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Posted to user@hive.apache.org by Zoltán Tóth-Czifra <zo...@softonic.com> on 2012/05/15 15:11:07 UTC
Date format - any easier way
Hi guys,
Thanks you very much in advance for your help.
My problem in short is getting the date for yesterday in a YYYYMMDD format. As I use this format for partitions, I need this format in quite some queries.
So far I have this:
concat(
year( date_sub( to_date( from_unixtime( unix_timestamp() ) ), 1 ) ),
CASE
WHEN month( date_sub( to_date( from_unixtime( unix_timestamp() ) ), 1 ) ) < 10
THEN concat( '0', month( date_sub( to_date( from_unixtime( unix_timestamp() ) ), 1 ) ) )
ELSE trim( month( date_sub( to_date( from_unixtime( unix_timestamp() ) ), 1 ) ) )
END,
CASE
WHEN day( date_sub( to_date( from_unixtime( unix_timestamp() ) ), 1 ) ) < 10
THEN concat( '0', day( date_sub( to_date( from_unixtime( unix_timestamp() ) ), 1 ) ) )
ELSE trim(day( date_sub( to_date( from_unixtime( unix_timestamp() ) ), 1 ) ) )
END
);
...but it seems to be a bit crazy, especially if you have to repeat it in hundreds of queries. Is there any other (better) way to get this format from yesterday? - there has to be. As I can't use local user variables nor macros whatsoever, I need to repeat myself a lot here. If there is no other way, probably I need to change my partitions.
Any ideas are appreciated. Thank you!
Zoltan
Re: Date format - any easier way
Posted by Nitin Pawar <ni...@gmail.com>.
may be something like this will work
can you try using concat(split(date_sub(),"-")))
split returns the array and then you can concat them as you want
if this does not work for you, writing a simple UDF is easy as well
Thanks,
nitin
On Tue, May 15, 2012 at 6:56 PM, Zoltán Tóth-Czifra <
zoltan.tothczifra@softonic.com> wrote:
> Nitin,
>
> Thank you. As you see below I know and use this function. My problem is
> that it doesn't give YYYYMMDD format, but YYYY-MM-DD instead, and
> formatting is not trivial as you can see it too.
>
> ------------------------------
> *From:* Nitin Pawar [nitinpawar432@gmail.com]
> *Sent:* Tuesday, May 15, 2012 3:24 PM
> *To:* user@hive.apache.org
> *Subject:* Re: Date format - any easier way
>
> you may want to have a look at this function
>
> date_sub(string startdate, int days) Subtract a number of days to
> startdate: date_sub('2008-12-31', 1) = '2008-12-30'
> On Tue, May 15, 2012 at 6:41 PM, Zoltán Tóth-Czifra <
> zoltan.tothczifra@softonic.com> wrote:
>
>> Hi guys,
>>
>> Thanks you very much in advance for your help.
>>
>> My problem in short is getting the date for yesterday in a YYYYMMDD
>> format. As I use this format for partitions, I need this format in quite
>> some queries.
>>
>> So far I have this:
>>
>> concat(
>> year( date_sub( to_date( from_unixtime( unix_timestamp() ) ), 1 ) ),
>> CASE
>> WHEN month( date_sub( to_date( from_unixtime( unix_timestamp() ) ), 1 ) )
>> < 10
>> THEN concat( '0', month( date_sub( to_date( from_unixtime(
>> unix_timestamp() ) ), 1 ) ) )
>> ELSE trim( month( date_sub( to_date( from_unixtime( unix_timestamp() ) ),
>> 1 ) ) )
>> END,
>> CASE
>> WHEN day( date_sub( to_date( from_unixtime( unix_timestamp() ) ), 1 ) ) <
>> 10
>> THEN concat( '0', day( date_sub( to_date( from_unixtime( unix_timestamp()
>> ) ), 1 ) ) )
>> ELSE trim(day( date_sub( to_date( from_unixtime( unix_timestamp() ) ), 1
>> ) ) )
>> END
>> );
>>
>>
>> ...but it seems to be a bit crazy, especially if you have to repeat it
>> in hundreds of queries. Is there any other (better) way to get this format
>> from yesterday? - there has to be. As I can't use local user variables nor
>> macros whatsoever, I need to repeat myself a lot here. If there is no other
>> way, probably I need to change my partitions.
