要求获取服务器用户密码，我通过zenmap看了它的弱点，说
|_http-frontpage-login: false
| http-vuln-cve2011-3192:
| VULNERABLE:
| Apache byterange filter DoS
| State: VULNERABLE
| IDs: OSVDB:74721 CVE:CVE-2011-3192
| The Apache web server is vulnerable to a denial of service attack when numerous
| overlapping byte ranges are requested.
| Disclosure date: 2011-08-19
| References:
|
d50K9s2c8@1M7q4)9K6b7g2)9J5c8W2)9J5c8X3!0K6N6X3c8T1i4K6u0W2L8%4u0Y4i4K6u0r3y4K6b7%4x3U0p5`.
|
9ebK9s2c8@1M7q4)9K6b7g2)9J5c8W2)9J5c8Y4y4W2j5$3I4A6M7%4c8K6i4K6u0W2L8%4u0Y4i4K6u0r3k6Y4g2D9L8r3c8A6M7$3y4D9L8%4y4#2M7X3g2Q4x3V1j5J5x3o6p5I4i4K6u0r3b7i4g2Y4i4K6u0r3x3e0M7#2
|
b36K9s2c8@1M7q4)9K6b7g2)9J5c8W2)9J5c8X3y4$3k6g2)9J5k6h3#2A6N6s2u0W2i4K6u0W2L8%4u0Y4i4K6u0r3j5$3N6A6i4K6u0V1j5X3W2F1i4K6u0r3j5%4k6W2L8X3q4E0k6g2)9J5k6h3y4Y4K9g2)9K6c8X3&6S2L8h3g2Q4x3@1c8o6g2V1g2Q4x3X3b7J5x3o6p5I4i4K6u0V1x3K6p5&6x3R3`.`.
|
a42K9s2c8@1M7q4)9K6b7g2)9J5c8W2)9J5c8X3&6W2M7%4y4#2M7#2)9J5k6h3!0J5k6#2)9J5c8Y4m8D9N6h3N6A6L8Y4y4Q4x3V1k6A6L8X3c8W2P5q4)9J5k6i4m8Z5M7q4)9K6c8Y4k6A6k6i4N6Q4x3@1c8K6K9h3&6Y4L8r3g2Q4x3U0k6A6k6q4)9K6c8o6f1#2z5e0M7$3
|_
b5cK9s2c8@1M7s2y4Q4x3@1q4Q4x3V1k6Q4x3V1k6U0N6X3g2Q4x3X3g2E0K9i4c8J5k6g2)9J5k6h3!0J5k6#2)9J5c8X3y4Y4K9g2)9J5k6r3u0A6L8W2)9J5c8X3y4$3k6h3&6S2L8h3g2Q4x3X3g2U0k6$3W2Q4x3@1k6F1j5h3#2W2i4K6y4p5b7#2k6q4i4K6u0V1x3U0l9I4x3g2)9J5k6o6x3I4z5e0t1`.
| ssl-dh-params:
| VULNERABLE:
| Diffie-Hellman Key Exchange Insufficient Group Strength
| State: VULNERABLE
| Transport Layer Security (TLS) services that use Diffie-Hellman groups of
| insufficient strength, especially those using one of a few commonly shared
| groups, may be susceptible to passive eavesdropping attacks.
| Check results:
| WEAK DH GROUP 1
| Cipher Suite: TLS_DHE_RSA_WITH_CAMELLIA_256_CBC_SHA
| Modulus Type: Safe prime
| Modulus Source: nginx/1024-bit MODP group with safe prime modulus
| Modulus Length: 1024
| Generator Length: 8
| Public Key Length: 1024
| References:
|_
499K9s2c8@1M7s2y4Q4x3@1q4Q4x3V1k6Q4x3V1k6%4k6h3q4C8k6r3S2Q4x3X3g2G2M7X3M7`.
嗯，我看了 weakdh.org的reference，但是其中没有涉及怎么攻击的？请问怎么获取相关密码？求具体步骤
PS: 已经知道了用户名:att acker
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