用于检查网站上关键字的脚本

一种简单的Python方法是:

import requests

urls = ['https://www.google.com'] # Fill this however
for url in urls:
    resp = requests.get(url)
    if 'Sorry, not found!' in resp.text:
        print(url + ' had no page') # or something

jQuery .我认为没有人能单独用javascript做到这一点。无论如何,您都必须使用jQuery。

首先,您应该在Chrome控制台中试用:

1.添加此扩展以消除CORS策略错误 Chrome Extension 。确保在Chrome中启用它->更多工具->扩展

2.现在我们必须运行get(),不能像$那样调用它。get(),通常在.js文件中使用。因此,我们需要通过在控制台中运行以下行将其转换为控制台:

var jq = document.createElement('script');
jq.src = "https://ajax.googleapis.com/ajax/libs/jquery/2.1.4/jquery.min.js";
document.getElementsByTagName('head')[0].appendChild(jq);

3.Fire get请求:

var rsp = jQuery.get("https://www.google.com/");

if (rsp.responseText && rsp.responseText.includes("was not found")) { //In your js file replace with Sorry! not found
console.log("The Url is Invalid"); 
}
else {
console.log("could be a valid url"); //this must get printed
}

尝试无效的url:

var rsp = jQuery.get("https://www.goesfsfsfsffogle.com/");

if (rsp.responseText && rsp.responseText.includes("was not found")) { //In your js file replace with Sorry! not found
console.log("The Url is Invalid"); //this must get printed
}
else {
console.log("could be a valid url"); 
}

在jQuery项目文件中运行:

var urls = ["https://www.google.com/"];
var url;
for ( url in urls ){
var rsp = $.get(url);
//A wait should be added here for rsp to get populated
//console.log("readyState="+rsp.readyState);
if (rsp.responseText && rsp.responseText.includes("Sorry! not found")) 
{  
console.log("The Url is Invalid"); 
}
else {
console.log("Its a valid url"); 
}
}

同样,如果rsp不包含readyState==4,则表示尚未收到异步响应。在这种情况下,我们需要在if检查之前添加等待。