如何防止同一帐户重复登录系统-ASP教程,ASP应用

如何防止控制客户端使其用同一帐户重复登录系统.

解决思路：

维护一online表，查看有登陆，就不允许再次登陆，以sessionid作为唯一标识符号，也可以产生一个ｇｕｉｄ发到ｃｏｏｋｉｅ中，以区分不同的ｃｌｉｅｎｔ，再佐以ｊｓ，可以达到更好的效果，比如离开后自动离线

解决代码：

public virtual void application_start(object sender, eventargs e)

{

// reset the mailer indicator

application["mailerstatus"] = "all mailings complete";

// initialize a datatable for users online

datatable objusertable = new datatable();

objusertable.columns.add("sessionid",system.type.gettype("system.guid"));

objusertable.columns.add("peopleid",system.type.gettype("system.int32"));

objusertable.columns.add("showdetail",system.type.gettype("system.boolean"));

datacolumn[] pk = new datacolumn[1];

pk[0] = objusertable.columns[0];

objusertable.primarykey = pk;

application["usertable"] = objusertable;

}

/**////

/// the session_start event adds user session information to

/// application["usertable"].

///

public virtual void session_start(object sender, eventargs e)

{

application.lock();

//application.lock ();

datatable objusertable = (datatable)application["usertable"];

datarow objrow = objusertable.newrow();

guid objguid = guid.newguid();

objrow[0] = objguid;

session["pfsessionid"] = objrow[0];

objrow[1] = 0;

objrow[2] = false;

objusertable.rows.add(objrow);

application["usertable"] = objusertable;

application.unlock();

}

/**////

/// the session_end event deletes user session information from

/// application["usertable"].

///

public virtual void session_end(object sender, eventargs e)

{

application.lock();

datatable objusertable = (datatable)application["usertable"];

objusertable.rows.find((guid)session["pfsessionid"]).delete();

application["usertable"] = objusertable;

application.unlock();

}