import javax.script.ScriptEngine;
import javax.script.ScriptEngineManager;
import java.io.FileReader;
public class Main {
public static void main(String[] args) {
ScriptEngineManager manager = new ScriptEngineManager();
ScriptEngine engine = manager.getEngineByName("js");
try {
FileReader reader = new FileReader("C:/yourfile.js");
engine.put("urlfromjava", "http://www.something.com/?asvb");
engine.eval(reader);
reader.close();
} catch (Exception e) {
e.printStackTrace();
}
}
}
眼下,yourfile.js包含此行执行JavaScript在Java中 - 打开一个URL,并获取链接
function urlget(url)
{
print("URL:"+url);
var loc = window.open(url);
var link = document.getElementsByTagName('a')["61"].href;
return ("\nLink is: \n"+link);
}
var x = urlget(urlfromjava);
print(x);
我得到的错误
"javax.script.ScriptException: sun.org.mozilla.javascript.internal.EcmaError: ReferenceError: "window" is not defined"
如何打开一个URL并从java获得它的链接?
哇!看起来像我已经回答了相同的:) – Tapos 2011-05-22 09:54:21
“伟大的头脑相似” – iruediger 2011-05-22 11:40:44
我喜欢的答案,除了,w3schools是像维基百科或随机的网页搜索结果一样多的“文档”。所以这个答案的前两行是不正确的。 – Kit10 2014-01-09 15:22:53