Question

如何从可用产品版本列表中返回最佳匹配/下一个可用产品版本ID？

以下是基于表

中的样本数据的逻辑

查找小于10.10.20的最佳匹配版本，并返回其versionID eg1：GetVersion（“10.10.20”）应该返回5（因为在表中没有“10,10,20”major.minor.build组合可用，所以它应该寻找最匹配的版本这里的下一个可用版本是10.7.1 ie。，versionID 5

eg2：GetVersion（“7.0.0”）应该返回3（因为在表中没有“7,0,0”major.minor.build组合可用，所以它应该寻找下一个可用的匹配版本。下一个可用版本是6.2.1即。，版本ID 3

eg3：GetVersion（“7.5.1”）应返回4，此处完全匹配可用，因此应返回versionid 4

 [Serializable]
    public class ProductVersions
    {
        [Key]
        public int Version_Id { get; set; }
        public int Major { get; set; }
        public int Minor { get; set; }
        public int Build { get; set; }
    }

以下是我的ProductVersions表中的一些示例数据

    [version_id , Major,Minor,Build]
        1           3      0    1
        2           4     10    5
        3           6     2     1
        4           7     5     1
        5           10    7     1
        6           11    10   10

这是我的方法，预计将返回最佳可用产品版本

private int GetVersion(string versionNumber)
   {
    int version-id=0;

    version-id= //retrieve best matching version

     return version-id
    }

Answer 1

您可以使用内置版本类，因为它已经实现了您基本上正在寻找的<=运算符，并且还可以为您处理字符串解析：

var data = new List<Version>()
{
     new Version(3,0,1),
     new Version(4,10,5),
     new Version(6,2,1),
     new Version(7,5,1),
     new Version(10,7,1),
     new Version(11,10,10)
};

var case1 = new Version("10.10.20");
// match1 is 5; the index of a List is 0-based, so we add 1
var match1 = data.FindLastIndex(d => d <= case1) + 1;

var case2 = new Version("7.0.0");
// match2 is 3
var match2 = data.FindLastIndex(d => d <= case2) + 1;

var case3 = new Version("7.5.1");
// match3 is 4
var match3 = data.FindLastIndex(d => d <= case3) + 1;

将ProductVersions的序列转换为Version个对象列表应该是微不足道的。

如果您出于某种原因不想使用Version类，您可以自己实现<=（以及所有其他缺少的运算符）：

public class ProductVersions
{
   //TODO error checking
   public int Version_Id { get; set; }
   public int Major { get; set; }
   public int Minor { get; set; }
   public int Build { get; set; }

   public ProductVersions(int major, int minor, int build)
   {
        Major=major;
        Minor=minor;
        Build=build;
   }

   public ProductVersions(string version)
   {
        var tmp = version.Split('.');
        Major = Int32.Parse(tmp[0]);
        Minor = Int32.Parse(tmp[1]);
        Build = Int32.Parse(tmp[2]);
   }

   public static bool operator == (ProductVersions a, ProductVersions b)
   {
        return a.Major==b.Major && a.Minor==b.Minor && a.Build==b.Build;
   }

   public static bool operator != (ProductVersions a, ProductVersions b)
   {
        return !(a==b);
   }

   public static bool operator <= (ProductVersions a, ProductVersions b)
   {
        if (a == b)
            return true;
        return a < b;
   }

   public static bool operator >= (ProductVersions a, ProductVersions b)
   {
        if (a == b)
            return true;
        return a > b;
   }

   public static bool operator < (ProductVersions a, ProductVersions b)
   {
        if(a.Major==b.Major)
            if(a.Minor==b.Minor)
                return a.Build < b.Build;
            else
                return a.Minor < b.Minor;
        else
            return a.Major < b.Major;
   }

   public static bool operator > (ProductVersions a, ProductVersions b)
   {
        if(a.Major==b.Major)
            if(a.Minor==b.Minor)
                return a.Build > b.Build;
            else
                return a.Minor > b.Minor;
        else
            return a.Major > b.Major;
   }

一个简单的测试：

var data = new List<ProductVersions>()
{
     new ProductVersions(3,0,1)    { Version_Id = 1},
     new ProductVersions(4,10,5)   { Version_Id = 2},
     new ProductVersions(6,2,1)    { Version_Id = 3},
     new ProductVersions(7,5,1)    { Version_Id = 4},
     new ProductVersions(10,7,1)   { Version_Id = 5},
     new ProductVersions(11,10,10) { Version_Id = 6}
};

