从对象列表中创建逗号分隔列表 [英] Create comma separated list from list of objects

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问题描述

我有对象的列表类型Person。这个类有许多属性,我需要他们所有的逗号分隔的列表,所以我可以使用它以后的Csv文件。

I have list of objects type Person. This class has many properties and I need them all in a comma separated list so I can use it later for Csv file.

我已经管理这个与foreach和添加每个属性,用逗号手动分隔它等。

I've managed this with foreach and adding each property, separating it with commas manual, etc.

const string commaSeparator = ",";
foreach (var item in individualsInformation)
{
    csv.AppendLine(item.ReferenceNumber + commaSeparator + item.FirstName + commaSeparator +
                   item.Surname + commaSeparator + item.MiddleName + commaSeparator +
                   item.Address1 + commaSeparator + item.Address2 + commaSeparator + 
                   item.Address3 + commaSeparator + item.Address4 + commaSeparator + 
                   item.City + commaSeparator + item.PostalCode + commaSeparator +
                   item.Country + commaSeparator + item.DateOfBirth.ToString() + commaSeparator +
                   item.ID + commaSeparator + item.Gender + commaSeparator +
                   item.Component + commaSeparator + item.NationalID + commaSeparator + 
                   item.SubSystemID + commaSeparator + item.System);
}

然后我意识到有很多有效的方法,通过使用字符串。加入

Then I've realized that there is much efficient way, by using string.Join

这当然不行:

string joined = string.Join(",", listOfPersons);

如果我选择这样的属性:

And if I go by selecting property like this:

string joined = string.Join(",", listOfPersons(x => x.Id);

当然只有那个属性的逗号分隔列表。

I get comma separated list only for that property of course.

推荐答案

如果可能,我会避免反射。

I would avoid reflection if possible.

您可以轻松实现编译时安全和可读代码:

You can achieve it with compile time safety and readable code easily:

IEnumerable<string> personTexts = listOfPersons
    .Select(p => String.Join(",", p.ReferenceNumber, p.FirstName, p.Surname, ...));
string joined = String.Join(Environment.NewLine, personTexts);

您可以完全控制应使用哪个属性以及顺序。

You have full control over which property should be used and in which order.

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