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Java Forum / General / August 2003StringTokenizer Question?
I have a method to unpack a String via a given String delimiter. It works fine but it used the 'String' Class 'indexOf' and 'substring' Methods within a 'for' loop. To make this method more efficient I tried to substitute the 'StringTokenizer' Class. Easy enough but it looks like the following input String returns 3 token where I would expect 4 tokens: "AA,BB,,1," with delimiter = "," Also, the following String returns 3 Tokens where I would expect 4: ",BB,CC,1," with delimiter = "," With my code I use an 'if' condition before I unpack to ensure that the string ends with the delimiter but for 'StringTokenizer' I see that the following input String returns 4 tokens: "AA,BB,CC,1" with delimiter = "," I assume that a token must have 1 or more characters and that I must return to my original code. Is there a way to make 'StringTokenizer' recognize consecutive delimiters and return a null String [e.g. ""] as a token? Thanks Jim. I can't see any way to do it with StringTokenizer. However, you could use String.split() > I have a method to unpack a String via a given String delimiter. It works > fine but it used the 'String' Class 'indexOf' and 'substring' Methods within [quoted text clipped - 25 lines] > > Jim. > To make this method more efficient I tried to substitute the > 'StringTokenizer' Class. Easy enough but it looks like the following input > String returns 3 token where I would expect 4 tokens: > > "AA,BB,,1," with delimiter = "," Thet's because you haven't told it to return "empty" tokens - check the constructors. > Also, the following String returns 3 Tokens where I would expect 4: > > ",BB,CC,1," with delimiter = "," Same thing, except you should expect five. :) >Also, the following String returns 3 Tokens where I would expect 4: StringTokenizer(String str, String delim, boolean returnDelims) Try setting the last parameter of the constructor to true. -- Canadian Mind Products, Roedy Green. Coaching, problem solving, economical contract programming. See http://mindprod.com/jgloss/jgloss.html for The Java Glossary. > I have a method to unpack a String via a given String delimiter. It > works fine but it used the 'String' Class 'indexOf' and 'substring' > Methods within a 'for' loop. > To make this method more efficient I tried to substitute the > 'StringTokenizer' Class. Easy enough but it looks like the following > input > String returns 3 token where I would expect 4 tokens: > "AA,BB,,1," with delimiter = "," > Also, the following String returns 3 Tokens where I would expect 4: > ",BB,CC,1," with delimiter = "," > With my code I use an 'if' condition before I unpack to ensure that > the string ends with the delimiter but for 'StringTokenizer' I see > that the following input String returns 4 tokens: > "AA,BB,CC,1" with delimiter = "," > I assume that a token must have 1 or more characters and that I must > return to my original code. > Is there a way to make 'StringTokenizer' recognize consecutive > delimiters and return a null String [e.g. ""] as a token? If you use string.split(regex) then it will return the empty matches for you. I.e, you'd get 4 from your first example. StringTokenizer is (almost) deprecated now for most uses. > If you use string.split(regex) then it will return the empty matches for > you. I.e, you'd get 4 from your first example. > > StringTokenizer is (almost) deprecated now for most uses. Really? That's news to me. I am perfectly comfortable with Java 1.4 regexes, but I prefer StringTokenizer for those tasks to which it is suited. In practice, StringTokenizer is suited to a substantial number of the most frequently occurring String parsing tasks. Regexes are more powerful, but also heavier, and their API is more complex. John Bollinger jobollin@idniana.edu > Is there a way to make 'StringTokenizer' recognize consecutive delimiters > and return a null String [e.g. ""] as a token? Thanks for the many responses. In my original post I neglected to state that I am still using Java 1.3.1. Therefore, I believe I can not use String.split(Regex) yet. To be honest, I did not know about 'Regex' until I received your responses. I have since read a tutorial and look forward to using it somewhere in my App. I am avoiding using Java 1.4.1 because whenever I go to a new upgrade I have had to make many changes to my many existing code and I can not afford the time right now. This is most likely do to poor programming practices on my part since I think that a new version should be upword compatable. Thanks again, Jim >Therefore, I believe I can not use String.split(Regex) yet. For a quick introduction to what Regexes can do for you, see http://mindprod.com/jgloss/regex.html -- Canadian Mind Products, Roedy Green. Coaching, problem solving, economical contract programming. See http://mindprod.com/jgloss/jgloss.html for The Java Glossary. I have three classes that will help you out: 1) A StringTokenizer class that has the ability to return empty tokens: http://ostermiller.org/utils/StringTokenizer.html 2) A StringHelper class with a split method that does not use regular expressions: http://ostermiller.org/utils/StringHelper.html 3) A Complete Comma Separated Value (CSV) parsing class that you could use to parse a string: http://ostermiller.org/utils/CSV.html Stephen Free MagazinesGet these publications absolutely FREE for up to 12 months. There are no hidden fees and no obligation. Simply choose a title, complete the application form and submit it. Read more ...
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