我正在尋找一種以互動方式創建具有靈活數量元素的 List-Object 的方法。目前，我有固定數量的元素；我可以像這樣創建物件：
private List<DropItem> _items = new()
{
new DropItem(){ Name = "Cam 0", Identifier = "Drop Zone 1", cam_idx = 0 },
new DropItem(){ Name = "Cam 1", Identifier = "Drop Zone 1", cam_idx = 1 },
new DropItem(){ Name = "Cam 2", Identifier = "Drop Zone 1", cam_idx = 2 },
new DropItem(){ Name = "Cam 3", Identifier = "Drop Zone 1", cam_idx = 3 },
new DropItem(){ Name = "Cam 4", Identifier = "Drop Zone 2", cam_idx = 4 },
new DropItem(){ Name = "Cam 5", Identifier = "Drop Zone 2", cam_idx = 5 },
new DropItem(){ Name = "Cam 6", Identifier = "Drop Zone 2", cam_idx = 6 },
new DropItem(){ Name = "Cam 7", Identifier = "Drop Zone 2", cam_idx = 7 },
};
有沒有辦法直接用任意數量的元素構造這樣一個物件？（在 new() 中撰寫一個回圈的簡單嘗試 - 構造不出所料地不起作用）
uj5u.com熱心網友回復：
Enumerable.Range()應該在這里幫助你
int count = 10;
List<DropItem> _items = Enumerable.Range(0, count).Select(x => new DropItem()
{
Name = "Cam " x,
Identifier = "Drop Zone" ((x < count / 2) ? 1 : 2),
cam_idx = x
}).ToList();
https://dotnetfiddle.net/ePpi5T
uj5u.com熱心網友回復：
為此目的使用linq和Enumerable ：
Enumerable.Range(1, 12) // 12 cand be other number
.Select(x =>
new DropItem
{
Name = "Cam " x,
cam_idx = x
... // whatever you want
}).ToList();
uj5u.com熱心網友回復：
int itemsCount = 8;
for (int i = 0; i < itemsCount; i )
{
_items.Add(new DropItem() { Name = $"Cam {i}", Identifier = $"Drop Zone {i / 4 1}", cam_idx = i});
}
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