i need to check a string that should contain only ABCDEFG characters, in any sequence and with only 7 characters. Please let me know the correct way of using regular expression.
as corrently i am using
String abs = "ABPID";
if(!Pattern.matches("[[ABCDEFG]", abs))
System.out.println("Error");
i am using the following code which works when i use the String abcdefg but for other cases it fails. please help me out.
Exactly 7 characters
"^[ABCDEFG]{7}$"
1 to 7 characters
"^[ABCDEFG]{1,7}$"
^(?:A()|B()|C()|D()|E()|F()|G()){7}\1\2\3\4\5\6\7$
"^(?:A()|B()|C()|D()|E()|F()|G()){7}\\1\\2\\3\\4\\5\\6\\7$" in java, for once char onceTo see if a string is a permutation of ABCDEFG, it's easy with negative lookahead and capturing group to enforce no duplicates:
^(?!.*(.).*\1)[A-G]{7}$
You don't need the anchors if you use String.matches() in Java. Here's a test harness:
String[] tests = {
"ABCDEFG", // true
"GBADFEC", // true
"ABCADFG", // false
};
for (String test : tests) {
System.out.format("%s %b%n", test,
test.matches("(?!.*(.).*\\1)[A-G]{7}")
);
}
Basically, [A-G]{7}, but also (?!.*(.).*\1). That is, no character is repeated.
Here's a test harness for the assertion to play around with:
String[] tests = {
"abcdeb", // "(b)"
"abcdefg", // "abcdefg"
"aba", // "(a)"
"abcdefgxxxhijyyy" // "(y)"
};
for (String test : tests) {
System.out.println(test.replaceAll("(?=.*(.).*\\1).*", "($1)"));
}
The way it works is by trying to match .*(.).*\1, that is, with .* in between, a captured character (.) that appears again \1.
(.), and use backreferences to see if it appears again \1. You want to allow these two occurrences to be separated by .*. It's understandable, it doesn't have to be memorized.