用Expression的嵌套类创建谓词

Question

我有这个:

public class Company
{
    public int Id { get; set; }
    public string Name { get; set; }
}
public class City
{
    public int Id { get; set; }
    public string Name { get; set; }
    public int ZipCode { get; set; }
}
public class Person
{
    public int Id { get; set; }
    public string FirstName { get; set; }
    public string LastName { get; set; }
    public int? Age { get; set; }
    public City City { get; set; }
    public Company Company { get; set; }
}

我想在某些情况下生成这样的谓词:

var result = listPerson.Where(x => x.Age == 10).ToList<>();

或者:

var result  = listPerson.Where( x => x.Company.Name == 1234).ToList();

或者:

var result  = listPerson.Where( x => x.City.ZipCode == "MyZipCode").ToList();

或者:

var result  = listPerson.Where( x => x.Company.Name == "MyCompanyName").ToList();

然后我创建了一个"PredicateBuilder"，这是工作(我得到的类型，如果可空或不，我建立谓词)，当我这样做:

BuildPredicate<Person>("Age", 10); I get this : x => x.Age == 10

但是我不知道如何管理当有一个嵌套的属性像这样:

BuildPredicate<Person>("City.ZipCode", "MyZipCode"); 
I'd like get this : x => x.City.ZipCode == "MyZipCode"

或者:

BuildPredicate<Person>("City.Name", "MyName"); 
I'd like get this : x => x.City.Name == "MyName"

或者:

BuildPredicate<Person>("Company.Name", "MyCompanyName"); 
I'd like get this : x => x.Company.Name == "MyCompanyName"

Answer 1

(不打算复制Jon - OP联系我提供答案)

下面的代码似乎可以正常工作:

static Expression<Func<T,bool>> BuildPredicate<T>(string member, object value) {
    var p = Expression.Parameter(typeof(T));
    Expression body = p;
    foreach (var subMember in member.Split('.')) {
        body = Expression.PropertyOrField(body, subMember);
    }
    return Expression.Lambda<Func<T, bool>>(Expression.Equal(
        body, Expression.Constant(value, body.Type)), p);
}

与Jon的答案之间唯一的功能区别是它通过告诉Expression.Constant期望的类型是什么来稍微更好地处理null 。作为用法示范:

static void Main() {
    var pred = BuildPredicate<Person>("City.Name", "MyCity");
    var people = new[] {
        new Person { City = new City { Name = "Somewhere Else"} },
        new Person { City = new City { Name = "MyCity"} },
    };
    var person = people.AsQueryable().Single(pred);
}

Answer 2

你只需要分割你的表达式点，然后迭代它，使用Expression.Property多次。像这样:

string[] properties = path.Split('.');
var parameter = Expression.Parameter(typeof(T), "x");
var lhs = parameter;
foreach (var property in properties)
{
    lhs = Expression.Property(lhs, property);
}
// I've assumed that the target is a string, given the question. If that's
// not the case, look at Marc's answer.
var rhs  = Expression.Constant(targetValue, typeof(string));
var predicate = Expression.Equals(lhs, rhs);
var lambda = Expression.Lambda<Func<T, bool>>(predicate, parameter);