Javascript 节点回调-为什么不起作用?
我目前正在尝试理解节点和回调,并尝试了该代码的各种解决方案以使其工作,但是param2返回时未定义。有人能告诉我为什么吗?如何解决这个问题?谢谢Javascript 节点回调-为什么不起作用?,javascript,node.js,callback,Javascript,Node.js,Callback,我目前正在尝试理解节点和回调,并尝试了该代码的各种解决方案以使其工作,但是param2返回时未定义。有人能告诉我为什么吗?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(u
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);
在getPage函数中,定义param2并传递它。在getPage函数中,定义param2并传递它,这是因为在调用回调函数的范围中,
param2
未定义。要定义param2
返回,请确保在getpage()中初始化param2
这是因为在调用回调函数的范围中,param2
未定义。要定义param2
返回,请确保在getpage()中初始化param2
谢谢我现在觉得有点傻。我早该知道的!无论如何,再次谢谢你!谢谢我现在觉得有点傻。我早该知道的!无论如何,再次谢谢你!
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);