启动C#函数作为后台进程

时间:2018-11-01 09:17:52

标签: c# .net

我想在不暂停的情况下启动程序。例如

private void button1_Click(object sender, EventArgs e)
{            
    test();
}

protected void test()
{            
    label1.Text = "";
    for (int i = 0; i < 10; i++)
    {
        label1.Text += i;
        Thread.Sleep(500);
    }
}

当我运行该程序时,表单将在循环结束时挂起。 如何使函数test()在不暂停表单的情况下运行,例如作为后台运行?

2 个答案:

答案 0 :(得分:3)

您可以使用Asynchronous programming,因此使您的方法async像这样:

protected async Task test()
{
    label1.Text = "";
    for (int i = 0; i < 10; i++)
    {
        label1.Text += i;
        await Task.Delay(500);
    }
}

最好一直使用它:

private async void button1_Click(object sender, EventArgs e)
{
    await test();
}

答案 1 :(得分:1)

尝试异步等待方法(我也将[main] sessionManager = org.apache.shiro.web.session.mgt.DefaultWebSessionManager ;cacheManager = org.apache.shiro.cache.MemoryConstrainedCacheManager cacheManager = se.cehis.sob.cache.EfsobCacheManager sessionDAO = se.cehis.efsob.web.authorization.shiro.CustomShiroSessionDAO securityManager.sessionManager = $sessionManager securityManager.sessionManager.sessionIdCookie.name = e_session_id securityManager.sessionManager.sessionIdCookie.path = / ;securityManager.sessionManager.sessionIdCookie.secure = true securityManager.cacheManager = $cacheManager securityManager.sessionManager.sessionDAO = $sessionDAO ;jdbcRealm=se.cehis.efsob.common.security.JdbcRealmImpl jdbcRealm=org.apache.shiro.realm.jdbc.JdbcRealm jdbcRealm.authenticationQuery = SELECT ID FROM ACL_USER WHERE ID = ? jdbcRealm.userRolesQuery = SELECT ROLE_NAME FROM ACL_ROLES where USER_ID = ? jdbcRealm.permissionsQuery = SELECT PERMISSION_NAME FROM ACL_PERMISSION WHERE ROLE_NAME = ? jdbcRealm.permissionsLookupEnabled = true ds = org.apache.shiro.jndi.JndiObjectFactory ds.resourceName = java:jboss/datasources/efobDS jdbcRealm.dataSource= $ds 替换为Thread.Sleep):

Task.Delay

如果您要执行大量CPU工作,则应执行以下操作:private async void button1_Click(object sender, EventArgs e) { await test(); } protected async Task test() { label1.Text = ""; for (int i=0;i<10;i++) { label1.Text += i; await Task.Delay(TimeSpan.FromMilliseconds(500)); //Thread.Sleep(500); } }

您可能会发现此enter image description here简介很有帮助。

相关问题