CODEWITHSUNDEEP
androidstringnewline
Igx33
Igx33

Reputation: 171

\n (Newline) deleting Android

There was some XML parsed text that looked like this:

06:00 Vesti<br>07:15 Something Else<br>09:10 Movie<a href="..."> ... <br>15:45 Something..

and there was a lot of it..

Well, I have done this:

String mim =ses.replaceAll("(?s)\\<.*?\\>", " \n");

there was no other way to show text nicely. Now, after few showings, and some time, I need that same text separated into alone strings like this:

06:00 Vesti

... or 

07:15 Something Else

I've tried something like this, but it does not work: 

char[] rast = description.toCharArray();
    int brojac = 0;
    for(int q=0; q<description.length(); q++){
        if(rast[q]=='\\' && rast[q+1]=='n' ) brojac++;
    }
    String[] niz = new String[brojac];

    int bf1=0;
    int bf2=0;
    int bf3=0;
    int oo=0;

    for(int q=0; q<description.length(); q++){
        if(rast[q]=='\\'&& rast[q+1]=='n'){
            bf3=bf1;
            bf1=q;

            String lol = description.substring(bf3, bf1);
            niz[oo]=lol;
            oo++;
        }
    }

I know that in description.substring(bf3,bf1) are not set as they should be but I think that this:

if(rast[q]=='\\' && rast[q+1]=='n)

does not work that way.. is there any other solution?

Note. there is no other way to get that resource. , It must be through this.

Upvotes: 1

Views: 238

Answers (2)

Jens
Jens

Reputation: 17077

Calling Html.fromHtml(String) will properly translate the <br> into \n.

String html = "06:00 Vesti<br>07:15 Something Else<br>09:10 Movie<a href=\"...\"> ... <br>15:45 Something..";
String str = Html.fromHtml(html).toString();
String[] arr = str.split("\n");

Then, just split it on a line basis - no need for regexps (which you shouldn't be using to parse HTML in the first case).

Edit: Turning everything into a bunch of Dates

// Used to find the HH:mm, in case the input is wonky
Pattern p = Pattern.compile("([0-2][0-9]:[0-5][0-9])");
SimpleDateFormat fmt = new SimpleDateFormat("HH:mm");
SortedMap<Date, String> programs = new TreeMap<Date, String>();
for (String row : arr) {
    Matcher m = p.matcher(row);
    if (m.find()) {
        // We found a time in this row
        ParsePosition pp = new ParsePosition(m.start(0));
        Date when = fmt.parse(row, pp);
        String title = row.substring(pp.getIndex()).trim();
        programs.put(when, title);
    }
}
// Now programs contain the sorted list of programs. Unfortunately, since
// SimpleDateFormat is stupid, they're all placed back in 1970 :-D.
// This would give you an ordered printout of all programs *AFTER* 08:00
Date filter = fmt.parse("08:00");
SortedMap<Date, String> after0800 = programs.tailMap(filter);
// Since this is a SortedMap, after0800.values() will return the program names in order.
// You can also iterate over each entry like so:
for (Map.Entry<Date,String> program : after0800.entrySet()) {
    // You can use the SimpleDateFormat to pretty-print the HH:mm again.
    System.out.println("When:" + fmt.format(program.getKey()));
    System.out.println("Title:" + program.getValue());            
}

Upvotes: 1

Michael Allen
Michael Allen

Reputation: 5828

Use regex:

List<String> results = new ArrayList<String>(); 

Pattern pattern = Pattern.compile("(\d+:\d+ \w+)<?");
Matcher matcher = pattern.matcher("06:00 Vesti<br>07:15 Something Else<br>09:10 Movie<a href="..."> ... <br>15:45 Something..");

while(matcher.find()) {
    results.add(matcher.group(0));
}

results will end up as a list of strings:

results = List[
    "06:00 Vesti", 
    "07:15 Something Else", 
    "09:10 Movie", 
    "15:45 Something.."]

See Rexgex Java Tutorial for an idea of how javas regex library works.

Upvotes: 0

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