[英]How to set properties of a List<Class> in C#
我有一个这样定义的类:
[DataContract]
public class Response2
{
[DataMember(Name = "done")]
public bool done;
[DataMember(Name = "records")]
public List<Response3> r3entry;
}
[DataContract]
public class Response3
{
[DataMember(Name = "Id")]
public string strId { get; set; }
[DataMember(Name = "Name")]
public string strName { get; set; }
}
现在,我想发生的是从ANOTHER类中获取值并填充...类似这样:
string propertyRequest2 = CreatePropertyRequest2();
Response2 propResponse2 = MakeRequest2(propertyRequest2, sfToken);
List<Response> listAllData = new List<Response>();
foreach (var responseEntry in propResponse2.r3entry)
{
listAllData.Add(new Response() { strId = responseEntry.strId, strName = responseEntry.strName } );
// NOTE .strId IS ALWAYS UNIQUE IN BOTH CLASSES
// - I know this is NOT the right syntax... will fix later.
Where listAllData.strId = responseEntry.strId
{
listAllData.property = propertyResponse2(.strId=responseEntry.strId).property
}
}
我确信(至少)代码的最后部分对大多数阅读此书的人来说很痛苦,但我会修复,所以它并不那么糟糕。 我只是不知道解释是否清楚。 以防万一我错了,这里的意思更像是这样:
// WE HAVE A LIST OF CLASSES WITH PROPERTIES
// ASSUME PROPERTIES ARE ID, ITEM, NAME
LIST1 = { ("1", "A", "APPLE"), ("2", "B", "BANANA"), ("3", "C", "COCONUT")}
// NOW WE HAVE ANOTHER LIST THAT HAS THE SAME ID BUT DIFF DATA
// ASSUME PROPERTIES ARE ID, COLOR
LIST2 = { ("1", "RED"), ("2", "YELLOW"), ("3", "BROWN) }
// AND THEN I WANT TO CREATE A NEW LIST WITH BOTH SETS OF DATA COMBINED
// ASSUME PROPERTIES ARE ID, ITEM, NAME, COLOR
LIST3 = { ("1", "A", "APPLE", "RED"), ("2", "B", "BANANA"), ("3", "C", "COCONUT", "BROWN") }
有关如何执行此操作的任何想法?
内连投影:
var list = from l1 in LIST1
join l2 in LIST2 on l1.ID equals l2.ID
select new {
l1.ID,
l1.Item,
l1.Name,
l2.Color
}
.ToList()
不清楚确切的目标是什么,Zip函数作用于两个列表并返回这两个列表的乘积。 您可以在这里了解更多信息:
但是实际上,如果您有两个清单,可以说:
int[] numbers = { 1, 2, 3, 4 };
string[] words = { "one", "two", "three" };
// The following example concatenates corresponding elements of the
// two input sequences.
var numbersAndWords = numbers.Zip(words, (first, second) => first + " " + second);
因此,在您的情况下,可能是这样的:
Response2[] response2 = //some list
Response3[] response3= // some other list
var response2Andresponse3 = response2.Zip(response3, (res2,res3) => //something you want to do with them
这会将response2的第一个元素与response3的第一个元素配对,依此类推。 我们在这里还假设它们具有相同的长度,以便您没有未配对的属性。
您将必须找出一种将其发送到您可能会发现有用的东西的方法,但是您将获得一个列表,其中它们逐个元素地对应。
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