android中用于不带空格的密码的正则表达式

Question

I have a EditText field in android. The password should contain minimum 4 characters and maximum 20 characters without spaces at the start or in the middle. Example blankspaceRaghu and RaghublankspaceRaghu should not be alllowed.

Pass= password.getText().toString();    

Pattern p= Pattern.compile("[^\\S]+[a-z,A-Z]+");

Matcher m = p.matcher(Pass);

Answer 1

Pass= mEditTextPassword.getText().toString();
Pattern p = Pattern.compile("^[A-Za-z0-9]{4,20}$"); 
Matcher m = p.matcher(Pass);

Answer 2

Try this pattern:

"\\A\\w{4,20}\\z"

\\\\A stands for "beginning of the input".

\\\\w is any alphanumeric (az, AZ, 0-9) character. When you don't allow digits, replace it with [a-zA-Z]

{4,20} means the previous things between 4 and 20 times in a row.

\\\\z is the end of the string. You didn't mentioned if you allow spaces AFTER the password string. When you want to allow this, insert \\\\s* (nothing or any number of whitespaces) before \\\\z .

More about regular expression syntax can be found in the documentation of java.util.regex.Pattern

Answer 3

I think you want to match with a RegEx of \\\\A\\\\S{4,20}\\\\Z . That's any non-whitespace characters, with a length between 4 and 20, matching the whole string.

Answer 4

Updated - Spaces are allowed at the end....

Pass= password.getText().toString();    
Pattern p= Pattern.compile("((?!\\s)\\A)(\\s|(?<!\\s)\\S){4,20}\\Z");
Matcher m = p.matcher(Pass);

This matches

• Match start of string as long as it is not followed by a space
• Match 4 to 20 occurrences of either a space or a non space provided it is not prefixed by a space
• Match end of string

Answer 5

You could try negative matching - that way you don't have to fuss around with negated character classes.

Match the password against /\\s/ . If it matches, you have spaces somewhere in it and it should be rejected. If not, you're free to go.