木偶师只从URL数组中获得第一项。

Question

我想从arrayUrl输入每个URL，得到价格并移到下一个，问题是:我不能得到超过一个项目，它总是返回第一个产品的价格，索引0，或者1，2.如果我手动更改为for(let i = "Number"; i<2; i++)，它会认为我的循环中断了，但我不知道如何修复它。

let manufacturers = [2]
// 'Norton', 'Vonder', 'Bosch', 'black+decker', 'Dewalt', 'Stanley', 'Makita', 'Gedore', '3m&common_filter[2012]=2082', 'Dremel', 'Toyama', 'Minipa'] (Ignore)
let lastpage = [3]
// , 15, 4, 3, 5, 6, 14, 2, 2, 2, 3, 2] //Tem que botar um número acima (Ignore)
var scrape = async () => {
    const browser = await puppeteer.launch({ headless: true });
    const page = await browser.newPage();
    page.setDefaultNavigationTimeout(0);
    results = await getProducts(page)
    return results
}
async function getProducts(page) {
    var arrayUrl = await (ConcatUrls(page))
    products = []
    for(let i = 0; i < 2; i++) { //i<2 Just to limit to 2 products per page, for testing purpose
    await page.goto(arrayUrl[i])
    return page.evaluate(() => {
    data = []
    price = document.querySelector('#valor-padrao-prod').getAttribute('value')
    data.push({price})
    return data
}
)}
    return products
}
async function ConcatUrls(page) {
    let urls = [];
    await page.goto('https://www.ferramentaskennedy.com.br/busca?marcas=99&q=vonder');
    for (let i = 0; i < manufacturers.length; i++) {
        var lastPageNumber = lastpage[i];
        for (let index = 1; index < lastPageNumber; index++) {
            await page.waitFor(1000);
            urls = urls.concat(await getUrl(page));
            if (index < lastPageNumber - 1) {
                await page.click('#app > div.categoria-produtos.pr-3.pl-3.p-md-0 > div.content > div > div > div.col-12.p-0 > ul > li:nth-child(8) > a');
            }
        }
    }
    return urls
}

async function getUrl(page) {
    await page.waitForSelector('#app > div.categoria-produtos.pr-3.pl-3.p-md-0 > div.content > div > div > div.col-12.p-0 > div > div > div > div.infos > div.title.mt-2.margin-card-sem-selo > h2');
    let url = await page.$$eval('#app > div.categoria-produtos.pr-3.pl-3.p-md-0 > div.content > div > div > div.col-12.p-0 > div > div > div > div.infos > div.title.mt-2.margin-card-sem-selo', links => {
    links = links.map(el => el.querySelector('h2 > a').href)
        return links;
    });
    return url
}

scrape().then((value) => {
    // console.log('Items scraped: ' + value.length);
    console.table(value);
})```

Answer 1

问题似乎在这里(见评论)：

async function getProducts(page) {
    var arrayUrl = await (ConcatUrls(page))
    products = []
    for(let i = 0; i < 2; i++) {
      await page.goto(arrayUrl[i])
      return page.evaluate(() => { // Breaks the loop and returns just the first result
        data = []
        price = document.querySelector('#valor-padrao-prod').getAttribute('value')
        data.push({price})
        return data
      })
    }
    return products // Not returned
}

也许你的意思是这样的：

async function getProducts(page) {
    var arrayUrl = await (ConcatUrls(page))
    products = []
    for(let i = 0; i < 2; i++) {
      await page.goto(arrayUrl[i])
      const price = await page.evaluate(() => {
        return document.querySelector('#valor-padrao-prod').getAttribute('value')
      })
      products.push({price})
    }
    return products
}