获取共享打印机的打印队列详细信息

Question

我试图在共享打印机上打印文档;我需要获取打印队列详细信息。以下代码始终从“Microsoft XPS文档”获取队列作为作业数= 0.但我的默认打印机配置为“HP LaserJet P1505n”

LocalPrintServer server = new LocalPrintServer()
PrintQueueCollection queueCollection = server.GetPrintQueues();
PrintQueue printQueue = null;
foreach (PrintQueue pq in queueCollection)
{
 Logger.LogInfo("PrintQueue1", "Printer1 Queue Name " + pq.FullName);
 printQueue = pq;
 numberOfJobs = printQueue.NumberOfJobs;
 Logger.LogInfo("numberOfJobs1"+ numberOfJobs);
}

如何从特定的共享打印机获取打印队列详细信息？我也尝试过跟随

PrintServer server = new PrintServer(@"\\192.168.100.168\HP LaserJet P1505n");

但得到的错误是：

Win32错误：文件名，目录名称或卷标语法不正确

我在这里失踪了什么？

Answer 1

如何从特定的共享打印机获取打印队列详细信息？

尝试这样的事情：

// string.Empty or null for local printers
string printServerName = @"\\server";
string printQueueName= "printer";

PrintServer ps = string.IsNullOrEmpty(printServerName)
    // for local printers
    ? new PrintServer()
    // for shared printers
    : new PrintServer(printServerName);
PrintQueue pq = ps.GetPrintQueue(printQueueName);

Console.WriteLine(pq.FullName);
Console.WriteLine(pq.NumberOfJobs);
// output is printer uri (\\server\printer) and 0.

也可以使用服务器IP地址（如字符串）代替服务器名称。

string printServerName = @"\\192.168.1.111"; // for example

即 用于本地打印机PDFCreator设置

string printServerName = null;
string printerName = "PDFCreator";

对于服务器S集上的共享打印机P.

string printServerName = @"\\S";
string printerName = "P";

Answer 2

MSDN上的System.Printing.PrintServer类文档

试试这个：

var myServer = @"\\192.168.100.168";

使用此代码：

// Create a PrintServer
// "theServer" must be a print server to which the user has full print access.
// var myServer = @"\\theServer"
PrintServer myPrintServer = new PrintServer(myServer);

// List the print server's queues
PrintQueueCollection myPrintQueues = myPrintServer.GetPrintQueues();
String printQueueNames = "My Print Queues:\n\n";
foreach (PrintQueue pq in myPrintQueues)
{
    printQueueNames += "\t" + pq.Name + "\n";
}
Console.WriteLine(printQueueNames);
Console.WriteLine("\nPress Return to continue.");
Console.ReadLine();