Java

Question

我正在开发一个程序，我希望用户在其中定义一个简单的函数，例如 randomInt(0,10) 要么 randomString(10) 而不是静态参数。解析和处理此类函数的最佳方法是什么？

我还没有找到任何此类问题的示例，解析器不必非常高效，不会经常调用它，但主要是我想关注良好的代码可读性和可扩展性。

用户输入示例：

"This is user randomString(5) and he is randomInt(18,60) years old!"

预期输出：

"This is user phiob and he is 45 years old!"

"This is user sdfrt and he is 30 years old!"

Answer 1

您可以使用如下内容：警告，我还没有测试过。刚开始的东西

public String parseInput(String input){
    String[] inputArray = input.split(" ");
    String output = "";
    for(String in : inputArray){ //run through each word of the user input
        if(in.contains("randomString(")){ //if the user is calling randomString
            String params = in.replace("randomString(", ""); //strip away function to get to params
            params = in.replace("(", ""); //strip away function to get to params
            String[] paramsArray = params.split(",");  //these are string integers, and could be converted
            //send off these split apart parameters to your randomString method
            String out = randomString(paramsArray); //method parses string integers, outputs string
            output += out + " ";
        }else if(in.contains("randomInt(")){ //if the user is calling randomInt
            String params = in.replace("randomInt(", ""); //strip away function to get to params
            params = in.replace("(", ""); //strip away function to get to params
            String[] paramsArray = params.split(","); //these are string integers, and could be converted
            //send off these split apart parameters to your randomInt method
            String out = randomInt(paramsArray); //method parses string integers, outputs string
            output += out + " ";
        }else{ //if the user is just entering text
            output += in + " "; //concat the output with what the user wrote plus a space
        }

    }
    return output;
}

Answer 2

一种选择是使用Spring SPEL。但它迫使您稍微更改表达式并使用 Spring library:

表达式可以如下所示：

'This is user ' + randomString(5) + ' and he is ' + randomInt(18,60) + ' years old!'

或者这个：

This is user #{randomString(5)} and he is #{randomInt(18,60)} years old!

或者您可以通过自定义 TemplateParserContext.

来实现您自己的

这是代码：

import org.springframework.expression.Expression;
import org.springframework.expression.ExpressionParser;
import org.springframework.expression.spel.standard.SpelExpressionParser;
import org.springframework.expression.spel.support.StandardEvaluationContext;

public class SomeTest {

        @Test
        public void test() {
            ExpressionParser parser = new SpelExpressionParser();

            Expression exp = parser.parseExpression(
                "This is user #{randomString(5)} and he is #{randomInt(18,60)} years old!", 
                new TemplateParserContext() );

            //alternative
            //Expression exp = parser.parseExpression(
            //        "'This is user ' + randomString(5) + ' and he is ' + randomInt(18,60) + ' years old!'");
            //    String message = (String) exp.getValue( new StandardEvaluationContext(this) );


            String message = (String) exp.getValue( new StandardEvaluationContext(this) );
        }

        public String randomString(int i) {
            return "rs-" + i;
        }

        public String randomInt(int i, int j) {
            return "ri-" + i + ":" + "j";
        }

    }

您传递给 StandardEvaluationContext 的任何对象都应该具有这些方法。我将它们放在同样运行表达式的 class 中。

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