这个Twitter脚本目前可以在推文时发布所选用户的推文,但不是在推文下面实际回复,而是将其作为独立的新推文推文。如何让它实际回复而不是发新的推文?我使用的是Twit API:https://github.com/ttezel/twit
以下是我所拥有的:
console.log('The bot is starting');
var Twit = require('twit');
var config = require('./config');
var T = new Twit(config);
//Setting up a user stream
var stream = T.stream('user');
stream.on('tweet', tweetEvent);
function tweetEvent(eventMsg) {
var replyto = eventMsg.user.screen_name;
var text = eventMsg.text;
var from = eventMsg.user.screen_name;
console.log(replyto + ' '+ from);
if (replyto =='other user's handle') {
var newtweet = '@' + from + ' Hello!';
tweetIt(newtweet);
}
}
function tweetIt(txt) {
var tweet = {
status: txt
}
T.post('statuses/update', tweet, tweeted);
function tweeted(err, data, response) {
if (err) {
console.log("Something went wrong!");
} else {
console.log("It worked!");
}
}
}
答案 0 :(得分:1)
要使用Twitter API在推文会话中显示回复,您需要以下内容:
// the status update or tweet ID in which we will reply
var nameID = tweet.id_str;
还需要in_reply_to_status_id
状态中的参数tweet
。请参阅下面的代码更新,它现在应该保留对话:
console.log('The bot is starting');
var Twit = require('twit');
var config = require('./config');
var T = new Twit(config);
//Setting up a user stream
var stream = T.stream('user');
stream.on('tweet', tweetEvent);
function tweetEvent(tweet) {
var reply_to = tweet.in_reply_to_screen_name;
var text = tweet.text;
var from = tweet.user.screen_name;
var nameID = tweet.id_str;
// params just to see what is going on with the tweets
var params = {reply_to, text, from, nameID};
console.log(params);
if (reply_to === 'YOUR_USERNAME') {
var new_tweet = '@' + from + ' Hello!';
var tweet = {
status: new_tweet,
in_reply_to_status_id: nameID
}
T.post('statuses/update', tweet, tweeted);
function tweeted(err, data, response) {
if (err) {
console.log("Something went wrong!");
} else {
console.log("It worked!");
}
}
}
}