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Insecure Cryptographic Storage Challenge 2(不安全加密存储 2)

题介绍

本题使用了不安全的加密函数,直接可以获得明文数据

功能实现

发现本题并没有发送网络请求,因此查看html源代码,发现是javascript实现的加密

    var input = $("#resultKeyAttempt").val();
    theKey = "kpoisaijdieyjaf";
    var theAlphabet =   "ABCDEFGHIJKLMNOPQRSTUVWXYZ" + "abcdefghijklmnopqrstuvwxyz";

    // Validate theKey:
    theKey = theKey.toUpperCase();
    var theKeysLength = theKey.length;
    var i;
    var adjustedKey = "";
    for(i = 0; i < theKeysLength; i ++)
    {
        var currentKeyChar = theAlphabet.indexOf(theKey.charAt(i));
        if(currentKeyChar < 0)
            continue;
        adjustedKey += theAlphabet.charAt(currentKeyChar);
    }
    theKey = adjustedKey;
    theKeysLength = theKey.length;

    // Transform input:
    var inputLength = input.length;
    var output = "";
    var theKeysCurrentIndex = 0;
    for(i = 0; i < inputLength; i ++)
    {
        var currentChar = input.charAt(i);
        var currentCharValue = theAlphabet.indexOf(currentChar);
        if(currentCharValue < 0)
        {
            output += currentChar;
            continue;
        }
        var lowercase = currentCharValue >= 26 ? true : false;
        currentCharValue += theAlphabet.indexOf(theKey.charAt(theKeysCurrentIndex));
        currentCharValue += 26;
        if(lowercase)
            currentCharValue = currentCharValue % 26 + 26;
        else
            currentCharValue %= 26;
        output += theAlphabet.charAt(currentCharValue);
        theKeysCurrentIndex =(theKeysCurrentIndex + 1) % theKeysLength;
    }
    // Check result for validity
    $("#resultDiv").hide("fast", function(){
        if(output == "DwsDagmwhziArpmogWaSmmckwhMoEsmgmxlivpDttfjbjdxqBwxbKbCwgwgUyam")
            $('#resultDiv').html("<p>Yeah, that's correct</p>");
        else
            $('#resultDiv').html("<p>No, that's not correct</p>");
        $("#resultDiv").show("slow");
    });
从代码知道,最终得到的输出结果为 DwsDagmwhziArpmogWaSmmckwhMoEsmgmxlivpDttfjbjdxqBwxbKbCwgwgUyam
密钥是 theKey 变量值

解题步骤

把密文变成明文输入,将

currentCharValue += theAlphabet.indexOf(theKey.charAt(theKeysCurrentIndex));
修改为
currentCharValue -= theAlphabet.indexOf(theKey.charAt(theKeysCurrentIndex));
即可

总结

不要使用不安全加密算法,密钥一定要妥善保管

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