【发布时间】:2021-08-27 19:29:01
【问题描述】:
我正在创建一个小程序,当您使用 selenium 在 google 上搜索内容时返回所有链接标题的名称
代码如下:
const {Builder,By, Key, until} = require('selenium-webdriver');
driver = new Builder().forBrowser("firefox").build();
var search_query='tigers';
(async()=>{
await driver.get(`https://www.google.com/search?q=${search_query}`);
await driver.findElements(By.css('h3')).then((search_results)=>{
for (var i = 0; i < search_results.length; i++)
search_results[i].getText().then((text)=>{console.log(text);})
});
console.log('...Task Complete!')
})();
当你运行它时,输出如下:-
...Task Complete! Tiger - Wikipedia Top stories Tigers (2014 film) - Wikipedia Detroit Tigers (@tigers) · Twitter Tiger | Species | WWF Videos Official Detroit Tigers Website | MLB.com Tiger | WWF Tiger - National Geographic Kids Tiger guide: species facts, how they hunt and where to see in ... Related searches Images Description
显然应该在整个功能完成后记录...Task Complete!
我不明白我做错了什么我使用了适当的await 关键字来保持代码流有序,是then() 中的错误吗?
【问题讨论】:
标签: javascript node.js selenium selenium-webdriver