前面的文章封装了查询条件 自己去组装条件,但是对 And Or 这种组合支持很差,但是也不是不能支持,只是要写更多的代码看起来很臃肿
根据 Where(Expression<Func<T, bool>>) 我们直接来处理这个,在处理这个之前其实看了下
Expression这个对象的处理,本生里面是包含了 AndAlso 、 Or 的处理 先来看看这个会遇到什么问题?为什么不行?
比如:
Expression.AndAlso(first,second)
来一段之前的扩展
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))); }
本质上没什么不同,最后连接都能拿到相关的表达式
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>()); }
按照如上写法多写几个条件,2个查询条件,2个值,感觉没问题, 但是运行到Where的时候报错误 表到时Parameter参数的个数对应不上表达式参数的个数,参数丢失了?
参数的值跟随表达式,在组合的时候需要重新组合参数,如果直接组合表达式,参数不会发生变化所以需要处理下参数问题,对(Expression<Func<T, bool>>) 的扩展就迎刃而解了
正确的处理方式:
public static class ExpressionExtensions { /// <summary> /// 添加And条件 /// </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条件 /// </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> /// 合并表达式以及参数 /// </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); } } }
使用方法就简单多了
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)); }
这里值得注意的是 一定要重新赋值到 filter ,按理说扩展了Expression<Func<T, bool>> 也返回了 Expression<Func<T, bool>> 好像可以不用重新赋值,然而这里并不是这样
如果我们直接
filter.And(c => c.UserName.Contains(username));
这样添加 会发现之中都是第一个参数的条件 都是 true,这是为什么呢?
Expression<Func<IdentityUser, bool>> filter = u => true;
下面看下这段代码其实给之前出现错误的原因是一样的?
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)); 是对 T 类中做的反射,本生合并两个带 T 的应该是没问题的,只是因为
与 Expression<Func<IdentityUser, bool>> filter = u => true; 组合后
Expression.Lambda<Func<T, bool>>( func(left, right), parameter);
一直都是True,导致最后的条件都是返回 True 查询条件就无效了,所以需要重新引用赋值 filter