How to parse a string that contains value only needed partially to an arraylist?

Question

I have a string that looks like this:

String a = "name:string, age:int, address:string,city:string, zip:int";

The string is separated by , and between : it is the fieldName and type. There might be space between each value but any space should be ignored.

What I want is to convert this string into an arrayList that only contents fieldName.

So the above example will become

List<String> result = new ArrayList<>; and in the result there should be values of name, age, address, city, zip. Everything else should be ignored.

How do i do this with regex in Java?

I can achieve this by first do the split by , then split the string array by :, and take the first element of the array then added to a list.

The reason i m asking is because I want to know if there is a more elegant way to do this with regex.

@Andreas i added what i have tried in the description. The whole reason to ask this is to check if i can do it in one step instead of multiple — OPK, Aug 20 '19 at 20:22
That's not how the original question read. It read as "I want this, tell me how to do it". --- You still don't **show** the *code* of what you have tried. Don't explain the code, show it! — Andreas, Aug 20 '19 at 20:25

Christian Emerson · Answer 1 · 2019-08-20T20:29:17.240

0

The following regex will give you the results you want

[a-z]+[^, ](?=:)

The first parentheses is called a look ahead, so anything that is before a : will be selected, the first set of brackets says we want any character a-z (note these are case sensitive so add +[A-Z] if caps may be used), the final set of brackets says no , or space

to use this in java you would write something along the lines of

Pattern.compile("[a-z]+[^, ](?=:)")
       .matcher(a)
       .find();

I use this site a lot to test regex's it's a really helpful tool

edited Aug 20 '19 at 20:29

answered Aug 20 '19 at 20:18

Christian Emerson

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1
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That will match all the types (`string, int, string, string, int`), but OP wants the names (`name, age, address, city, zip`). – Andreas Aug 20 '19 at 20:20
Sorry I misread, I have updated my answer – Christian Emerson Aug 20 '19 at 20:22
i still don't think this works. just tested it. – OPK Aug 20 '19 at 20:24
What results are you getting with that? And where are you testing? – Christian Emerson Aug 20 '19 at 20:25
The code works now, but only if you use a regex `find()` loop, not if you use `split()`. Neither question **nor** answer shows how the regex is ***used***. Both question and answer are incomplete. `unspecific question + unspecific question = mass confusion` – Andreas Aug 20 '19 at 20:27
That code does nothing. – Andreas Aug 20 '19 at 20:29
Yeah I'm going to leave this one be lol, I added how I would use it in Java and hopefully that helps – Christian Emerson Aug 20 '19 at 20:30

score 0 · Accepted Answer · answered Aug 20 '19 at 20:35

0

String[] b = a.split(":[^,]+(?:, *)?");

Or:

String[] b = Pattern.compile("[^, ]+(?=:)").matcher(a).results()
                    .map(r -> r.group()).toArray(String[]::new);

answered Aug 20 '19 at 20:35

Andreas

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@OPK [Reference - What does this regex mean?](https://stackoverflow.com/q/22937618/5221149) – Andreas Aug 20 '19 at 21:47

How to parse a string that contains value only needed partially to an arraylist?

2 Answers2