如何设置PAM模块控制Linux密码策略

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如何设置PAM模块控制Linux密码策略 时间：2015-06-28 00:00:00 来源：IT猫扑网作者：网管联盟 我要评论(0)

　　我们在使用linux系统设置密码的时候，经常遇到这样的问题，系统提示：您的密码太简单，或者您的密码是字典的一部分。那么系统是如何实现对用户的密码的复杂度的检查的呢？

　　系统对密码的控制是有两部分(我知道的)组成：

　　1 cracklib

　　2 login.defs

　　声明：login.defs主要是控制密码的有效期。对密码进行时间管理。此处不细谈

　　pam_cracklib.so 才是控制密码复杂度的关键文件

　　redhat公司专门开发了cracklib这个安装包来判断密码的复杂度

　　可以rpm -ql cracklib查看

　　密码的复杂度的判断是通过pam模块控制来实现的，具体的模块是pam_cracklibpam_cracklib 的参数介绍：

　　debug

　　This option makes the module write information to syslog(3) indicating the behavior of the module (this option does not write password information to the log file).

　　type=XXX

　　The default action is for the module to use the following prompts when requesting passwords: "New UNIX password: " and "Retype UNIX password: ". The default word UNIX can be replaced with this option.

　　retry=N

　　Prompt user at most N times before returning with error. The default is 1

　　difok=N

　　This argument will change the default of 5 for the number of characters in the new password that must not be present in the old password. In addition, if 1/2 of the characters in the new password are different then the new password will be accepted anyway.

　　difignore=N

　　How many characters should the password have before difok will be ignored. The default is 23.

　　minlen=N

　　The minimum acceptable size for the new password (plus one if credits are not disabled which is the default). In addition to the number of characters in the new password, credit (of +1 in length) is given for each different kind of character (other, upper, lower and digit). The default for this parameter is 9 which is good for a old style UNIX password all of the same type of character but may be too low to exploit the added security of a md5 system. Note that there is a pair of length limits in Cracklib itself, a "way too short" limit of 4 which is hard coded in and a defined limit (6) that will be checked without reference to minlen. If you want to allow passwords as short as 5 characters you should not use this module.

　　dcredit=N

　　(N >= 0) This is the maximum credit for having digits in the new password. If you have less than or N digits, each digit will count +1 towards meeting the current minlen value. The default for dcredit is 1 which is the recommended value for minlen less than 10.

　　(N < 0) This is the minimum number of digits that must be met for a new password.

　　ucredit=N

　　(N >= 0) This is the maximum credit for having upper case letters in the new password. If you have less than or N upper case letters each letter will count +1 towards meeting the current minlen value. The default for ucredit is 1 which is the recommended value for minlen less than 10.

　　(N > 0) This is the minimum number of upper case letters that must be met for a new password.

　　lcredit=N

　　(N >= 0) This is the maximum credit for having lower case letters in the new password. If you have less than or N lower case letters, each letter will count +1 towards meeting the current minlen value. The default for lcredit is 1 which is the recommended value for minlen less than 10.

　　(N < 0) This is the minimum number of lower case letters that must be met for a new password.

#p#副标题#e#

　　ocredit=N

　　(N >= 0) This is the maximum credit for having other characters in the new password. If you have less than or N other characters, each character will count +1 towards meeting the current minlen value. The default for ocredit is 1 which is the recommended value for minlen less than 10.

　　(N < 0) This is the minimum number of other characters that must be met for a new password.

　　use_authtok

　　This argument is used to force the module to not prompt the user for a new password but use the one provided by the previously stacked password module.

　　dictpath=/path/to/dict

　　Path to the cracklib dictionaries.

　　dictpath=/path/to/dict //注：密码字典，这个是验证用户的密码是否是字典一部分的关键。

　　Path to the cracklib dictionaries.

　　cracklib密码强度检测过程

　　首先检查密码是否是字典的一部分，如果不是，则进行下面的检查

　　密码强度检测过程

　　These checks are:

　　Palindrome

　　Is the new password a palindrome of the old one?

　　新密码是否旧密码的回文

　　Case Change Only

　　Is the new password the the old one with only a change of case?

　　新密码是否只是就密码改变了大小写

　　Similar

　　Is the new password too much like the old one?

　　新密码是否和旧密码很相似

　　This is primarily controlled by one argument, difok which is a number of characters that if different between the old and new are enough to accept the new password, this defaults to 10 or 1/2 the size of the new password whichever is smaller.

　　To avoid the lockup associated with trying to change a long and complicated password, difignore is available. This argument can be used to specify the minimum length a new password needs to be before the difok value is ignored. The default value for difignore is 23.

　　Simple

　　Is the new password too small?

　　新密码是否太短

　　This is controlled by 5 arguments minlen, dcredit, ucredit, lcredit, and ocredit. See the section on the arguments for the details of how these work and there defaults.

　　Rotated

　　Is the new password a rotated version of the old password?

　　新密码的字符是否是旧密码字符的一个循环

　　例如旧密码：123

　　新密码：231

　　Already used

　　Was the password used in the past?

　　这个密码以前是否使用过

　　Previously used passwords are to be found in /etc/security/opasswd.

　　那么系统是如何实现这个控制的呢？

　　在系统的配置文件/etc/pam.d/system-auth 中有这样一行

　　password requisite pam_cracklib.so try_first_pass retry=3

　　我们可以根据pam_cracklib的参数这样配置这个pam模块来达到我们想要的目的

　　password required /lib/security/pam_cracklib.so retry=3 type= minlen=8 difok=3 dictpath=/path/to/dict

关键词标签：Linux密码策略

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