之前我在检索数据服务器端的数量时询问了一个问题,并在网站上提供了解决方案.建议是使用linq,但是因为我比较新,所以我需要一点深入的帮助.
使用John的解决方案:
class Guy
{
public int age; public string name;
public Guy( int age, string name ) {
this.age = age;
this.name = name;
}
}
class Program
{
static void Main( string[] args ) {
var GuyArray = new Guy[] {
new Guy(22,"John"),new Guy(25,"John"),new Guy(27,"John"),new Guy(29,"John"),new Guy(12,"Jack"),new Guy(32,"Jack"),new Guy(52,"Jack"),new Guy(100,"Abe")};
var peeps = from f in GuyArray group f by f.name into g select new { name = g.Key, count = g.Count() };
foreach ( var record in peeps ) {
Console.WriteLine( record.name + " : " + record.count );
}
}
}
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我可以按照约翰的建议使用上面的约翰,杰克和安倍的事件计数.但是如果问题稍微复杂一点怎么办呢
var GuyArray = new Guy[] {
new Guy(22,"John", "happy"),new Guy(25,"John", "sad"),new Guy(27,"John", "ok"),
new Guy(29,"John", "happy"),new Guy(12,"Jack", "happy"),new Guy(32,"Jack", "happy"),
new Guy(52,"Jack", "happy"),new Guy(100,"Abe", "ok")};
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上面的代码非常适合检索不同名称的出现次数,但如果我需要名称的出现次数以及每个人的快乐,悲伤或确定的出现次数,该怎么办呢.即输出是:名称,名称计数,满意的名字计数,悲伤的名称数,可以的名称数.如果linq不是最好的解决方案,我准备听取所有替代方案.非常感谢您的帮助.
坦率地说,目前尚不清楚你是否想要幸福的总人数,或者想要通过名字感到满意的总人数(也是为了悲伤,好吧).我会给你一个解决方案,可以给你两个.
var nameGroups = from guy in GuyArray
group guy by guy.name into g
select new {
name = g.Key,
count = g.Count(),
happy = g.Count(x => x.status == "happy"),
sad = g.Count(x => x.status == "sad"),
ok = g.Count(x => x.status == "ok")
};
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然后:
foreach(nameGroup in nameGroups) {
Console.WriteLine("Name = {0}, Count = {1}, Happy count = {2}, Sad count = {3}, Okay count = {4}", nameGroup.name, nameGroup.count, nameGroup.happy, nameGroup.sad, nameGroup.ok);
}
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如果你想要完全快乐,悲伤,好的计数,你可以说:
Console.WriteLine(nameGroups.Sum(nameGroup => nameGroup.happy));
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等等
另外,你应该做一个枚举
public enum Mood {
Happy,
Sad,
Okay
}
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然后
class Guy {
public int Age { get; set; }
public string Name { get; set; }
public Mood Mood { get; set; }
}
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这样你就可以写:
var people = from guy in guyArray
group guy by guy.Name into g
select new {
Name = g.Key,
Count = g.Count(),
HappyCount = g.Count(x => x.Mood == Mood.Happy),
SadCount = g.Count(x => x.Mood == Mood.Sad),
OkayCount = g.Count(x => x.Mood == Mood.Okay)
};
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