c# - 获取具有文件夹 ID 的子文件 - 客户端对象模型

Question

我想获取一些文件夹 - 子文件。我在列表中有所有文件夹 SharePoint ID。我的代码可以工作，但性能很差，因为有很多 context.ExecuteQuery;

我想用 Caml 查询实现它。

using (var context = new ClientContext("http://xxx/haberlesme/"))
{
    var web = context.Web;
    var list = context.Web.Lists.GetById(new Guid(target));

    int[] siraliIdArray;
    //siraliIdArray = loadSharePointIDList(); think like this

    for (var j = 0; j < siraliIdArray.Count; j++)
    {
        var folderName = listItemCollection[j]["Title"].ToString();//Folder Name
        var currentFolder = web.GetFolderByServerRelativeUrl("/haberlesme/Notice/" + folderName);
        var currentFiles = currentFolder.Files;

        context.Load(currentFiles);

        //I don't want to execute for n folder n times. I want to execute n folder 1 time.
        context.ExecuteQuery();

        var ek = new LDTOTebligEk();

        //I don't want to add one - one 
        foreach (var file1 in currentFiles)
        {
            ek.DokumanPath = urlPrefix + folderName + "/" + file1.Name;
            ek.DokumanAd = file1.Name;

            ekler.Add(ek);

        }
    }
}

例如，我有 100 个文件夹，但我想在一次执行

中获取 10 个文件夹子文件夹

Answer 1

由于 CSOM API 支持 Request Batching :

The CSOM programming model is built around request batching. When you work with the CSOM, you can perform a series of data operations on the ClientContext object. These operations are submitted to the server in a single request when you call the ClientContext.BeginExecuteQuery method.

您可以重构您的代码，如下所示:

var folders = new Dictionary<string,Microsoft.SharePoint.Client.Folder>();
var folderNames = new[] {"Orders","Requests"};
foreach (var folderName in  folderNames)
{
   var folderKey = string.Format("/Shared Documents/{0}", folderName);
   folders[folderKey] = context.Web.GetFolderByServerRelativeUrl(folderKey);
   context.Load(folders[folderKey],f => f.Files);
}
context.ExecuteQuery();  //execute request only once




//print all files
var allFiles = folders.SelectMany(folder => folder.Value.Files);
foreach (var file in allFiles)
{
   Console.WriteLine(file.Name);
}