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问标识一个值的javascript语句
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Stack Overflow用户

提问于 2020-11-17 18:58:36

回答 1查看 59关注 0票数 0

我有一个问题，让我的可信ip地址的工作-我可以让IP1的工作，但它应该如何，但其余的只是转移回持有页-我错过了什么？

代码：

  addEventListener("fetch", event => {
      event.respondWith(fetchAndReplace(event.request))
    })
     
    async function fetchAndReplace(request) {
     
      let modifiedHeaders = new Headers()
     
      modifiedHeaders.set('Content-Type', 'text/html')
      modifiedHeaders.append('Pragma', 'no-cache')
    
      // Allow users through if they are bypass flag and set a cookie
      if(request.url.includes("bypass"))
      {
        // Forward request to origin, get response.
        let response = await fetch(request)
    
        // Copy Response object so that we can edit headers.
        response = new Response(response.body, response)
        
        // set a cookie for the session
        response.headers.set("Set-Cookie", "cf:maintenance-mode=bypass");
        
        // Return on to client.
        return response
      }
    
      // Allow users through if they have the bypass cookie
      // Check for cookie.
      let cookies = request.headers.get('Cookie') || ""
      if (cookies.includes("cf:maintenance-mode=bypass")) {
        // User has been here before. Just pass request through.
        return fetch(request)
      }
    
      //Return maint page if you're not calling from a trusted IP
      
      var connectionIP = request.headers.get("cf-connecting-ip")
    
    if (connectionIP != "ip1" || connectionIP != "ip2" || connectionIP != "ip3") { 
        // Return modified response.
        return new Response(maintPage, {
          headers: modifiedHeaders
        })
      }
       //Allow users from trusted into site
      else
      {
        //Fire all other requests directly to our WebServers
        return fetch(request)
      }
    }
     
    let maintPage = `
     
    <!doctype html>
    <title>Site Maintenance</title>
    <style>
      body { 
            text-align: center; 
            padding: 150px; 
            background: url('data:image/jpeg;base64,<base64EncodedImage>'); 
            background-size: cover;
            -webkit-background-size: cover;
            -moz-background-size: cover;
            -o-background-size: cover;
          }
     
        .content {
            background-color: rgba(255, 255, 255, 0.75); 
            background-size: 100%;      
            color: inherit;
            padding-top: 1px;
            padding-bottom: 10px;
            padding-left: 100px;
            padding-right: 100px;
            border-radius: 15px;        
        }
     
      h1 { font-size: 40pt;}
      body { font: 20px Helvetica, sans-serif; color: #333; }
      article { display: block; text-align: left; width: 75%; margin: 0 auto; }
      a:hover { color: #333; text-decoration: none; }  
     
     
    </style>
     
    <article>
     
            <div class="background">
                <div class="content">
            <h1>We&rsquo;ll be back soon!</h1>        
                <p>We're very sorry for the inconvenience but we&rsquo;re performing maintenance. Please check back soon...</p>
            </div>
        </div>
     
    </article>`;

javascript

listener

cloudflare

cloudflare-workers

回答 1

Stack Overflow用户

发布于 2020-11-17 19:16:02

你有错误的逻辑运算符。

假设connectionIP是ip1。

connectionIP != "ip1" || connectionIP != "ip2" || connectionIP != "ip3"
-> "ip1" != "ip1" || "ip1" != "ip2" || "ip1" != "ip3"
-> false || true || true
-> true

// Or if it's ip2...
-> "ip2" != "ip1" || "ip2" != "ip2" || "ip2" != "ip3"
-> true || false || true
-> true

// Or if it's something else...
-> "different" != "ip1" || "different" != "ip2" || "different" != "ip3"
-> true || true || true
-> true

不管connectionIP是什么，它总是要计算为true，因为它总是不等于列表中的值之一。

你真正想要的是&&

connectionIP !== "ip1" && connectionIP !== "ip2" && connectionIP !== "ip3"
-> "ip1" !== "ip1" && "ip1" !== "ip2" && "ip1" !== "ip3"
-> false && true && true
-> false

// Or if it's ip2...
-> "ip2" !== "ip1" && "ip2" !== "ip2" && "ip2" !== "ip3"
-> true && false && true
-> false

// Or if it's something else...
-> "different" !== "ip1" && "different" !== "ip2" && "different" !== "ip3"
-> true && true && true
-> true

如果使用Set，则逻辑可能更易于遵循和更易于维护。

const internalAddresses = new Set(["ip1", "ip2", "ip3"]);

// ...

return internalAddresses.has(connectionIP)
  ? fetch(request) 
  : new Response(maintPage, { headers: modifiedHeaders })

票数 3

页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持

原文链接：

https://stackoverflow.com/questions/64881610

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