公开的函数 querySelector 在 Puppeteer 中不起作用

Question

document.querySelectorAll('.summary').innerText;

这会在下面的 sn-p 中引发错误，在我的 Puppeteer 页面暴露的函数 docTest 中说“document.querySelector 不是函数”。

我想将特定节点传递给每个方法并在evaluate 中获取结果。

与document.getElemenetbyId 相同。

const puppeteer = require('puppeteer');
//var querySelectorAll = require('query-selector');


let docTest = (document) => {
var summary = document.querySelectorAll(.summary).innerText;
console.log(summary);
return summary;
}

let scrape = async () => {

const browser = await puppeteer.launch({
    headless: false
});
const page = await browser.newPage();

await page.goto('http://localhost.com:80/static.html');
await page.waitFor(5000)
await page.exposeFunction('docTest', docTest);

var result = await page.evaluate(() => {
    var resultworking = document.querySelector("tr");
    console.log(resultworking);
    var summary  = docTest(document);
    console.log(resultworking);
    return summary;

});
console.log(result);

await page.waitFor(7000);
browser.close();
return {
    result
}
};

scrape().then((value) => {
console.log(value); // Success!
});

Answer 1

我也有同样的问题。问题是page.evaluate() 函数回调必须是async 函数，而您的函数docTest() 在page.evaluate() 内部调用时将返回Promise。要修复它，只需将 async 和 await 关键字添加到您的代码中：

await page.exposeFunction('docTest', docTest);

var result = await page.evaluate(async () => {
    var summary = await docTest(document);
    console.log(summary);
    return summary;
});

请记住page.exposeFunction() 将使您的函数返回Promise，然后，您需要使用async 和await。发生这种情况是因为您的函数不是running inside your browser，而是在您的nodejs 应用程序中。

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