java valid regex for any url starting with my domain, protocol, view and the rest of the url

Question

I need a regex string to match URL starting with "https" "www.mydomain.com", "myView?" "rest of the url" The regex is for use in the jsf views no for code.

Valid urls:

https://www.mydomain.com/myView?z
https://www.mydomain.com/myView?a=hello&b=world&c=how&d=are&e=you
https://www.mydomain.com/myView?a=hello&b=world&c=how&d=are&e=(you)

Invalid urls:

https://www.mydomain.com/myView?
https://www.mydomain.com/myView
https://www.mydomain.com/myViewfdgdfgfd

Test code:

//Pattern to check if this is a valid URL address, not work
Pattern p = Pattern.compile("^(https://www.mydomain.com/myView?)");
Matcher m;
m=p.matcher(urlAddress);

Have you tried anything? Doesn't seem like you put much effort into it. You should really look into regex.. And for no effort shown (this is a code request), -1 — Vince, Nov 28 '14 at 18:31
Also, [URL](https://docs.oracle.com/javase/7/docs/api/java/net/URL.html) class already implements all the logic required to parse URL's. Consider using that instead of regular expressions. — Marko Živanović, Nov 28 '14 at 18:32
You practically wrote most of the regex yourself. Add a start anchor, escape the metacharacters and use `\S+` after the `?` — hwnd, Nov 28 '14 at 18:33
You also never mention if the regex should account for `www` — Vince, Nov 28 '14 at 18:36
Use the `URI` class; it makes this kind of tests much easier. — fge, Nov 28 '14 at 18:40

score 0 · Accepted Answer · edited May 23 '17 at 11:57

It's a simple case. This only have two parts, the start of the expression and the rest.

Start expression:

^https://www.mydomain.com/myView\\?

Rest of the url (Valid url chars):

[a-zA-Z0-9~!';@#\\^\\$%&\\*\\(\\)-_\\+=\\[\\]\\{\\}\\|\\\\,\\.\\?\\s]

Test code:

    String regex = "^https://www.mydomain.com/myView\\?[a-zA-Z0-9~!';@#\\^\\$%&\\*\\(\\)-_\\+=\\[\\]\\{\\}\\|\\\\,\\.\\?\\s]+";

    String validUrlchars = "[a-zA-Z0-9~!';@#\\^\\$%&\\*\\(\\)-_\\+=\\[\\]\\{\\}\\|\\\\,\\.\\?\\s]";

    String url1 = "https://www.mydomain.com/myView?z";
    String url2 = "https://www.mydomain.com/myView?";
    String url3 = "https://www.mydomain.com/myView";
    String url4 = "https://www.mydomain.com/myViewfdgdfgfd";
    String url5 = "https://www.mydomain.com/myView?a=hello&b=world&c=how&d=are&e=you";
    String url6 = "https://www.mydomain.com/myView?a=hello&b=world&c=how&d=are&e=(you)";

    Pattern p = Pattern.compile(regex);
    Matcher m;
    m = p.matcher(url1);
    System.out.println(m.matches() + " " + url1);
    m = p.matcher(url2);
    System.out.println(m.matches() + " " + url2);
    m = p.matcher(url3);
    System.out.println(m.matches() + " " + url3);
    m = p.matcher(url4);
    System.out.println(m.matches() + " " + url4);
    m = p.matcher(url5);
    System.out.println(m.matches() + " " + url5);
    m = p.matcher(url6);
    System.out.println(m.matches() + " " + url6);

Output:

true https://www.mydomain.com/myView?z
false https://www.mydomain.com/myView?
false https://www.mydomain.com/myView
false https://www.mydomain.com/myViewfdgdfgfd
true https://www.mydomain.com/myView?a=hello&b=world&c=how&d=are&e=you
true https://www.mydomain.com/myView?a=hello&b=world&c=how&d=are&e=(you)

java valid regex for any url starting with my domain, protocol, view and the rest of the url

1 Answers1