需求是将一个string 表达式 转换成 逻辑 表达式 并得到结果。
例如: bool result = (key1==val1) || (key2!=val2) && (key3==val3). 其中 keyN, valN 均为变量。
基本的思路是先做Express string 验证,然后进行解析。
验证可以用正则表达式。
解析最基本的方法是用 表达式树 (Express Tree). PostFix.
C# linq 提供了一个解决方案就是 Linq 语法树。
https://www.codeproject.com/Articles/74018/How-to-Parse-and-Convert-a-Delegate-into-an-Expre
目前较好的方法是用Microsoft 的 Dynamic Linq (http://msdn.microsoft.com/en-us/vcsharp/bb894665.aspx)
static void Main(string[] args) { string[] exps = new string[] { @" ""keyA"" !=""valA"" ", @"keyA !=""valA""", @"(keyA ==""valA"")", @"keyA ==""val1"" || key2 ==""val2""", @"key1 ==""val1"" && key2 ==""val2""", @" key1 !=""val1"" || key2 ==""val2""&& key3 != ""val3""", @"(key1 ==""val1"" || key2 ==""val2"")&& key3 != ""val3""", }; string pattern = @"(?<key>[\w\d_]+)""*\s*(?<equality>==|!=)\s*""*(?<value>[\w\d_]+)"; foreach (var exp in exps) { // Parses to get the parameters. string formattedExp = exp; var matches = Regex.Matches(exp, pattern); ParameterExpression[] parameters = new ParameterExpression[matches.Count]; for (int i =0; i < matches.Count; i++) { Debug.Write(" key[" + matches[i].Groups["key"].Value); Debug.Write("] equality[" + matches[i].Groups["equality"].Value); Debug.Write("] value[" + matches[i].Groups["value"].Value +"]\t\n"); parameters[i] = Expression.Parameter(typeof(string), matches[i].Groups["key"].Value); // Removing " if it has in keys. if (exp.Contains(string.Format(@"""{0}""", matches[i].Groups["key"].Value))) { formattedExp = exp.Replace(string.Format(@"""{0}""", matches[i].Groups["key"].Value), matches[i].Groups["key"].Value); } } var e = System.Linq.Dynamic.DynamicExpression.ParseLambda(parameters, null, formattedExp); string[] GotValues = new string[matches.Count]; if (GotValues.Length == 2) { GotValues[0] = "val1"; GotValues[1] = "val2"; } if (GotValues.Length == 3) { GotValues[0] = "val1"; GotValues[1] = "val2"; GotValues[2] = "val3"; } var result = e.Compile().DynamicInvoke(GotValues); Debug.WriteLine("{0} ---> {1}", e.ToString(), result); } }
输出如下:
keyA => (keyA != "valA") ---> True
key[keyA] equality[!=] value[valA]
keyA => (keyA != "valA") ---> True
key[keyA] equality[==] value[valA]
keyA => (keyA == "valA") ---> False
key[keyA] equality[==] value[val1]
key[key2] equality[==] value[val2]
(keyA, key2) => ((keyA == "val1") OrElse (key2 == "val2")) ---> True
key[key1] equality[==] value[val1]
key[key2] equality[==] value[val2]
(key1, key2) => ((key1 == "val1") AndAlso (key2 == "val2")) ---> True
key[key1] equality[!=] value[val1]
key[key2] equality[==] value[val2]
key[key3] equality[!=] value[val3]
(key1, key2, key3) => ((key1 != "val1") OrElse ((key2 == "val2") AndAlso (key3 != "val3"))) ---> False
key[key1] equality[==] value[val1]
key[key2] equality[==] value[val2]
key[key3] equality[!=] value[val3]
(key1, key2, key3) => (((key1 == "val1") OrElse (key2 == "val2")) AndAlso (key3 != "val3")) ---> False
参考:
https://stackoverflow.com/questions/821365/how-to-convert-a-string-to-its-equivalent-linq-expression-tree