How do I get a ScriptEngine by language name and version?

This example show you how you can obtain a script engine for a specific language name and specific language version. In the code below we try to obtain script engine instance for ECMAScript version 1.6.

package org.kodejava.example.script;

import javax.script.ScriptEngineManager;
import javax.script.ScriptEngineFactory;
import javax.script.ScriptEngine;
import javax.script.ScriptException;
import java.util.List;

public class ScriptEngineSearch {
    public static void main(String[] args) {
        String languageName = "ECMAScript";
        String languageVersion = "1.6";

        //
        // Creating a ScriptEngineManager and get the list of available
        // engine factories.
        //
        ScriptEngineManager manager = new ScriptEngineManager();
        List factories = manager.getEngineFactories();

        //
        // We obtain a ScriptEngine from the available factories where
        // the language name is ECMAScript and the version is 1.6.
        // ECMAScript is the standard name for JavaScript programming
        // language. If we found the desired language we then get the
        // ScriptEngine by calling factory's getScriptEngine() method.
        //
        ScriptEngine engine = null;
        for (ScriptEngineFactory factory : factories) {
            String language = factory.getLanguageName();
            String version = factory.getLanguageVersion();

            if (language.equals(languageName)
                    && version.equals(languageVersion)) {
                engine = factory.getScriptEngine();
                break;
            }
        }

        if (engine != null) {
            try {
                engine.eval("print('Hello There')");
            } catch (ScriptException e) {
                e.printStackTrace();
            }
        }
    }
}

Wayan

Programmer, runner, recreational diver, live in the island of Bali, Indonesia. Mostly programming in Java, Spring Framework, Hibernate / JPA. Support me by donating.

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