的SelectMany [英] SelectMany
问题描述
你好，
请看这个样本。
- 创建w员工类：
public int id;
public string firstName;
public string lastName;
{
Employee [] emp = new Employee < font size = 2> [] {
new Employee < font size = 2> {id = 1，firstName = " Joe" ，lastName = < font color ="＃a31515"size = 2>" Rattz" }，
new 员工 {id = 2，firstName = " William" ; ，lastName = " Gates" }，
new Employee {id = 3，firstName = " Anders" ，lastName = " Hejlsberg" }，
new Employee {id = 3，firstName = " Anders" ，lastName = "Hejlsberg" }，
new 员工 {id = 4，firstName = " David" ，lastName = " Lightman" }，
new Employee {id = 101，firstName = " Kevin" ，lastName = " Flynn" }，
new 员工 {id = 102，firstName = " Michael" ，lastName = " Bolton" }，
};
return （emp）;
}
- 创建EmployeeOptionEntry类：
public int id;
public long optionsCount;
public DateTime dateAwarded;
public static EmployeeOptionEntry [] GetEmployeeOptionEntries（）
{
EmployeeOptionEntry [] empOptions = new EmployeeOptionEntry [] {
new EmployeeOptionEntry {id = 1，optionsCount = 2，dateAwarded = DateTime .Parse（ " 1999/12/31" ）}，< br> new EmployeeOptionEntry {id = 2 ，optionsCount = 10000，dateAwarded = DateTime .Parse（ " 1992/06/30" ）}，
new EmployeeOptionEntry {id = 2，optionsCount = 60000，dateAwarded = < font color ="＃2b91af"size = 2> DateTime .Parse（ " 1994/01/01" ）}，
new EmployeeOptionEntry {id = 3，optionsCount = 5000，dateAwarded = DateTime .Parse（ " 1997/09/30" ）}，
new EmployeeOptionEntry {id = 2，optionsCount = 10000，dateAwarded = < font color ="＃2b91af"size = 2> DateTime .Parse（ " 2003/04/04" ）}，
new EmployeeOptionEntry {id = 3，optionsCount = 7500，dateAwarded = DateTime 。解析（ " 1998/09/30" ）}，
new EmployeeOptionEntry {id = 3，optionsCount = 7500，dateAwarded = DateTime .Parse（ "1998/09/30" ; ）}，
new EmployeeOptionEntry {id = 4，optionsCount = 1500，dateAwarded = DateTime 。解析（ " 1997/12/31" ）}，
new EmployeeOptionEntry {id = 101，optio nsCount = 2，dateAwarded = DateTime .Parse（ " 1998/12/31" ）}
};
return （empOptions）;
}
- 在Main类中：
private static void Main（ string [] args）
{
员工 [] employee = 员工 .GetEmployeesArray（）;
EmployeeOptionEntry [] employeeOption = EmployeeOptionEntry .GetEmployeeOptionEntries（）;
{
id = e.id，
name = string .Format（ " ; {0} {1}" ，e.firstName，e.lastName），
options = o！= null ？o.optionsCount：0}））。SelectMany（r => r） ;
< font color ="＃2b91af"size = 2>控制台 .WriteLine（item）;
}
- 它给了我这个结果：
{id = 1，name = joe Rattaz，option = 2}
{id = 2 ，name = William Gates，option = 10000}
{id = 2，name = William Gates，option = 10000}
{id = 2，name = William Gates，option = 60000}
{id = 3，name = Anders Hejlsberg，option = 5000}
{id = 3，name = Anders Hejlsberg，option = 7500}
{id = 3，name = Anders Hejlsberg，option = 7500}
{id = 3，name = Anders Hejlsberg，option = 5000}
{id = 3，name = Anders Hejlsberg，option = 7500}
{id = 3，na me = Anders Hejlsberg，选项= 7500}
{id = 4，name = David Lighman，option = 2}
{id = 101，name = Kevyn Flynn，option = 2}
但我不知道结果。因此，任何人都可以将问题描述给我以获得此结果。
>解决方案
你好赫马，很抱歉，如果我不理解你的问题。要获得上面的结果，您可以按如下方式编写LINQ查询：
在o上的employeeOptions中加入o .Id等于e.Id
选择新的{e.Id， 名称 = e.FirstName + QUOT; " + e.LastName，o.OptionsCount}
Hello ,
Plase look at this sample .
- create w Employee class :
public int id;
public string firstName;
public string lastName; public static Employee[] GetEmployeesArrayList()
{
Employee[] emp = new Employee[] {
new Employee { id = 1, firstName = "Joe", lastName = "Rattz" },
new Employee { id = 2, firstName = "William", lastName = "Gates" },
new Employee { id = 3, firstName = "Anders", lastName = "Hejlsberg" },
new Employee { id = 3, firstName = "Anders", lastName = "Hejlsberg" },
new Employee { id = 4, firstName = "David", lastName = "Lightman" },
new Employee { id = 101, firstName = "Kevin", lastName = "Flynn" },
new Employee {id = 102,firstName = "Michael",lastName = "Bolton" },
};
return (emp);
}
- create EmployeeOptionEntry class :
public int id;
public long optionsCount;
public DateTime dateAwarded;public static EmployeeOptionEntry[] GetEmployeeOptionEntries()
{
EmployeeOptionEntry[] empOptions = new EmployeeOptionEntry[] {
new EmployeeOptionEntry {id = 1,optionsCount = 2,dateAwarded = DateTime.Parse("1999/12/31") },
new EmployeeOptionEntry {id = 2,optionsCount = 10000,dateAwarded = DateTime.Parse("1992/06/30") },
new EmployeeOptionEntry {id = 2,optionsCount = 60000,dateAwarded = DateTime.Parse("1994/01/01") },
new EmployeeOptionEntry {id = 3,optionsCount = 5000,dateAwarded = DateTime.Parse("1997/09/30") },
new EmployeeOptionEntry {id = 2,optionsCount = 10000,dateAwarded = DateTime.Parse("2003/04/01") },
new EmployeeOptionEntry {id = 3,optionsCount = 7500,dateAwarded = DateTime.Parse("1998/09/30") },
new EmployeeOptionEntry {id = 3,optionsCount = 7500,dateAwarded = DateTime.Parse("1998/09/30") },
new EmployeeOptionEntry {id = 4,optionsCount = 1500,dateAwarded = DateTime.Parse("1997/12/31") },
new EmployeeOptionEntry {id = 101,optionsCount = 2,dateAwarded = DateTime.Parse("1998/12/31") }
};
return (empOptions);
}- in the Main class :
private static void Main(string[] args)
{
Employee[] employee = Employee.GetEmployeesArray();
EmployeeOptionEntry[] employeeOption = EmployeeOptionEntry.GetEmployeeOptionEntries();
var employeeOptions1 = employee.GroupJoin(employeeOption, e => e.id, o => o.id, (e, os) => os.Select(o => new
{
id = e.id,
name = string.Format("{0} {1}", e.firstName, e.lastName),
options = o != null ? o.optionsCount : 0})).SelectMany(r => r); foreach (var item in employeeOptions1)
Console.WriteLine(item);}
- it gives me this result :
{ id=1 , name = joe Rattaz , option =2}
{ id=2 , name = William Gates , option =10000}
{ id=2 , name = William Gates , option =10000}
{ id=2 , name = William Gates , option =60000}
{ id=3 , name = Anders Hejlsberg, option =5000}
{ id=3 , name = Anders Hejlsberg, option =7500}
{ id=3 , name = Anders Hejlsberg, option =7500}
{ id=3 , name = Anders Hejlsberg, option =5000}
{ id=3 , name = Anders Hejlsberg, option =7500}
{ id=3 , name = Anders Hejlsberg, option =7500}
{ id=4 , name = David Lighman, option =2}
{ id=101 , name = Kevyn Flynn, option =2}
But I dont know the result . So could anyone descripe me the quesrto to get this result .
thanks
hema
ASk解决方案Hello hema, Sorry if I did not understand your question well. To get the results above you could write a LINQ query as follows:
from e in employees join o in employeeOptions on o.Id equals e.Id select new {e.Id, Name = e.FirstName + " " + e.LastName, o.OptionsCount}
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