我正在创建一个小程序,当您使用 selenium 在 google 上搜索某些内容时,它会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=${search_query}`);
await driver.findElements(By.css('h3')).then((search_results)=>{
for (var i = 0; i < search_results.length; i++)
search_results[i].getText().then((text)=>{console.log(text);})
});
console.log('...Task Complete!')
})();
当你运行它时,输出如下:-
<块引用>...Task Complete!
Tiger - Wikipedia
Top stories
Tigers (2014 film) - Wikipedia
Detroit Tigers (@tigers) · Twitter
Tiger | Species | WWF
Videos
Official Detroit Tigers Website | MLB.com
Tiger | WWF
Tiger - National Geographic Kids
Tiger guide: species facts, how they hunt and where to see in ...
Related searches
Images
Description
显然应该在整个函数完成后记录...Task Complete!
我不明白我做错了什么我使用了适当的 await
关键字来保持代码流的有序,是 then()
中的错误吗?
答案 0 :(得分:2)
由于您的 .then(()=>...)
不返回 Promise,因此开头的 await
关键字没有任何作用。 Node 已经开始承诺获取 h3、获取它们的文本内容并记录它们,但是您放错位置的 await
并没有告诉 Node 等待所有这些完成。您需要 await
获取元素,然后同步循环遍历所有元素,await
处理文本,然后同步打印文本,最后同步打印 "...Task Complete!"
const { Builder, By, Key, until } = require('selenium-webdriver');
const driver = new Builder().forBrowser("firefox").build();
const search_query = 'tigers';
(async () => {
await driver.get(`https://www.google.com/search?q=${search_query}`);
const h3Elements = await driver.findElements(By.css('h3'));
for (const h3 of h3Elements) {
const text = await h3.getText();
console.log(text);
}
console.log('...Task Complete!')
})();
如果你想减少额外的变量,你可以这样做:
const { Builder, By, Key, until } = require('selenium-webdriver');
const driver = new Builder().forBrowser("firefox").build();
const search_query = 'tigers';
(async () => {
await driver.get(`https://www.google.com/search?q=${search_query}`);
const h3Elements = await driver.findElements(By.css('h3'));
for (const h3 of h3Elements) {
console.log(await h3.getText());
}
console.log('...Task Complete!')
})();