如何在30秒后删除文件/或在作业完成后删除文件一次

时间:2014-10-29 01:06:16

标签: c# wpf visual-studio-2010 visual-studio delete-file

我正在开发一个C#项目,我需要在30秒后删除该文件。因此,一旦文件发送到机器,我需要软件计数到30秒,同时显示一个启动窗体,一旦30秒越过关闭启动画面,然后删除文件。

我添加了一个名为“image”的启动画面。所以现在发生的事情是,数据仅在启动屏幕关闭后发送到打印机。我需要多线程的工作。我的意思是数据应该在一面打印,而启动画面应该同时显示。有没有办法可以出来!! ..请帮帮我。

所以在我的情况下,我将文件复制到bin / debug文件夹。然后将数据发送到机器同时显示启动画面30秒并关闭启动画面然后我需要删除文件..

代码:

 private void button4_Click(object sender, EventArgs e)
    {
        //string filePath = image_print();
       // MessageBox.Show(filePath, "path");
        string s = image_print() + Print_image();
        if (String.IsNullOrEmpty(s) || img_path.Text == "")
        {
            return;
        }
        else
        {
         //here its coming to the splash screen code, But data is transferred to the machine only after the splash screen is close :-(
           this.Hide();
        omg = new image();
        omg.ShowDialog();
        this.Show();
         //splash screen closed and then data is transferred.. which i don't need.. i need simultaneous job to be done at the same time..
         PrintFactory.sendTextToLPT1(s);
        }
    }

    private string image_print()
    {
        OpenFileDialog ofd = new OpenFileDialog();
        string path = "";
        string full_path = "";
        string filename_noext = "";
        ofd.InitialDirectory = @"C:\ZTOOLS\FONTS";
        ofd.Filter = "GRF files (*.grf)|*.grf";
        ofd.FilterIndex = 2;
        ofd.RestoreDirectory = true;
        if (ofd.ShowDialog() == DialogResult.OK)
        {
            filename_noext = System.IO.Path.GetFileName(ofd.FileName);
            path = Path.GetFullPath(ofd.FileName);
            img_path.Text = filename_noext;
            //MessageBox.Show(filename_noext, "Filename"); - - -> switching.grf
            // MessageBox.Show(full_path, "path");
            //move file from location to debug
            string replacepath = @"\\bin\Debug";
            string fileName = System.IO.Path.GetFileName(path);
            string newpath = System.IO.Path.Combine(replacepath, fileName);
           // string newpath = string.Empty;
            if (!System.IO.File.Exists(filename_noext))
                System.IO.File.Copy(path, newpath);
            filename_noext = img_path.Text;
         MessageBox.Show(filename_noext, "path");
        }

        if (string.IsNullOrEmpty(img_path.Text))
            return "";//

        StreamReader test2 = new StreamReader(img_path.Text);
        string s = test2.ReadToEnd();
        return s;
    }


    private string Print_image()
    {
        //some codes
            return s;
    } 

图片形式:我有以下代码

public partial class image : Form
{
    string filePath;
    public image()
    {
        InitializeComponent();
       // this.filePath = FileToDeletePath;

        System.Timers.Timer timer1 = new System.Timers.Timer();
        timer1.Interval = 30000;
        timer1.Elapsed += timer1_Elapsed;
        timer1.Start();
    }

    private void image_Load(object sender, EventArgs e)
    {

    }

    void timer1_Elapsed(object sender, ElapsedEventArgs e)
    {
        //delete the file using "filePath"
        string Filename = img_path.Text; // here i cannot pass the old string file name with  extension to this form.. Any ways please help me out

        if (string.IsNullOrEmpty(Filename))
            return;

        if (Filename.ToCharArray().Intersect(Path.GetInvalidFileNameChars()).Any())
            return;

        File.Delete(Path.Combine(@"\\bin\Debug", Filename));
    }

}

3 个答案:

答案 0 :(得分:2)

像这样的东西????

Task waitfordelete = Task.Run(() =>
{
    image im = new image();
});

答案 1 :(得分:2)

假设:窗口image应显示为对话框(模态),且仅在调用PrintFactory.sendTextToLPT1时才会显示。

如果这是正确的,那么这样的事情对你有用:

// Don't forget, you need to dispose modal dialogs
image omg = new image();

// Ensure the dialog has been shown before starting task. That
// way the task knows for sure the dialog's been opened and can
// be closed.
omg.Loaded += (sender, e) =>
{
    // Run the print task in a separate task
    Task.Run(() =>
    {
        PrintFactory.sendTextToLPT1(s);
        // But get back onto the main GUI thread to close the dialog
        Dispatcher.Invoke(() => omg.Close());
    });
};

this.Hide();
omg.ShowDialog();
this.Show();

提前为任何拼写错误/语法错误/等道歉。希望上述内容足以表达一般概念。

答案 2 :(得分:1)

Narzul Peter 给出的答案都是正确的。你可以实现任何一个。但是,我知道您的下一个问题是如何在代码中实现该方法。

您可以使用ThreadTask类对象来分隔流程。因此,当一个进程正在运行时,其他进程可以在那时执行其进程。登录中有两个进程。第一个是将文件发送到打印机,第二个是显示对话框30秒,然后删除文件。您应该创建另一个线程来调用任何一个进程,以便其他进程可以异步执行。

1st:为打印文件创建单独的进程。

Task waitfordelete = Task.Run(() =>
{
    PrintFactory.sendTextToLPT1(s);
});

this.Hide();
omg = new image();
omg.ShowDialog();
this.Show();

2nd:为show对话框创建单独的进程并删除该文件。但是,我认为你可能会在这种方法中得到错误。 您无法从其他线程更改UI

Task waitfordelete = Task.Run(() =>
{
    Dispatcher.Invoke(() => this.ShowSplashScreen());
});

PrintFactory.sendTextToLPT1(s);


private void ShowSplashScreen()
{
    this.Hide();
    omg = new image(); 
    omg.ShowDialog();
    this.Show();        
}

如果您不想使用该线程或任务,那么只需处理Image表单的关闭事件

this.Hide();
omg = new image(); 
omg.Show();
PrintFactory.sendTextToLPT1(s);
omg.FormClosed += (object sender, EventArgs e) => { 
    File.Delete(Path.Combine(Application.StartupPath, Path.GetFileName(img_path.Text));
    this.Show();    
};

并以timer_tick形式修改Image事件中的代码,并在删除文件语句后添加this.Close()

void timer1_Elapsed(object sender, ElapsedEventArgs e)
{
    ....
    //File.Delete(Path.Combine(@"\\bin\Debug", Filename)); comment this line
    this.Close();
}

我在这里找到的另一个隐藏的问题。 在这里,我无法传递旧的字符串文件名,扩展名为此表单..任何方式请帮助我

 void timer1_Elapsed(object sender, ElapsedEventArgs e)
{
    //delete the file using "filePath"
    string Filename = img_path.Text; // here i cannot pass the old string file name with  extension to this form.. Any ways please help me out

为此,您可以在Image类中创建属性,并从父窗体中分配文件名。

Image omg = new Image()
omg.FileName = Path.Combine(Application.StartupPath, Path.GetFileName(img_path.Text));
omg.Show();

并且Image形式的属性将像这样创建

public class Image : Form
{
    public string FileName { get; set; }
    public Image()
    {
    }

    void timer1_Elapsed(object sender, ElapsedEventArgs e)
    {
        ....
        File.Delete(Path.Combine(Application.StartupPath, this.Filename));
        this.Close();
    }        
}

注意:使用Application.StartupPath而不是\\bin\debug