Question

我在尝试使用.Compile（）

从表达式树中获取对象的值时遇到问题

对象非常简单。

var userModel = new UserModel { Email = "John@Doe.com"};

给我问题的方法看起来像这样。

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，但此代码需要在iOS上运行，因此无法使用.Compile()

Answer 1

的 TLDR;
只要您没有使用Emit或Compile，就可以使用反射。在问题中，正在为FieldInfo提取值，但不会为PropertyInfo提取该值。确保你可以同时获得。

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(); }

冗长的版本

因此，评论指出了我正确的方向。获得PropertyInfo后我有点挣扎，但最后，这就是我想出来的。

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); } }

基本上，如果Member.NodeType是Parameter，那么我将把它用作SQL字段。 [FieldName]

否则，我将它用作SQL参数@FieldName ...我知道后退。

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，我得到适当的值，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类似于
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");

如何在没有.Compile（）的情况下从MemberExpression获取属性值？

1 个答案: