[英]Use array of keywords and loop through script in Playwright
所以,我試圖用 Playwright 的幾個搜索短語來抓取幾個搜索引擎。 使用一個查詢運行腳本是有效的。
在職的:
const { chromium } = require('playwright');
(async () => {
const browser = await chromium.launch({ headless: false, slowMo: 250 });
const context = await browser.newContext()
const page = await context.newPage();
const keyWord = ('Arsenal');
await page.goto('https://duckduckgo.com/');
await page.fill('//input[@name="q"]',keyWord);
await page.keyboard.press('Enter');
const getOne = (' (//h2[@class="result__title"])[9] ');
await page.waitForSelector(getOne)
const pushOne = await page.$(getOne);
const One = await pushOne.evaluate(element => element.innerText);
console.log(One);
await page.goto('https://yandex.com/');
await page.fill('//input[@aria-label="Request"]', keyWord);
await page.keyboard.press('Enter');
const getTwo = (' //li[@data-first-snippet] //div[@class="organic__url-text"] ');
await page.waitForSelector(getTwo)
const pushTwo = await page.$(getTwo);
const Two = await pushTwo.evaluate(element => element.innerText);
console.log(Two);
await browser.close()
})()
但是,當我使用帶有短語 (keyWordlist) 的數組時,我無法運行腳本。 已經四處尋找使用帶有“For”和“Foreach”循環的數組,但無法修復它。 我想通過不同的搜索引擎運行不同的關鍵字並列出結果。 對於將獲得 6 個結果的兩個搜索引擎中的 3 個關鍵字。
const { chromium } = require('playwright');
(async () => {
const browser = await chromium.launch({ headless: false, slowMo: 250 });
const context = await browser.newContext()
const page = await context.newPage();
let kewWordlist = ['Arsenal', 'Liverpool', 'Ajax']
for (var i=0; i<=kewWordlist.length; i++) {
// for (const i in kewWordlist){
async () => {
const keyWord = kewWordlist[i];
await page.goto('https://duckduckgo.com/');
await page.fill('//input[@name="q"]',keyWord);
// await page.fill('//input[@name="q"]',[i]);
// await page.fill('//input[@name="q"]',`${keyWord}`);
await page.keyboard.press('Enter');
const getOne = (' (//h2[@class="result__title"])[9] ');
await page.waitForSelector(getOne)
const pushOne = await page.$(getOne);
const One = await pushOne.evaluate(element => element.innerText);
console.log(One);
// await page.goto('https://yandex.com/');
// await page.fill('//input[@aria-label="Request"]', keyWord);
// await page.keyboard.press('Enter');
// const getTwo = (' //li[@data-first-snippet] //div[@class="organic__url-text"] ');
// await page.waitForSelector(getTwo)
// const pushTwo = await page.$(getTwo);
// const Two = await pushTwo.evaluate(element => element.innerText);
// console.log(Two);
}}
await browser.close()
})()
如果有人對如何解決這個問題有一些指示,非常有義務。
也許結果選擇器需要一些調整,但我認為這就是您要找的:
test.only('search search engines', async({page, context}) => {
const search = [
{
name: 'yandex',
url: 'https://yandex.com/',
elementFill: '//input[@aria-label="Request"]',
elementResult: '//li[@data-first-snippet] //div[@class="organic__url-text"]'
},
{
name: 'google',
url: 'https://www.google.nl',
elementFill: '//input[@name="q"]',
elementResult: '(//h2[@class="result__title"])[9]'
},
{
name: '',
url: 'https://duckduckgo.com/',
elementFill: '//input[@name="q"]',
elementResult: '(//h2[@class="result__title"])[9]'
}
]
const kewWordlist = ['Arsenal', 'Liverpool', 'Ajax']
for (let i = 0; i < search.length; i++) {
const searchName = search[i].name
const searchResult = search[i].elementResult
const searchFill = search[i].elementFill
const searchPage = await context.newPage()
await searchPage.waitForLoadState()