// ensure data is sorted by version
data.Sort((a,b) => a > b ? 1 : a < b ? -1 : 0);

var case1 = new ProductVersions("10.10.20");
// match1 is 5
var match1 = data.Last(d => d <= case1).Version_Id;

var case2 = new ProductVersions("7.0.0");
// match2 is 3
var match2 = data.Last(d => d <= case2).Version_Id;

var case3 = new ProductVersions("7.5.1");
// match3 is 4
var match3 = data.Last(d => d <= case3).Version_Id;

Answer 2

我喜欢Dominic使用版本类的答案（为什么发明它存在？）但是如果您想知道如何在不使用Version类的情况下执行此操作并且您认为列表已经排序（因此您不需要需要像他那样对它进行排序。

（TL; DR）

 // assume verArray is already ordered (this would need to be sorted otherwise.)
 // this where checks for less than or equal to.
 int result = verArray.Where(v => (v.Major < major) ||
                                  (v.Major == major && v.Minor < minor) ||
                                  (v.Major == major && v.Minor == minor && v.Build <= build))
                      .Last().Version_Id;

完整的代码和测试：

public ProductVersions[]verArray = {
      new ProductVersions() { Version_Id = 1, Major = 3, Minor = 0, Build = 1 },
      new ProductVersions() { Version_Id = 2, Major = 4, Minor = 10, Build = 5 },
      new ProductVersions() { Version_Id = 3, Major = 6, Minor = 2, Build = 1 },
      new ProductVersions() { Version_Id = 4, Major = 7, Minor = 5, Build = 1 },
      new ProductVersions() { Version_Id = 5, Major = 10, Minor = 7, Build = 1 },
      new ProductVersions() { Version_Id = 6, Major = 11, Minor = 10, Build = 10 },
  };

void Main()
{
  string test = "10.10.20";
  Console.WriteLine(test + " gives "+GetVersion(test));

  test = "7.0.0";
  Console.WriteLine(test + " gives "+GetVersion(test));

  test = "7.5.1";
  Console.WriteLine(test + " gives "+GetVersion(test));
}

private int GetVersion(string versionNumber)
{
  string [] input = versionNumber.Split(".".ToCharArray());

  int major = int.Parse(input[0]);
  int minor = int.Parse(input[1]);
  int build = int.Parse(input[2]);

  // assume verArray is already ordered (this would need to be sorted otherwise.
  int result = verArray.Where(v => (v.Major < major) ||
                                   (v.Major == major && v.Minor < minor) ||
                                   (v.Major == major && v.Minor == minor && v.Build <= build))
                       .Last().Version_Id;

  return result;
}

public class ProductVersions
{
  public int Version_Id { get; set; }
  public int Major { get; set; }
  public int Minor { get; set; }
  public int Build { get; set; }
}

返回以下内容：

10.10.20 gives 5
7.0.0 gives 3
7.5.1 gives 4

Answer 3

我刚检查了你给的样品。这段代码适合我。

private int getver(int m, int n, int b)
{
    List<ProductVersions> pv = new List<ProductVersions>();
    pv.Add(new ProductVersions { Version_Id = 3, Major = 6, Minor = 2, Build = 1 });
    pv.Add(new ProductVersions { Version_Id = 4, Major = 7, Minor = 5, Build = 1 });
    pv.Add(new ProductVersions { Version_Id = 5, Major = 10, Minor = 7, Build = 1 });
    pv.Add(new ProductVersions { Version_Id = 6, Major = 11, Minor = 10, Build = 10 });

    int mm = m;
    if (m == 0)
        mm = int.MaxValue;
    int nn = n;
    if (n == 0)
        nn = int.MaxValue;
    int bb = b;
    if (b == 0)
        bb = int.MaxValue;
    var v = pv.FindAll(mj => mj.Major <= m).FindAll(mn => n == 0 ? mn.Major <= mm - 1 && mn.Minor <= nn : mn.Minor <= n).FindAll(bl => b == 0 ? bl.Minor <= nn - 1 && bl.Build <= bb : bl.Build <= b).Last().Version_Id;

    return v;
}

我正在尝试根据主要，次要和构建级别的标准缩小列表并获取列表的最后一个实体。这里我假设列表是根据这些值排序的。

Answer 4

我发现这是一种快速而简单的方法，但有一些限制。

    var v = from p in pv
            select new { version = p.Version_Id, val = p.Major * 100000000 + p.Minor * 10000 + p.Build };
    int vver=v.ToList().FindAll(pp => pp.val <= m * 100000000 + n * 10000 + b).Last().version;

从可用产品版本列表中找到最匹配的产品版本

4 个答案: