Removing Dates from Strings

[Bertram Hurtig]

Hi,

I have a big file - each line may start with date and time (german date
formatting). To be able to sort and compare these lines, I want to
remove the Date and time sub strings.

For the date and time, I got this regex:
\d\d\.\d\d.\d\d\d\d\s\d\d\:\d\d:\d\d

I know there are classes like Pattern and Matcher, but this only tells
me if the String contains a date + time - but no idea how to also get
the relevant index positions to be able to remove the substrings found.

I know I could "manually" do this writting my own parsing method,
but I would prefer to have it nice, short (and performance optimized) -
and I don't want to reinvent the wheel.... ;-)

Any ideas?

Thanks in advance,


Betram

 

[Christian]

Hi,

I have a big file - each line may start with date and time (german date
formatting). To be able to sort and compare these lines, I want to
remove the Date and time sub strings.

For the date and time, I got this regex:
\d\d\.\d\d.\d\d\d\d\s\d\d\:\d\d:\d\d

I know there are classes like Pattern and Matcher, but this only tells
me if the String contains a date + time - but no idea how to also get
the relevant index positions to be able to remove the substrings found.

I know I could "manually" do this writting my own parsing method,
but I would prefer to have it nice, short (and performance optimized) -
and I don't want to reinvent the wheel.... ;-)

Any ideas?

Thanks in advance,


Betram

use capturing Groups and you are done.. just surrround interesting parts
with () then you can retrieve them ..

 

[Jeff Higgins]

Bertram Hurtig wrote

Hi,

I have a big file - each line may start with date and time (german date
formatting). To be able to sort and compare these lines, I want to remove
the Date and time sub strings.

For the date and time, I got this regex:
\d\d\.\d\d.\d\d\d\d\s\d\d\:\d\d:\d\d

I know there are classes like Pattern and Matcher, but this only tells me
if the String contains a date + time - but no idea how to also get the
relevant index positions to be able to remove the substrings found.

I know I could "manually" do this writting my own parsing method,
but I would prefer to have it nice, short (and performance optimized) -
and I don't want to reinvent the wheel.... ;-)

Any ideas?

Maybe java.text.SimpleDateFormat.parse(String text, ParsePosition pos).

 

[Stefan Ram]

formatting). To be able to sort and compare these lines, I want to
remove the Date and time sub strings.

public class Main
{
public static void main
( final java.lang.String[] args )
{
final java.util.Scanner source =new java.util.Scanner
( "12.12.2007 10:39:59 abc def\n" +
"12.12.2007 10:39:59 abc def\n" );

while( source.hasNextLine() )
{ java.lang.System.out.println
( source.nextLine().replaceAll( "(?:\\S+\\s+){2}", "" )); }}}

abc def
abc def

 

[Jeff Higgins]

Jeff Higgins wrote

Bertram Hurtig wrote
Maybe java.text.SimpleDateFormat.parse(String text, ParsePosition pos).

import java.text.ParsePosition;
import java.text.SimpleDateFormat;

public class Main
{
public static void main(String[] args)
{
String[] source =
{
"12.12.2007 10:39:59 abc def",
"abc def",
"12.12.2007 10:39:59 abc def ghi",
"12.12.2007 10:39:59 abc 12.12.2007 10:39:59",
"12.12.2007 10:39:59 12.12.2007 10:39:59 abc",
"abc def ghi 12.12.2007 10:39:59"
};
SimpleDateFormat format =
new SimpleDateFormat("dd.MM.yyyy HH:mm:ss ");
ParsePosition pos = new ParsePosition(0);
for(String s : source)
{
if(format.parse(s, pos) != null)
System.out.println(s.substring(pos.getIndex()));
else
System.out.println(s);
pos.setIndex(0);
}
}
}

abc def
abc def
abc def ghi
abc 12.12.2007 10:39:59
12.12.2007 10:39:59 abc
abc def ghi 12.12.2007 10:39:59

 

[Jeff Higgins]

Jeff Higgins wrote

Jeff Higgins wrote

import java.text.NumberFormat;
import java.text.ParsePosition;
import java.text.SimpleDateFormat;
import java.util.Arrays;
import java.util.Comparator;

public class Main
{
public static void main(String[] args)
{
String[] source =
{
"12.12.2007 10:39:59 776854 3.1416e0",
"144209 456332.987e3",
"12.12.2007 10:39:59 567445 1e3",
"12.12.2007 10:39:59 999222 19",
"334978, 332.987e3",
"999224 789.3e-15"
};

Arrays.sort(source, new IgnoreDateComparator());
for(String s : source)
System.out.println(s);
}

static class IgnoreDateComparator
implements Comparator<String>
{
@Override
public int compare(String s1, String s2)
{
SimpleDateFormat df =
new SimpleDateFormat("dd.MM.yyyy HH:mm:ss ");
ParsePosition s1Pos = new ParsePosition(0);
ParsePosition s2Pos = new ParsePosition(0);
Long s1Long, s2Long;
NumberFormat nf = NumberFormat.getIntegerInstance();

df.parse(s1, s1Pos);
df.parse(s2, s2Pos);
if(s1Pos.getIndex() > 0)
s1Pos.setIndex(s1Pos.getIndex());
if(s2Pos.getIndex() > 0)
s2Pos.setIndex(s2Pos.getIndex());
s1Long = (Long)nf.parse(s1, s1Pos);
s2Long = (Long)nf.parse(s2, s2Pos);
return s1Long.compareTo(s2Long);
}
}
}

144209 456332.987e3
334978, 332.987e3
12.12.2007 10:39:59 567445 1e3
12.12.2007 10:39:59 776854 3.1416e0
12.12.2007 10:39:59 999222 19
999224 789.3e-15

 

[Roedy Green]

I know there are classes like Pattern and Matcher, but this only tells
me if the String contains a date + time - but no idea how to also get
the relevant index positions to be able to remove the substrings found.

the key in what are called groups to get the pieces. See
http://mindprod.com/jgloss/regex.html