# How do I arrange multi-dimensional array by the String date value (MM/dd/yyyy) of the sub-arrays?

This question is very urgent.  I'm finalizing something that needs to be done in a few hours.  While I'm tinkering with it, I figured I would post it here.

I've got an multi-dimentsional arary that would expand out as follows:

People Array {
0 =>  Array {
0 => String:"Jon Doe"
1 => String:"08/21/1980"
2 => String:"OtherInfo1"
3 => String:"OtherInfo2"
4 => String:"OtherInfo3"
}
1 =>  Array {
0 => String:"Jane Doe"
1 => String:"09/11/1978"
2 => String:"OtherInfo1"
3 => String:"OtherInfo2"
4 => String:"OtherInfo3"
}
2 =>  Array {
0 => String:"Billy Doe"
1 => String:"05/18/1992"
2 => String:"OtherInfo1"
3 => String:"OtherInfo2"
4 => String:"OtherInfo3"
}
}

Assume the data is already verified.  How do I put the People Array in order by Date (People[x][1])?  The data is received by the user and the length the the People array is undertermined.  So I'm thinking something along the lines of loop, but just not sure one the data best way to do the conversions.

Thanks in advance for any assistance.
LVL 2
###### Who is Participating?

x

Commented:
The fact that you're writing procedural code (as opposed to OO) makes this trickier but the following should work:
``````Arrays.sort(array, new Comparator() {
DateFormat df = new SimpleDateFormat("MM/dd/yyyy");
public int compare(Object o1, Object o2) {
String[] s1 = (String[])o1;
String[] s2 = (String[])o2;
try {
return df.parse(s1[1]).compareTo(df.parse(s2[1]);
} catch (ParseException e) {
e.printStackTrace();
return Integer.MIN_VALUE;
}
}
});
``````
0

Commented:
Use Arrays.sort with a Comparator
0

Author Commented:
Man I butchered the heck out of the last paragraph.  Here's a re-write:

Assume the data is already verified.  How do I put the People Array in order by Date (People[x][1])?  The data is received by the user and the length the the People array is undetermined.  So I'm thinking something along the lines of loop, but I am just not sure what the best way to do the data comparisons.

~Ray
http://www.vertigion.com
0

Author Commented:
CEHJ:  Can you expand on that?   Maybe a link to an example?
0

Author Commented:
http://java.sun.com/javase/6/docs/api/java/util/Arrays.html#sort(T[],%20java.util.Comparator)

I read through it and the link to the comparator (http://java.sun.com/javase/6/docs/api/java/util/Comparator.html), but I'm still not clear on how to compare them.  That's why I posted this (phishing for an example).

Thanks again for the time!
0

Author Commented:
Forgot the closing ")" on the following line (fixed in my version:
return df.parse(s1[1]).compareTo(df.parse(s2[1]));

It is not working my particular code, but I think that it's something I need to work out.  I just need to fix a bug in my array creation which involves dynamically extending it.  I was under the impression that if I declared an array as shown below that it would dynamically expand the array as I in creased the value of the size.

The code below doesn't work ... but I was told that the logic of increasing the value of the "peopleCnt" should increase the length of the array.  The code below fails after the first pass.

Any thoughts?
``````int peopleCnt = 1;
String people[][];
people = new String[peopleCnt][];

for (<condition>)
{
people[peopleCnt-1][0] = "<this String>";
people[peopleCnt-1][1] = "<this String>";
people[peopleCnt-1][2] = "<this String>";
people[peopleCnt-1][3] = "<this String>";
people[peopleCnt-1][4] = "<this String>";
peopleCnt++;
}
``````
0

Author Commented:
Correction ... line 3:
``````int peopleCnt = 1;
String people[][];
people = new String[peopleCnt][5];

for (<condition>)
{
people[peopleCnt-1][0] = "<this String>";
people[peopleCnt-1][1] = "<this String>";
people[peopleCnt-1][2] = "<this String>";
people[peopleCnt-1][3] = "<this String>";
people[peopleCnt-1][4] = "<this String>";
peopleCnt++;
}
``````
0

Commented:
An array is inherently fixed size. You can't change it. You need to use a collection class if you want dynamic growth/shrinkage
0

Author Commented:
I'll figure out collections.  I believe I'm out of bounds on the existing question, so I'll award you the earned points.  Thank you very much for your assistance!
0

Commented:
:-)
0
Question has a verified solution.

Are you are experiencing a similar issue? Get a personalized answer when you ask a related question.

Have a better answer? Share it in a comment.