Question

我一直在编写一个脚本来检查反映的XSS漏洞。我检查你是否有＆＃34; http：//＆＃34;或＆＃34; https：//＆＃34;在您的网址和＆＃39; *＆＃39;在查询的地方。但是，当我把https://google.com/#q=*", it results in错误时！ MISSING＆＃39; http：//＆＃39;，或＆＃39; https：//＆＃39;！`。这是我的代码：

<!DOCTYPE html>

<html>

  <head>

    <title>Slingshot.XSS</title>

  </head>

  <body style="font-family:monospace;" align="center">

    <h2>Slingshot.XSS</h2>
    <h3>Slingshot.XSS is a script that launches pre-loaded XSS payloads at a target to test its vulnerabilities.</h3>
    <h4>Please report all issues to <a href="https://github.com/keeganjk/slingshot.xss/issues"></a> or contact me at keeganjkuhn@gmail.com.</h4>
    <a href="github.com/keeganjk/slingshot.xss" style="font-family:monospace" align="center">Source Code / Learn More</a>
    <br />

    <h4>Enter a URL with <b>*</b> in the place of query.</h4>
    <h5>Example: https://www.google.com/#q=*</h5>
    <input type="text" id="myText" placeholder="Enter a URL"> <button onclick="myFunction()">Submit</button>

    <p id="demo">No Submitted URL</p>

    <script>

      function myFunction() {

        var x = document.getElementById("myText").value;

        // Error check
        if ( !x.includes("*") && ( !x.includes("http://") || !x.includes("https://") ) ) {

            document.getElementById("demo").innerHTML = "ERROR! MISSING \'*\' IN PLACE OF QUERY, \'http://\', AND \'https://\'!";
            x = false;
            return 0;

        }

        if ( !x.includes("*") ) {

            document.getElementById("demo").innerHTML = "ERROR! MISSING \'*\' IN PLACE OF QUERY!";
            x = false;
            return 0;

        }

        if ( !x.includes("http://") || !x.includes("https://") ) {

            document.getElementById("demo").innerHTML = "ERROR! MISSING \'http://\', OR \'https://\'!";
            x = false;
            return 0;

        }

        document.getElementById("demo").innerHTML = x;

      }

    </script>

  </body>

</html>

我做错了什么？

Answer 1

您检查http是否不在，或者https不在。两者之一将始终为真。一个接一个地执行检查......例如

Answer 2

您需要正确编写if条件。

从

更改条件
if ( !x.includes("http://") || !x.includes("https://") ) {

以

if ( !(x.includes("http://") || x.includes("https://")) ) {

这样，只有在网址不包含http://或https://
时才会引发错误
完整代码：

＆＃13;
＆＃13;
<!DOCTYPE html> <html> <head> <title>Slingshot.XSS</title> </head> <body style="font-family:monospace;" align="center"> <h2>Slingshot.XSS</h2> <h3>Slingshot.XSS is a script that launches pre-loaded XSS payloads at a target to test its vulnerabilities.</h3> <h4>Please report all issues to <a href="https://github.com/keeganjk/slingshot.xss/issues"></a> or contact me at keeganjkuhn@gmail.com.</h4> <a href="github.com/keeganjk/slingshot.xss" style="font-family:monospace" align="center">Source Code / Learn More</a> <br /> <h4>Enter a URL with <b>*</b> in the place of query.</h4> <h5>Example: https://www.google.com/#q=*</h5> <input type="text" id="myText" placeholder="Enter a URL"> <button onclick="myFunction()">Submit</button> <p id="demo">No Submitted URL</p> <script> function myFunction() { var x = document.getElementById("myText").value; // Error check if ( !x.includes("*") && ( !x.includes("http://") || !x.includes("https://") ) ) { document.getElementById("demo").innerHTML = "ERROR! MISSING \'*\' IN PLACE OF QUERY, \'http://\', AND \'https://\'!"; x = false; return 0; } if ( !x.includes("*") ) { document.getElementById("demo").innerHTML = "ERROR! MISSING \'*\' IN PLACE OF QUERY!"; x = false; return 0; } if ( !(x.includes("http://") || x.includes("https://")) ) { document.getElementById("demo").innerHTML = "ERROR! MISSING \'http://\', OR \'https://\'!"; x = false; return 0; } document.getElementById("demo").innerHTML = x; } </script> </body> </html>
＆＃13;
＆＃13;
＆＃13;

Answer 3

我重构了您的函数，以展示在将验证逻辑与错误呈现分开时如何降低代码的复杂性。

function myFunction() {
    var errors = [];
    var x = document.getElementById("myText").value;

    if (!x.includes("http://") && !x.includes("https://")) {
      errors.push('missing HTTP or HTTPS');
    }

    if (!x.includes("*")) {
      errors.push('missing * in place of query')
    }

    // render the errors
    if (errors.length) {
      x = 'Error: ' + errors.join(', ') + '!';        
    }

    document.getElementById("demo").innerHTML = x;
}

＆＃13;

<!DOCTYPE html>

<html>

  <head>

    <title>Slingshot.XSS</title>

  </head>

  <body style="font-family:monospace;" align="center">

    <h2>Slingshot.XSS</h2>
    <h3>Slingshot.XSS is a script that launches pre-loaded XSS payloads at a target to test its vulnerabilities.</h3>
    <h4>Please report all issues to <a href="https://github.com/keeganjk/slingshot.xss/issues"></a> or contact me at keeganjkuhn@gmail.com.</h4>
    <a href="github.com/keeganjk/slingshot.xss" style="font-family:monospace" align="center">Source Code / Learn More</a>
    <br />

    <h4>Enter a URL with <b>*</b> in the place of query.</h4>
    <h5>Example: https://www.google.com/#q=*</h5>
    <input type="text" id="myText" placeholder="Enter a URL"> <button onclick="myFunction()">Submit</button>

    <p id="demo">No Submitted URL</p>

    <script>

      function myFunction() {
        var errors = [];
        var x = document.getElementById("myText").value;

        if (!x.includes("http://") && !x.includes("https://")) {
          errors.push('missing HTTP or HTTPS');
        }
        
        if (!x.includes("*")) {
          errors.push('missing * in place of query')
        }

        if (errors.length) {
          x = 'Error: ' + errors.join(', ') + '!';        
        }


        document.getElementById("demo").innerHTML = x;

      }
    </script>

＆＃13;

为什么不应该出现这种错误呢？

3 个答案: