我在尝试不使用 .Compile() 的情况下从表达式树中获取对象的值时遇到问题
这个对象很简单。
var userModel = new UserModel { Email = "John@Doe.com"};
给我问题的方法如下所示。
private void VisitMemberAccess(MemberExpression expression, MemberExpression left)
{
var key = left != null ? left.Member.Name : expression.Member.Name;
if (expression.Expression.NodeType.ToString() == "Parameter")
{
// add the string key
_strings.Add(string.Format("[{0}]", key));
}
else
{
// add the string parameter
_strings.Add(string.Format("@{0}", key));
// Potential NullReferenceException
var val = (expression.Member as FieldInfo).GetValue((expression.Expression as ConstantExpression).Value);
// add parameter value
Parameters.Add("@" + key, val);
}
}
我正在运行的测试非常简单
[Test] // PASS
public void ShouldVisitExpressionByGuidObject ()
{
// Setup
var id = new Guid( "CCAF57D9-88A4-4DCD-87C7-DB875E0D4E66" );
const string expectedString = "[Id] = @Id";
var expectedParameters = new Dictionary<string, object> { { "@Id", id } };
// Execute
var actualExpression = TestExpression<UserModel>( u => u.Id == id );
var actualParameters = actualExpression.Parameters;
var actualString = actualExpression.WhereExpression;
// Test
Assert.AreEqual( expectedString, actualString );
CollectionAssert.AreEquivalent( expectedParameters, actualParameters );
}
[Test] // FAIL [System.NullReferenceException : Object reference not set to an instance of an object.]
public void ShouldVisitExpressionByStringObject ()
{
// Setup
var expectedUser = new UserModel {Email = "john@doe.com"};
const string expectedString = "[Email] = @Email";
var expectedParameters = new Dictionary<string, object> { { "@Email", expectedUser.Email } };
// Execute
var actualExpression = TestExpression<UserModel>( u => u.Email == expectedUser.Email );
var actualParameters = actualExpression.Parameters;
var actualString = actualExpression.WhereExpression;
// Assert
Assert.AreEqual( expectedString, actualString );
CollectionAssert.AreEquivalent( expectedParameters, actualParameters );
}
我应该注意改变
var val = (expression.Member as FieldInfo).GetValue((expression.Expression as ConstantExpression).Value);
到
var val = Expression.Lambda( expression ).Compile().DynamicInvoke().ToString();
将允许测试通过,但是此代码需要在 iOS 上运行,因此不能使用 .Compile()
最佳答案
TLDR;
只要您不使用 Emit
或 Compile
,就可以使用反射。在问题中,正在为 FieldInfo
提取值,但没有为 PropertyInfo
提取值。确保您可以同时获得两者。
if ((expression.Member as PropertyInfo) != null)
{
// get the value from the PROPERTY
}
else if ((expression.Member as FieldInfo) != null)
{
// get the value from the FIELD
}
else
{
throw new InvalidMemberException();
}
长篇大论
所以评论为我指明了正确的方向。我在获取 PropertyInfo 时遇到了一些困难,但最终,这就是我想出的。
private void VisitMemberAccess(MemberExpression expression, MemberExpression left)
{
// To preserve Case between key/value pairs, we always want to use the LEFT side of the expression.
// therefore, if left is null, then expression is actually left.
// Doing this ensures that our `key` matches between parameter names and database fields
var key = left != null ? left.Member.Name : expression.Member.Name;
// If the NodeType is a `Parameter`, we want to add the key as a DB Field name to our string collection
// Otherwise, we want to add the key as a DB Parameter to our string collection
if (expression.Expression.NodeType.ToString() == "Parameter")
{
_strings.Add(string.Format("[{0}]", key));
}
else
{
_strings.Add(string.Format("@{0}", key));
// If the key is being added as a DB Parameter, then we have to also add the Parameter key/value pair to the collection
// Because we're working off of Model Objects that should only contain Properties or Fields,
// there should only be two options. PropertyInfo or FieldInfo... let's extract the VALUE accordingly
var value = new object();
if ((expression.Member as PropertyInfo) != null)
{
var exp = (MemberExpression) expression.Expression;
var constant = (ConstantExpression) exp.Expression;
var fieldInfoValue = ((FieldInfo) exp.Member).GetValue(constant.Value);
value = ((PropertyInfo) expression.Member).GetValue(fieldInfoValue, null);
}
else if ((expression.Member as FieldInfo) != null)
{
var fieldInfo = expression.Member as FieldInfo;
var constantExpression = expression.Expression as ConstantExpression;
if (fieldInfo != null & constantExpression != null)
{
value = fieldInfo.GetValue(constantExpression.Value);
}
}
else
{
throw new InvalidMemberException();
}
// Add the Parameter Key/Value pair.
Parameters.Add("@" + key, value);
}
}
本质上,如果 Member.NodeType
是一个 Parameter
,那么我将把它用作 SQL 字段。 [字段名]
否则,我将它用作 SQL 参数 @FieldName
...我知道倒退。
如果 Member.NodeType
不是参数,那么我会检查它是模型 Field
还是模型 Property
。从那里,我获得了适当的值,并将键/值对添加到字典中以用作 SQL 参数。
最终结果是我构建了一个看起来像这样的字符串
SELECT * FROM TableName WHERE
[FieldName] = @FieldName
然后传递参数
var parameters = new Dictionary<string, object> Parameters;
parameters.Add("@FieldName", "The value of the field");
关于c# - 如何在没有 .Compile() 的情况下从 MemberExpression 获取属性值?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/19718560/