Re: regular expressions

From:
John Ersatznom <j.ersatz@nowhere.invalid>
Newsgroups:
comp.lang.java.programmer
Date:
Sun, 17 Dec 2006 16:03:07 -0500
Message-ID:
<em4b9m$4e9$1@aioe.org>
Flo 'Irian' Schaetz wrote:

And thus, Covington Bradshaw spoke...

How to extract 123456789@@abcdefhghij from

@123456789@@abcdefghij@987654321@@stuvwxyz@

using regular expressions


Why regular expressions?

String myString = "@123456789@@abcdefghij@987654321@@stuvwxyz@";

int first = myString.indexOf("@")+1;
int third = myString.indexOf("@", myString.indexOf("@", first)+1);


Shouldn't that be

int third = myString.indexOf("@", myString.indexOf("@", first)+2);

?

I assume the idea is to parse @foo@@bar@baz@@quux@quuux... into the
"foo@@bar" records, so you actually want something like

public static List<String> getRecords (String input)
        throws FormatException {
    int index = 0;
    int len = input.length() - 1;
    List<String> result = new LinkedList<String>();
    while (index < len) {
        int nextStart = input.indexOf('@');
        if (nextStart == -1) throw new FormatException();
        int nextMid = input.indexOf("@@", nextStart + 1);
        if (nextMid == -1) throw new FormatException();
        index = input.indexOf("@", nextMid + 2);
        if (index == -1) throw new FormatException();
        result.add(input.substring(nextStart + 1, index));
    }
    return result;
}

Result:
foo@@bar, baz@@quux, ...

Or maybe you want key/value pairs?

public static Map<String, String> getRecords (String input)
        throws FormatException {
    int index = 0;
    int len = input.length() - 1
    Map<String, String> result = new HashMap<String, String>();
    while (index < len) {
        int nextStart = input.indexOf('@');
        if (nextStart == -1) throw new FormatException();
        int nextMid = input.indexOf("@@", nextStart + 1);
        if (nextMid == -1) throw new FormatException();
        index = input.indexOf("@", nextMid + 2);
        if (index == -1) throw new FormatException();
        String key = input.substring(nextStart + 1, nextMid);
        String value = input.substring(nextMid + 2, index);
        if (key.length() == 0 || value.length() == 0)
            throw new FormatException();
        // Optional if duplicate keys are bad.
        if (result.containsKey(key))
            throw new FormatException();
        // End optional
        result.put(key, value);
    }
    return result;
}

Result: foo -> bar; baz -> quux; ...

(Both of the above should throw the exception of your choice if the
input isn't empty and its format isn't exactly as given above: @
followed by however-many occurrences of foo@@bar@. The latter disallows
empty strings, e.g. @foo@@@baz@@quux@.)

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