Question

我正在开发一个程序，希望用户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。解析和处理此类功能的最佳方法是什么？

我还没有发现任何此类问题的示例，解析器不必具有超高效率，它不会经常被调用，但主要是我想着重于良好的代码可读性和可伸缩性。

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

一种选择是使用Spring SPEL。但这会迫使您稍微更改表达式并使用Spring库：

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来实现自己的功能。

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";
        }

    }

无论您传递给StandardEvaluationContext的任何对象都应具有这些方法。我把它们放在同样运行表达式的同一个类中。

Answer 2

您可以使用以下内容：警告，我尚未测试。只是一些入门

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;
}

Java-解析用户定义的函数

2 个答案: