正如您已经指出数组的长度可变(因为它是 SQL Server 层次结构 ID),您绝对需要创建自定义IComparer<byte[]>
实现。
逻辑很简单:
- 逐字节比较每个数组的第一个
n
字节,其中n
是两个数组中较小的一个的字节数。当检测到任何字节之间存在差异时,返回不同字节的比较结果。
- 如果第一个
n
字节相等,则返回两个数组长度的比较。
这样,给定一组数据,如下所示:
00 01 02
00 01
01
排序后,您将获得的结果是:
00 01
00 01 02
01
也就是说,这就是您的IComparer<byte[]>
实现的样子:
// I could be wrong in that this is called natural order.
class NaturalOrderByteArrayComparer : IComparer<byte[]>
{
public int Compare(byte[] x, byte[] y)
{
// Shortcuts: If both are null, they are the same.
if (x == null && y == null) return 0;
// If one is null and the other isn't, then the
// one that is null is "lesser".
if (x == null && y != null) return -1;
if (x != null && y == null) return 1;
// Both arrays are non-null. Find the shorter
// of the two lengths.
int bytesToCompare = Math.Min(x.Length, y.Length);
// Compare the bytes.
for (int index = 0; index < bytesToCompare; ++index)
{
// The x and y bytes.
byte xByte = x[index];
byte yByte = y[index];
// Compare result.
int compareResult = Comparer<byte>.Default.Compare(xByte, yByte);
// If not the same, then return the result of the
// comparison of the bytes, as they were the same
// up until now.
if (compareResult != 0) return compareResult;
// They are the same, continue.
}
// The first n bytes are the same. Compare lengths.
// If the lengths are the same, the arrays
// are the same.
if (x.Length == y.Length) return 0;
// Compare lengths.
return x.Length < y.Length ? -1 : 1;
}
}
顺便说一句,如果您的字节数组保证具有相同的长度,作为替代方案,您可以动态创建 order by 子句,按第一个元素排序,然后是第二个元素,等等,如下所示:
static IEnumerable<foo> OrderBySortField(this IEnumerable<foo> items,
int sortLength)
{
// Validate parameters.
if (items == null) throw new ArgumentNullException("items");
if (sortLength < 0) throw
new ArgumentOutOfRangeException("sortLength", sortLength,
"The sortLength parameter must be a non-negative value.");
// Shortcut, if sortLength is zero, return the sequence, as-is.
if (sortLength == 0) return items;
// The ordered enumerable.
IOrderedEnumerable<foo> ordered = items.OrderBy(i => i.sort[0]);
// Cycle from the second index on.
for (int index = 1; index < sortLength; index++)
{
// Copy the index.
int indexCopy = index;
// Sort by the next item in the array.
ordered = ordered.ThenBy(i => i.sort[indexCopy]);
}
// Return the ordered enumerable.
return ordered;
}
然后你可以简单地这样称呼它:
// You have to supply the length of the array you're sorting on.
List<foo> sortedFoo = FooSource().
OrderBySortField(sortLength).ToList();