await searchPage.goto(`${search[i].url}`)
for (let i = 0; i < kewWordlist.length; i++) {
await searchPage.fill(searchFill,kewWordlist[i])
await searchPage.keyboard.press('Enter')
await searchPage.waitForSelector(searchResult)
const result = await page.$(searchResult)
console.log(`${searchName}: ${result} `)
}
}
})
你的循環不工作的原因是你有一個異步 function 里面你從來沒有調用過。 您可以通過以下幾種方法 go 關於此:
您可以使用您的第一個版本,讓它接受要搜索的單詞,然后在數組的每個元素上運行它:
const searchOneKeyword = async (keyWord) => {
const browser = await chromium.launch({ headless: false, slowMo: 250 });
const context = await browser.newContext()
const page = await context.newPage();
// rest of code
}
const kewWordList = ['Arsenal', 'Liverpool', 'Ajax']
keyWordList.forEach((k) => {
searchOneKeyword(k)
})
或者,如果您想保留相同的瀏覽器實例,您可以在 function 中循環執行此操作:
const search = async (words) => {
const browser = await chromium.launch({ headless: false, slowMo: 250 });
const context = await browser.newContext()
const page = await context.newPage();
for (const keyWord of words) {
await page.goto('https://duckduckgo.com/');
await page.fill('//input[@name="q"]',keyWord);
await page.keyboard.press('Enter');
const getOne = (' (//h2[@class="result__title"])[9] ');
await page.waitForSelector(getOne)
const pushOne = await page.$(getOne);
const One = await pushOne.evaluate(element => element.innerText);
console.log(One);
// etc.
}
await browser.close()
}
search(keyWordList)
在這兩種情況下,您都在記錄,但從不返回任何內容,因此如果您之后需要另一個 function 中的數據,則必須更改它。 例子:
const search = async (words) => {
const browser = await chromium.launch({ headless: false, slowMo: 250 });
const context = await browser.newContext()
const page = await context.newPage();
const results = await Promise.all(words.map((keyWord) => {
await page.goto('https://duckduckgo.com/');
await page.fill('//input[@name="q"]',keyWord);
await page.keyboard.press('Enter');
// etc.
return [ One, Two ]
}))
await browser.close()
return results
}
search(keyWordList).then((results) => { console.log(results.flat()) })
我花了幾個小時試圖根據您的建議使腳本正常工作。 不幸的是沒有結果。 我收到諸如“等待僅在異步函數中有效”和“檢測到無法訪問的代碼”之類的錯誤。 搜索其他示例,尋找一些靈感,但沒有找到。 如果您或其他人有建議,請分享:這是我現在擁有的代碼:
const { chromium } = require('playwright');
let keyWordList = ['Arsenal', 'Liverpool', 'Ajax']
const search = async function words() {
const browser = await chromium.launch({ headless: false, slowMo: 250 });
const context = await browser.newContext()
const page = await context.newPage();
}
const results = await Promise.all(words.map(keyWord))
//DUCKDUCKGO
await page.goto('https://duckduckgo.com/');
await page.fill('//input[@name="q"]',keyWord);
await page.keyboard.press('Enter');
const getOne = (' (//h2[@class="result__title"])[9] ');
await page.waitForSelector(getOne)
const pushOne = await page.$(getOne);
const One = await pushOne.evaluate(element => element.innerText);
//YANDEX
await page.goto('https://yandex.com/');
await page.fill('//input[@aria-label="Request"]', keyWord);
await page.keyboard.press('Enter');
const getTwo = (' //li[@data-first-snippet] //div[@class="organic__url-text"] ');
await page.waitForSelector(getTwo)
const pushTwo = await page.$(getTwo);
const Two = await pushTwo.evaluate(element => element.innerText);
console.log(Two);
return [ One , Two ]
return results
search(keyWordList).then((results) => { console.log(results.flat())
await browser.close();
})
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