如何从linq到xml获取一个生成list的list<int>?

不要为每个项目创建新列表,只需创建一个列表:

LegalTextIds = (from legalText in panel.Elements("LegalText").Elements("Line")
                select (int)legalText.Attribute("id")).ToList()

使用 SelectMany 扩展方法:

List<List<int>> lists = new List<List<int>>()
    { 
        new List<int>(){1, 2},
        new List<int>(){3, 4}
    };

var result = lists.SelectMany(x => x);  // results in 1, 2, 3, 4

或者,对于您的特定情况:

var performancePanels = new
{
    Panels = (from panel in doc.Elements("PerformancePanel")
            select new
            {
                LegalTextIds = (from legalText in panel.Elements("LegalText").Elements("Line")
                             select new List<int>()
                             {
                                 (int)legalText.Attribute("id")
                             }).SelectMany(x => x)
            }).ToList()
};

这个怎么样?

List<int> GenListOfIntegers = 
          (from panel in doc.Elements("PerformancePanel").Elements("Line")
              select int.Parse(panel.Attribute("id").Value)).ToList<int>();