在LINQ中如何查詢條件不固定，如何合并兩個(gè)lambda表達(dá)式？其中一個(gè)方式是LINQ.Dynamic，關(guān)于LINQ.Dynamic的簡(jiǎn)單使用可以參考這篇文章，還有一種方法是利用Expression表達(dá)式樹，有關(guān)表達(dá)式樹的介紹，可以看這篇文章。

測(cè)試代碼如下：

public class Phone {public string Country { get; set; }public string City { get; set; }public string Name { get; set; } } public class Person {public string Name { get; set; }public string Gender { get; set; }public int Age { get; set; }public List<Phone> Phones { get; set; }} static void Main(string[] args) {List<Person> PersonLists = new List<Person>(){new Person { Name = "張三",Age = 20,Gender = "男",Phones = new List<Phone> {new Phone { Country = "中國(guó)", City = "北京", Name = "小米" },new Phone { Country = "中國(guó)",City = "北京",Name = "華為"},new Phone { Country = "中國(guó)",City = "北京",Name = "聯(lián)想"},new Phone { Country = "中國(guó)",City = "臺(tái)北",Name = "魅族"},}},new Person { Name = "松下",Age = 30,Gender = "男",Phones = new List<Phone> {new Phone { Country = "日本",City = "東京",Name = "索尼"},new Phone { Country = "日本",City = "大阪",Name = "夏普"},new Phone { Country = "日本",City = "東京",Name = "松下"},}},new Person { Name = "克里斯",Age = 40,Gender = "男",Phones = new List<Phone> {new Phone { Country = "美國(guó)",City = "加州",Name = "蘋果"},new Phone { Country = "美國(guó)",City = "華盛頓",Name = "三星"},new Phone { Country = "美國(guó)",City = "華盛頓",Name = "HTC"}}}};Expression<Func<Person, bool>> lambdaone = ex => ex.Name.Equals("張三");Expression<Func<Person, bool>> lambdatwo = ex => ex.Age == 30;ParameterExpression pa = Expression.Parameter(typeof(Person), "ex");Expression<Func<Person,bool>> newEx = Expression.Lambda<Func<Person,bool>>(Expression.Or(lambdaone.Body, lambdatwo.Body), pa);var Lists = PersonLists.Where(newEx.Compile());Console.WriteLine();Console.Read(); }

?在上述代碼中，我要構(gòu)建一個(gè)滿足兩個(gè)條件的過濾查詢：名字叫張三或者年齡是30。運(yùn)行結(jié)果如下：

結(jié)果報(bào)錯(cuò)了，說明沒有想象的那么簡(jiǎn)單。原因是即便我的命名是相同的，都是ex，但是兩個(gè)表達(dá)式本質(zhì)并不是相同的參數(shù)ex ,我只是引用了兩個(gè)不同parameter的lambda表達(dá)式的body，然后將這兩個(gè)body強(qiáng)行給了另一個(gè)parameter，具體信息可以這篇文藏

這時(shí)候我們需要重寫ExpressionVisitor，代碼如下：

public class MyExpressionVisitor : ExpressionVisitor{public ParameterExpression _Parameter { get; set; }public MyExpressionVisitor(ParameterExpression Parameter){_Parameter = Parameter;}protected override Expression VisitParameter(ParameterExpression p){return _Parameter;}public override Expression Visit(Expression node){return base.Visit(node);//Visit會(huì)根據(jù)VisitParameter()方法返回的Expression修改這里的node變量}}

修改調(diào)試代碼如下：

Expression<Func<Person, bool>> lambdaone = ex => ex.Name.Equals("張三"); Expression<Func<Person, bool>> lambdatwo = ex => ex.Age == 30; ParameterExpression pa = Expression.Parameter(typeof(Person), "ex"); MyExpressionVisitor visitor = new MyExpressionVisitor(pa);//統(tǒng)一參數(shù)類型 Expression bodyone = visitor.Visit(lambdaone.Body); Expression bodytwo = visitor.Visit(lambdatwo.Body); Expression<Func<Person, bool>> newEx = Expression.Lambda<Func<Person, bool>>(Expression.Or(bodyone, bodytwo), pa); var Lists = PersonLists.Where(newEx.Compile()); foreach (var List in Lists) {Console.WriteLine(List.Name); } Console.Read();

從運(yùn)行結(jié)果來看，是沒問題的

?

第二種方式是利用Expression.Invoke() ,代碼如下：

Expression<Func<Person, bool>> lambdaone = ex => ex.Name.Equals("張三"); Expression<Func<Person, bool>> lambdatwo = ex => ex.Age == 30; // 創(chuàng)建參數(shù)表達(dá)式 InvocationExpression invocation = Expression.Invoke(lambdaone, lambdatwo.Parameters.Cast<Expression>()); // 創(chuàng)建or運(yùn)算 BinaryExpression binary = Expression.Or(lambdatwo.Body, invocation); // 生成lambda表達(dá)式 var exp = Expression.Lambda<Func<Person, bool>>(binary, lambdatwo.Parameters); var Lists = PersonLists.Where(exp.Compile()); foreach (var List in Lists) {Console.WriteLine(List.Name); } Console.Read();

運(yùn)行結(jié)果相同：

?

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