前面的文章封裝了查詢條件 自己去組裝條件,但是對 And Or 這種組合支持很差,但是也不是不能支持,只是要寫更多的代碼看起來很臃腫
根據 Where(Expression<Func<T, bool>>) 我們直接來處理這個,在處理這個之前其實看了下
Expression這個對象的處理,本生裏面是包含了 AndAlso 、 Or 的處理 先來看看這個會遇到什麼問題?爲什麼不行?
比如:
Expression.AndAlso(first,second)
來一段之前的擴展
public static Expression AndExpression(this Expression expression, Expression right)
{
return Expression.AndAlso(expression, right);
}
public static Expression OrExpression(this Expression expression, Expression right)
{
return Expression.Or(expression, right);
}
public static Expression<Func<T,bool>> ToFilter<T>(this Expression expression)
{
return Expression.Lambda<Func<T, bool>>(expression, Expression.Parameter(typeof(T)));
}
本質上沒什麼不同,最後連接都能拿到相關的表達式
Expression filter= Expression.Constant(true, typeof(bool));
if (!string.IsNullOrEmpty(username))
{
filter = filter.AndExpression(new UosoConditions {
Key = "UserName",
Operator = UosoOperatorEnum.Contains,
Value = username,
ValueType = "string"
}.Parser<IdentityUser>());
}
按照如上寫法多寫幾個條件,2個查詢條件,2個值,感覺沒問題, 但是運行到Where的時候報錯誤 表到時Parameter參數的個數對應不上表達式參數的個數,參數丟失了?
參數的值跟隨表達式,在組合的時候需要重新組合參數,如果直接組合表達式,參數不會發生變化所以需要處理下參數問題,對(Expression<Func<T, bool>>) 的擴展就迎刃而解了
正確的處理方式:
public static class ExpressionExtensions
{
/// <summary>
/// 添加And條件
/// </summary>
/// <typeparam name="T"></typeparam>
/// <param name="first"></param>
/// <param name="second"></param>
/// <returns></returns>
public static Expression<Func<T, bool>> And<T>(
this Expression<Func<T, bool>> first,
Expression<Func<T, bool>> second)
{
return first.AndAlso<T>(second, Expression.AndAlso);
}
/// <summary>
/// 添加Or條件
/// </summary>
/// <typeparam name="T"></typeparam>
/// <param name="first"></param>
/// <param name="second"></param>
/// <returns></returns>
public static Expression<Func<T, bool>> Or<T>(
this Expression<Func<T, bool>> first,
Expression<Func<T, bool>> second)
{
return first.AndAlso<T>(second, Expression.OrElse);
}
/// <summary>
/// 合併表達式以及參數
/// </summary>
/// <typeparam name="T"></typeparam>
/// <param name="expr1"></param>
/// <param name="expr2"></param>
/// <param name="func"></param>
/// <returns></returns>
private static Expression<Func<T, bool>> AndAlso<T>(
this Expression<Func<T, bool>> expr1,
Expression<Func<T, bool>> expr2,
Func<Expression, Expression, BinaryExpression> func)
{
var parameter = Expression.Parameter(typeof(T));
var leftVisitor = new ReplaceExpressionVisitor(expr1.Parameters[0], parameter);
var left = leftVisitor.Visit(expr1.Body);
var rightVisitor = new ReplaceExpressionVisitor(expr2.Parameters[0], parameter);
var right = rightVisitor.Visit(expr2.Body);
return Expression.Lambda<Func<T, bool>>(
func(left, right), parameter);
}
private class ReplaceExpressionVisitor
: ExpressionVisitor
{
private readonly Expression _oldValue;
private readonly Expression _newValue;
public ReplaceExpressionVisitor(Expression oldValue, Expression newValue)
{
_oldValue = oldValue;
_newValue = newValue;
}
public override Expression Visit(Expression node)
{
if (node == _oldValue)
return _newValue;
return base.Visit(node);
}
}
}
使用方法就簡單多了
Expression<Func<IdentityUser, bool>> filter = u => true;
if (!string.IsNullOrEmpty(username))
{
filter = filter.And(c => c.UserName.Contains(username));
}
if (!string.IsNullOrEmpty(phone))
{
filter = filter.And(c => c.PhoneNumber.Contains(phone));
}
if (!string.IsNullOrEmpty(email))
{
filter = filter.And(c => c.Email.Contains(email));
}
這裏值得注意的是 一定要重新賦值到 filter ,按理說擴展了Expression<Func<T, bool>> 也返回了 Expression<Func<T, bool>> 好像可以不用重新賦值,然而這裏並不是這樣
如果我們直接
filter.And(c => c.UserName.Contains(username));
這樣添加 會發現之中都是第一個參數的條件 都是 true,這是爲什麼呢?
Expression<Func<IdentityUser, bool>> filter = u => true;
下面看下這段代碼其實給之前出現錯誤的原因是一樣的?
private static Expression<Func<T, bool>> AndAlso<T>(
this Expression<Func<T, bool>> expr1,
Expression<Func<T, bool>> expr2,
Func<Expression, Expression, BinaryExpression> func)
{
var parameter = Expression.Parameter(typeof(T));
var leftVisitor = new ReplaceExpressionVisitor(expr1.Parameters[0], parameter);
var left = leftVisitor.Visit(expr1.Body);
var rightVisitor = new ReplaceExpressionVisitor(expr2.Parameters[0], parameter);
var right = rightVisitor.Visit(expr2.Body);
return Expression.Lambda<Func<T, bool>>(
func(left, right), parameter);
}
var parameter = Expression.Parameter(typeof(T)); 是對 T 類中做的反射,本生合併兩個帶 T 的應該是沒問題的,只是因爲
與 Expression<Func<IdentityUser, bool>> filter = u => true; 組合後
Expression.Lambda<Func<T, bool>>( func(left, right), parameter);
一直都是True,導致最後的條件都是返回 True 查詢條件就無效了,所以需要重新引用賦值 filter