriter in Append mode. Write your data with WriteLine(data).
using (StreamWriter writer = new StreamWriter("emails.txt", true))
{
writer.WriteLine(email);
}
Seems like a very easy question with a very easy answer: Open existing file, append a single line
If you post the current code, we can modify that to append instead of overwrite.
Does anyone know of a way to (reasonably simple) create a file without actually opening/locking it? In File class, the methods for file creation always return a FileStream. What I want to do is to create a file, rename it (with File.Move) and then use it.
Now I have to:
Create it
Close
Rename
Reopen for use
Maybe you can try using File.WriteAllText Method (String, String)
with the file name and an empty string.
Creates a new file, writes the
specified string to the file, and then
closes the file. If the target file
already exists, it is overwritten.
using (File.Create(...)) { }
While this will briefly open your file (but close it again right away), the code should look quite unobtrusive.
Even if you did some P/Invoke call to a Win32 API function, you would get a file handle. I don't think there's a way to silently create a file without having it open right afterwards.
I think the real issue here is why you go about creating your file in the way you've planned. Creating a file in one place simply to move it to another location doesn't seem very efficient. Is there a particular reason for it?
What about using File.WriteAllBytes method?
// Summary:
// Creates a new file, writes the specified byte array to the file, and then
// closes the file. If the target file already exists, it is overwritten.
Another way is to use FileStream and Close it after creating the file. It will not lock the file. The code will look like:
FileStream fs = new FileStream(filePath, FileMode.Create);
fs.Flush(true);
fs.Close();
You just after this you can rename it as well or move it some other location.
Below is the Test program to test functionality.
using System;
using System.Collections.Generic;
using System.IO; using
System.Linq;
using System.Text;
namespace FileLocking {
class Program
{
static void Main(string[] args)
{
string str = #"C:\Test\TestFileLocking.Processing";
FileIOTest obj = new FileIOTest();
obj.CreateFile(str);
}
}
class FileIOTest
{
internal void CreateFile(string filePath)
{
try
{
//File.Create(filePath);
FileStream fs = new FileStream(filePath, FileMode.Create);
fs.Flush(true);
fs.Close();
TryToAccessFile(filePath);
}
catch (Exception ex)
{
Console.WriteLine(ex.Message);
}
}
void TryToAccessFile(string filePath)
{
try
{
string newFile = Path.ChangeExtension(filePath, ".locked");
File.Move(filePath, newFile);
}
catch (Exception ex)
{
Console.WriteLine(ex.Message);
}
}
} }
If you use File.Create(commented in above code) then it will give error saying file is being used by another process.
Incredibly grotty hack, probably the most complicated way to achieve your goal:
use Process class
processInfo = new ProcessStartInfo("cmd.exe", "/C " + Command);
processInfo.CreateNoWindow = true;
processInfo.UseShellExecute = false;
process = process.Start(processInfo);
process.WaitForExit();
where Command would be echo 2>> yourfile.txt