java - Get last n matches from String with regex -


i have string html withoutany kind of close tag (</.*?>) , without new line (\n):

<tr><td align=center>01/01/2001<td align=center>500,01<td align=center>0,99<td align=center>15

this repeat indefinitely , may have 1 or more td's values. @ moment using string.split("<tr><td align=center>") separate string , use 1 regex find date , 1 find value want.

something this:

string[] stringarray = text.split("<tr><td align=center>");          string[] array1 = arrays.copyofrange(stringarray, stringarray.length - /*0<n<21*/,                 stringarray.length);          (int = 0; < array1.length; i++) {             system.out.println(array1[i]);             m1 = pattern.compile("(\\d{2}\\/\\d{2}\\/\\d{4})").matcher(                     array1[i]);              //getting date             m1.find();             system.out.println(m1.group(1));              m1 = pattern.compile("<td align=center>(\\d+,*\\d*)").matcher(array1[i]);             while (m1.find()) {                 system.out.println(m1.group(/*0<n*/));             }         }

i want way string equivalent array1 (the last n positions of string) using regex.

i know can use bigger regex $ @ end last <tr>, want 19 <tr> before to.

i don't know if being clear here. let me know if can provide more details.

ps: yes values writen ',' instead of '.'... use replace later on.

with java regular expressions can't collect arbitrary number of matches single group, unless know exact/maximum number of groups you'd have apply regex multiple times , collect matches yourself.

btw, should check whether m1.find(); returns true before calling m1.group(1); otherwise you'd illegalstateexception if expression doesn't match.

as note, i'd compile date pattern outside loop, in initialization code.


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