类型错误：'x' 不是函数

Question

我正在开发一个 Express Web 应用程序，该应用程序在第一次加载页面时运行 JavaScript 抓取代码。

这里是节点网页抓取代码（scrape.js）：

const request = require('request-promise');
const cheerio = require('cheerio');
const fs = require('fs');
const data = require('../public/state_data.json');
const cases_data = require('../public/cases_data.json');

// retrieve wikipeida page
request('https://en.wikipedia.org/wiki/2020_coronavirus_pandemic_in_the_United_States', (error, response, html) => {
if(!error && response.statusCode == 200) {
    // create cheerio scraper 
    const $ = cheerio.load(html);

    // find, and loop through all the rows in the table
    var rows = $('.wikitable').find('tr');
    for(var i = 3; i < 59; i++) {
        // scrape state name and cases from table
        var state = $(rows[i]).children('th:nth-child(2)').text().split("\n");
        var cases = parseInt($(rows[i]).children('td').html().replace(",", ""));

        // update state_data.json file w/ proper cases and per capita
        for(var j = 0; j < data.length; j++) {
            if(data[j].state === state[0]) {
                // push new data to cases_data.json
                cases_data.push({
                    state: state[0],
                    latitude: data[j].latitude,
                    longitude: data[j].longitude,
                    cases: cases,
                    percapita: (cases / data[j].population)
                });
                    
                // write to new cases_data.json file w/ state name, cases and calculated per capita 
                fs.writeFile('../public/cases_data.json', JSON.stringify(cases_data, null, 2), function(err) {
                        if (err) throw err;
                });
            }
        }
    }
} else {
    console.log('request error')
}
});

这里是 express 应用 (app.js)：

const express = require('express');
const app = express();
const port = 3000;
const scrape = require('./scrape.js');

app.get('/', (req, res) => {
    scrape();
    res.render('index');
})

app.listen(port, () => console.log(`Example app listening at http://localhost:${port}`))

现在当我运行“node app.js”时，我得到了一个错误：

TypeError: scrape 不是函数

我尝试将 scrape.js 包装在一个函数中，但无济于事。有什么想法吗？

---

修复/解决方案：

我必须导出请求函数，如下代码所示：

module.exports = () => {
    request('https://en.wikipedia.org/wiki/2020_coronavirus_pandemic_in_the_United_States', (error, response, html)) => {
        ... remaining code ...
    }    
}

Answer 1

你需要从scrape.js导出一个函数，比如：

const request = require('request-promise');
const cheerio = require('cheerio');
const fs = require('fs');
const data = require('../public/state_data.json');
const cases_data = require('../public/cases_data.json');

module.export = () => {
    // retrieve wikipeida page
    request('https://en.wikipedia.org/wiki/2020_coronavirus_pandemic_in_the_United_States', (error, response, html) => {
        ...
    });
};

另外，我建议您使用回调或承诺来处理您的异步代码，例如：

• 承诺：

// scrape.js
module.export = () => {
    return new Promise((resolve, reject) => {
        // retrieve wikipeida page
        request('https://en.wikipedia.org/wiki/2020_coronavirus_pandemic_in_the_United_States', (error, response, html) => {
            if(!error && response.statusCode == 200) {
                ...
                resolve();
            } else {
                console.log('request error');
                reject(error);
            }
        });
    });
};

// app.js
app.get('/', (req, res) => {
    scrape()
        .then(() => {
            res.render('index');
        })
        .catch((error) => {
            // response with an error
        });
});

• 有回调：

// scrape.js
module.export = (cb) => {
    // retrieve wikipeida page
    request('https://en.wikipedia.org/wiki/2020_coronavirus_pandemic_in_the_United_States', (error, response, html) => {
        if(!error && response.statusCode == 200) {
            ...
            cb(null);
        } else {
            console.log('request error');
            cb(error);
        }
    });
};

// app.js
app.get('/', (req, res) => {
    scrape((error) => {
        if (error) {
            // response with an error;
            return;
        }

        res.render('index');
    });
})

Answer 2

导出函数：

// scrape.js
module.exports = () => {
    request('https://en.wikipedia.org/wiki/2020_coronavirus_pandemic_in_the_United_States', (error, response, html) => {
        // rest of function
    }
}