java 密码验证 - 添加额外要求
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/9962382/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
Password validation - adding additional requirments
提问by rick
I have some code that currently checks for minimum and maximum lentgh. I want to also require uppercase, lowercase, special char, and numeric. Any suggestions on what to add to this or where I can find some examples? I've been Googling and looking thru this forum and have been trying to add the additional password requirments and have been unsuccessful.
我有一些当前检查最小和最大 lentgh 的代码。我还想要求大写、小写、特殊字符和数字。关于添加什么或在哪里可以找到一些示例的任何建议？我一直在谷歌搜索并浏览这个论坛，并一直试图添加额外的密码要求，但没有成功。
This is what I want to require.
这就是我想要的要求。
At least eight characters in length No more than 20 characters in length at least lower-case letter and one upper-case at least one special character from: !@#$%^&*()~`-=_+[]{}|:";',./<>? at least one number [0-9] character Cannot match the account login name or email address
长度至少为 8 个字符长度不超过 20 个字符至少小写字母和一个大写字母至少一个特殊字符来自：!@#$%^&*()~`-=_+[]{ }|:";',./<>? 至少一个数字 [0-9] 字符 无法匹配帐户登录名或电子邮件地址
My current password validation code
我当前的密码验证码
public static final int MIN_PASSWORD_LENGTH = 8;
public static final int MAX_PASSWORD_LENGTH = 20;
public static boolean isAcceptablePassword(String password)
{
if(TextUtils.isEmpty(password))
return false;
int len = password.length();
if(len < MIN_PASSWORD_LENGTH || len > MAX_PASSWORD_LENGTH)
return false;
for(int i = 0; i < len; i++)
{
char c = password.charAt(i);
if (Character.isWhitespace(c))
return false;
}
return true;
}
回答by Luiggi Mendoza
When you're analyzing Stringdata, you should erase the whitespaces on the right and left. This is done by the Strimg#trimfunction like this:
分析String数据时，应擦除左右两侧的空格。这是由这样的Strimg#trim函数完成的：
password = password.trim();
To analize every character of the String, you can transform it to a char array, so it will be easier to fulfill your requirements:
要分析 String 的每个字符，您可以将其转换为 char 数组，因此更容易满足您的要求：
char[] arrPassword = password.toCharArray();
Now you can evaluate a char using these functions: Character#isUpperCase, Character#isLowerCase, Character#isDigit.
现在您可以使用以下函数计算字符：Character#isUpperCase, Character#isLowerCase, Character#isDigit。
Last but not least, you can have a String with the special characters you need to check, and check if the actual character you're evaluating is inside that String. This could be achieved using String#indexOfand String#valueOf, this las one to convert the char to a String type.
最后但并非最不重要的是，您可以拥有一个包含您需要检查的特殊字符的字符串，并检查您正在评估的实际字符是否在该字符串内。这可以使用String#indexOfand来实现String#valueOf，这是将 char 转换为 String 类型的方法。
Here is a code sample for all this explanation:
这是所有这些解释的代码示例：
public static final String SPECIAL_CHARACTERS = "!@#$%^&*()~`-=_+[]{}|:\";',./<>?";
public static final int MIN_PASSWORD_LENGTH = 8;
public static final int MAX_PASSWORD_LENGTH = 20;
public static boolean isAcceptablePassword(String password) {
if (TextUtils.isEmpty(password)) {
System.out.println("empty string.");
return false;
}
password = password.trim();
int len = password.length();
if(len < MIN_PASSWORD_LENGTH || len > MAX_PASSWORD_LENGTH) {
System.out.println("wrong size, it must have at least 8 characters and less than 20.");
return false;
}
char[] aC = password.toCharArray();
for(char c : aC) {
if (Character.isUpperCase(c)) {
System.out.println(c + " is uppercase.");
} else
if (Character.isLowerCase(c)) {
System.out.println(c + " is lowercase.");
} else
if (Character.isDigit(c)) {
System.out.println(c + " is digit.");
} else
if (SPECIAL_CHARACTERS.indexOf(String.valueOf(c)) >= 0) {
System.out.println(c + " is valid symbol.");
} else {
System.out.println(c + " is an invalid character in the password.");
return false;
}
}
return true;
}
The System.out.println(c + " is an invalid character in the password.");sentence is just to check the result of analyze the actual character.
这System.out.println(c + " is an invalid character in the password.");句话只是为了检查分析实际字符的结果。
回答by Adam
How about some good old regular expressions? This seems to work correctly, although might have made slip in the escaping for special char check
一些好的旧正则表达式怎么样？这似乎工作正常，尽管可能在转义特殊字符检查时出错
Pattern[] checks = {
Pattern.compile("[!@#\$%^&*()~`\-=_+\[\]{}|:\\";',\./<>?]"),
Pattern.compile("\d+"),
Pattern.compile("[A-Z]+"),
Pattern.compile("[a-z]+"),
Pattern.compile("^.{8,20}$") };
for (String test : new String[] { "password", "Password1",
"Password1&", "toolongtoolongtoolong" }) {
boolean ok = true;
for (Pattern check : checks) {
ok = ok && check.matcher(test).find();
}
System.out.println(test + " " + ok);
}
回答by Michael
Stephen is right with a bit of searching you would have found your answers easily around here. But the thread Stephen refers to is using a thirdparty library.
斯蒂芬是对的，稍微搜索一下，您就可以在这里轻松找到答案。但是斯蒂芬提到的线程正在使用第三方库。
If you want to implement this yourself then before starting the for-loop initialize 4 booleans for your requirements with false. While looping check for all four requirements until one is true. Set the corresponding boolean to true.
如果你想自己实现这个，那么在开始 for 循环之前用 false 为你的要求初始化 4 个布尔值。循环检查所有四个要求，直到其中一个为真。将相应的布尔值设置为 true。
How to check the 4 requirements:
如何检查4个要求：
- The length req you already implemented.
- Character(yourChar).isLowerCase()
- Character(yourChar).isUpperCase()
- Special character: see here: Java String Special character replacement- you can choose a similar approach
- 您已经实现的长度要求。
- 字符(yourChar).isLowerCase()
- 字符(yourChar).isUpperCase()
- 特殊字符：请参阅此处：Java 字符串特殊字符替换- 您可以选择类似的方法
After the loop check for the 4 booleans and react accordingly.
在循环检查 4 个布尔值后并做出相应的反应。
