javascript - Node 回调 - 为什么这不起作用？

Question

我目前正在尝试了解 Node 和回调，并尝试了对此代码的各种解决方案以使其正常工作，但是 param2 返回未定义。有人能告诉我为什么吗？以及如何解决这个问题？谢谢!

function getPage(callback) {

  url = 'http://www.google.com';

  if (url) {
    url = url;
  } else {
    console.log('There was an error. No URL submitted');
  }

  callback(url, param2);
}

function CB(url, param2) {
console.log(`The URL of the page requested was ${url} and the added argument          was ${param2}`);
}

 getPage(CB);

Answer 1

这是因为 param2 在您调用回调函数的范围内未定义。要让 param2 重新定义，请确保 param2 已在 getpage()

中初始化

function getPage(callback) {

  url = 'http://www.google.com';
  // **make sure param2 is defined**
  var param2 = "param2 value"

  if (url) {
    url = url;
  } else {
    console.log('There was an error. No URL submitted');
  }

  callback(url, param2);
}

function CB(url, param2) {
console.log(`The URL of the page requested was ${url} and the added argument was ${param2}`);
}

 getPage(CB);