Question

使用.Contains（）/。StartsWith（）/。EndsWith（）生成的SQL如下所示：

(( NVL(INSTR(x, y), 0) ) = 1)

有没有办法改用：

LIKE 'x%' or '%x%' or '%x'

因为在查询的执行计划中这两者之间存在巨大的成本差异（44 000 vs 30）。

Answer 1

当我环顾abit时，我发现了LIKE operator in LINQ，其中有一些很好的例子说明了如何做到这一点。我已经测试了下面的链接来自上面的链接

以下是 adobrzyc

发布的使用与lambda相同的扩展程序

    public static class LinqEx
    {
        private static readonly MethodInfo ContainsMethod = typeof(string).GetMethod("Contains");
        private static readonly MethodInfo StartsWithMethod = typeof(string).GetMethod("StartsWith", new[] { typeof(string) });
        private static readonly MethodInfo EndsWithMethod = typeof(string).GetMethod("EndsWith", new[] { typeof(string) });

        public static Expression<Func<TSource, bool>> LikeExpression<TSource, TMember>(Expression<Func<TSource, TMember>> property, string value)
        {
            var param = Expression.Parameter(typeof(TSource), "t");
            var propertyInfo = GetPropertyInfo(property);
            var member = Expression.Property(param, propertyInfo.Name);

            var startWith = value.StartsWith("%");
            var endsWith = value.EndsWith("%");

            if (startWith)
                value = value.Remove(0, 1);

            if (endsWith)
                value = value.Remove(value.Length - 1, 1);

            var constant = Expression.Constant(value);
            Expression exp;

            if (endsWith && startWith)
            {
                exp = Expression.Call(member, ContainsMethod, constant);
            }
            else if (startWith)
            {
                exp = Expression.Call(member, EndsWithMethod, constant);
            }
            else if (endsWith)
            {
                exp = Expression.Call(member, StartsWithMethod, constant);
            }
            else
            {
                exp = Expression.Equal(member, constant);
            }

            return Expression.Lambda<Func<TSource, bool>>(exp, param);
        }

        public static IQueryable<TSource> Like<TSource, TMember>(this IQueryable<TSource> source, Expression<Func<TSource, TMember>> parameter, string value)
        {
            return source.Where(LikeExpression(parameter, value));
        }

        private static PropertyInfo GetPropertyInfo(Expression expression)
        {
            var lambda = expression as LambdaExpression;
            if (lambda == null)
                throw new ArgumentNullException("expression");

            MemberExpression memberExpr = null;

            switch (lambda.Body.NodeType)
            {
                case ExpressionType.Convert:
                    memberExpr = ((UnaryExpression)lambda.Body).Operand as MemberExpression;
                    break;
                case ExpressionType.MemberAccess:
                    memberExpr = lambda.Body as MemberExpression;
                    break;
            }

            if (memberExpr == null)
                throw new InvalidOperationException("Specified expression is invalid. Unable to determine property info from expression.");


            var output = memberExpr.Member as PropertyInfo;

            if (output == null)
                throw new InvalidOperationException("Specified expression is invalid. Unable to determine property info from expression.");

            return output;
        }
    }

要使用它，您只需添加Like函数即可放置Contains函数。您可以在下面看到示例

            using (CustomerEntities customerContext = new CustomerEntities())
            {
                IQueryable<Customer> customer = customerContext.Customer.Like(x => x.psn, "%1%");    
            }

这将创建一个类似于此的SQL查询。

SELECT 
[Extent1].[psn] AS [psn]
FROM [dbo].[Customer] AS [Extent1]
WHERE [Extent1].[psn] LIKE '%1%'

LINQ使用“like”而不是“（（NVL（INSTR（x，y），0））= 1）”

1 个答案: