【发布时间】:2020-04-30 20:23:28
【问题描述】:
我正在使用 Express 创建一个端点,以便我可以通过 API 调用来访问它。当我第一次进行搜索时,一切都很好,但如果我再做一次,我会得到上一次的结果加上新搜索的结果。如何让搜索结果每次都重置?
这里是实际端点的链接:(将“covid”一词更改为您喜欢的任何搜索词,如果您至少执行两次,即使您完成了新的搜索,也会显示上次搜索的数据搜索)
https://laffy.herokuapp.com/search/covid
非常感谢您提供的任何帮助!
这是调用 twitterRouter 并使用 app.use 在 /search/:searchTerm 创建端点的 app.js 文件:
app.js
const createError = require('http-errors');
const express = require('express');
const path = require('path');
const indexRouter = require('./routes/index');
const twitterRouter = require('./routes/twitterCall.js');
const top20 = require('./routes/twitterTop20.js');
const app = express();
app.set('views', path.join(__dirname, 'views'));
// app.set('port', process.env.PORT || 3001);
app.use(express.urlencoded({ extended: false }));
app.use(express.static(path.join(__dirname, 'public')));
app.use('/', indexRouter);
//creates route to use search at /search/
app.use('/search/:searchTerm', twitterRouter.search);
//creates route to access to get the top 20 Twitter hashtags trending
app.use('/top20', top20);
// catch 404 and forward to error handler
app.use(function(req, res, next) {
next(createError(404));
});
// error handler
app.use(function(err, req, res, next) {
// set locals, only providing error in development
res.locals.message = err.message;
res.locals.error = req.app.get('env') === 'development' ? err : {};
// render the error page
res.status(err.status || 500);
res.render('error');
});
module.exports = app;
我的印象是使用 res.send() 会结束 API 搜索,但似乎并没有结束。
然后是实际的 API 调用以及它为端点生成数据的位置:
twitterCall.js
//twitter file that searchs for tweets specified in params.q
var Twitter = require('twitter');
var config = require('../config/config.js');
var express = require('express');
var router = express.Router();
var T = new Twitter(config);
var locationsToSend = [];
exports.search = (req, res) => {
if (req.body == null) {
res.status(404).send( {
message: "Search can not be blank"
})
}
var params = {
q: req.params.searchTerm,
count: 1000,
result_type: 'recent',
lang: 'en'
}
//Initiate your search using the above parameters
T.get('search/tweets', params, function(err, data, response) {
//if there is no error, proceed
if(!err){
// Loop through the returned tweets
for(let i = 0; i < data.statuses.length; i++){
if (data.statuses[i].user.location!==null && data.statuses[i].user.location!=="") {
locationsToSend.push({
id: data.statuses[i].id_str,
createdAt: data.statuses[i].created_at,
text: data.statuses[i].text,
name: data.statuses[i].user.screen_name,
location: data.statuses[i].user.location
});
}
}
res.send(locationsToSend);
} else {
console.log(err);
return res.status(404).send({
message: "error searching " + err
});
}
});
};
【问题讨论】: