Question

我正在尝试构建一个允许用户选择以前打开的Excel实例的表单。

从那里，我想提取与该打开的实例关联的.xlsx文件的名称。

我怀疑我必须使用COM才能将它从那里拿出来，但我还没有找到办法。

我想的代码如下：

Process[] openApplications = Process.GetProcessesByName("excel");

foreach(Process p in openApplications)
{
    ///Find the associated .xlsx path and file
    ///...Maybe from the handle id?
}

非常感谢任何帮助

Answer 1

您必须使用ActiveWorkbook属性。请参阅this。

        try
        {
            Process[] openApplications = Process.GetProcessesByName("excel");
            int proLen = openApplications.Length;
            if (proLen == 0)
            {
                Console.WriteLine("The process does NOT exist or has exited...");
                return 0;
            }

            foreach(Process p in openApplications)
             {
             //validate p for null/nothing
             //get the name of the workbook using
             //Use p.ActiveWorkbook.Name to get the file name.
             }        
            return 1;
        }
        catch (Exception ex)
        {

        }

Answer 2

首先，您必须添加对Microsoft Excel Object Library的引用以获取正在运行的MS Excel应用程序的实例：

var application = (Application)Marshal.GetActiveObject("Excel.Application");

然后，您可以通过ActiveWorkbook和FullName属性获取文件的完整路径。

Console.WriteLine(application.ActiveWorkbook.FullName);

如果您打开了多个MS Excel实例，则可以通过Workbooks属性迭代它们：

foreach (Workbook workbook in application.Workbooks)
{
    // do something with workbook.FullName
}

从进程ID获取打开的xlsx文件名

2 个答案: