px.*_*px. 32 random passwords ascii passphrase character
你如何生成密码?
像这样的东西?
cat /dev/urandom | strings
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aar*_*nsw 93
Mac OS X的"Keychain Access"应用程序使您可以访问不错的OS X密码生成器.单击命令-N并单击键图标.您可以选择密码样式(令人难忘,数字,字母数字,随机,FIPS-181)并选择长度.它还会警告您密码不足.
小智 22
使用这个&重击:)
cat /dev/urandom | tr -dc 'a-zA-Z0-9-!@#$%^&*()_+~' | fold -w 10 | head -n 1
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更改头数以生成密码数.
Dou*_*der 12
一个简短的python脚本,用于生成密码,最初来自python cookbook.
#!/usr/bin/env python
from random import choice
import getopt
import string
import sys
def GenPasswd():
chars = string.letters + string.digits
for i in range(8):
newpasswd = newpasswd + choice(chars)
return newpasswd
def GenPasswd2(length=8, chars=string.letters + string.digits):
return ''.join([choice(chars) for i in range(length)])
class Options(object):
pass
def main(argv):
(optionList,args) = getopt.getopt(argv[1:],"r:l:",["repeat=","length="])
options = Options()
options.repeat = 1
options.length = 8
for (key,value) in optionList:
if key == "-r" or key == "--repeat":
options.repeat = int(value)
elif key == "-l" or key == "--length":
options.length = int(value)
for i in xrange(options.repeat):
print GenPasswd2(options.length)
if __name__ == "__main__":
sys.exit(main(sys.argv))
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你的建议略有不同:
head -c 32 /dev/random | base64
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(可选)您可以修剪尾部=并使用echo以获取换行符:
echo $(head -c 32 /dev/random | base64 | head -c 32)
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这为您提供了更可预测的输出长度密码,同时仍然确保只有可打印的字符.