>>
>> Any ideas are appreciated. Thank you!
>>
>> Zoltan
>>
>
>
>
> --
> Nitin Pawar
>
>
--
Nitin Pawar
Re: Date format - any easier way
Posted by Philip Tromans <ph...@gmail.com>.
I knocked up the following when we were experimenting with Hive. I've been
meaning to go and tidy it up for a while, but using it with a separator of
"" (empty string) should have the desired effect. (Obviously the UDF throws
an exception if the array is empty, been meaning to fix that for a while...)
Cheers,
Phil.
import java.util.List;
import org.apache.hadoop.hive.ql.exec.Description;
import org.apache.hadoop.hive.ql.exec.UDFArgumentException;
import org.apache.hadoop.hive.ql.metadata.HiveException;
import org.apache.hadoop.hive.ql.udf.generic.GenericUDF;
import org.apache.hadoop.hive.serde2.objectinspector.ListObjectInspector;
import org.apache.hadoop.hive.serde2.objectinspector.ObjectInspector;
import
org.apache.hadoop.hive.serde2.objectinspector.ObjectInspector.Category;
import
org.apache.hadoop.hive.serde2.objectinspector.PrimitiveObjectInspector;
import
org.apache.hadoop.hive.serde2.objectinspector.primitive.PrimitiveObjectInspectorFactory;
@Description(name = "implode", value = "_FUNC_(list,separator) - joins the
elements of list together, separated by the given string."
+ " Returns a string. [Add an example here]")
public class GenericUDFImplode extends GenericUDF {
private ListObjectInspector listOI = null;
private PrimitiveObjectInspector glueOI = null;
@Override
public Object evaluate(DeferredObject[] params) throws HiveException {
return join(listOI.getList(params[0].get()),
glueOI.getPrimitiveJavaObject(params[1].get()).toString());
}
private String join(List<?> list, String separator) {
if (list == null) {
return null;
}
if (list.size() == 0) {
return "";
}
StringBuffer buf = new StringBuffer();
for (Object o : list) {
buf.append(o);
buf.append(separator);
}
return buf.substring(0, buf.length() - separator.length());
}
@Override
public String getDisplayString(String[] args) {
return "implode(" + args[0] + "," + args[1] + ")";
}
@Override
public ObjectInspector initialize(ObjectInspector[] params)
throws UDFArgumentException {
if (params[0].getCategory() != Category.LIST) {
throw new UDFArgumentException("Expected: List as argument 1 to implode()");
}
if (params[1].getCategory() != Category.PRIMITIVE) {
throw new UDFArgumentException("Expected: Primitive as argument 2 to
implode()");
}
listOI = (ListObjectInspector) params[0];
glueOI = (PrimitiveObjectInspector) params[1];
return PrimitiveObjectInspectorFactory.javaStringObjectInspector;
}
}
On 15 May 2012 15:33, Nitin Pawar <ni...@gmail.com> wrote:
> I will write an UDF for array concatenation and upload on GIT if anyone
> does not have it already
>
>
> On Tue, May 15, 2012 at 7:24 PM, Zoltán Tóth-Czifra <
> zoltan.tothczifra@softonic.com> wrote:
>
>> Matt, thanks!
>>
>> Luckily the order of the parts of the date is correct (reordering them
>> would bet he same craziness).
>>
>> Finally it is:
>>
>> regexp_replace(
>> date_sub(
>> to_date(
>> from_unixtime(
>> unix_timestamp()
>> )
>> ), 1
>> ), "[-]", ""
>> )
>>
>> Nitin, concat apparently doesn't take arrays, and I did not find any
>> other way to join arrays in HQL. However, it would be very handy.
>>
>> Thanks guys!
>>
>> ------------------------------
>> *From:* Tucker, Matt [Matt.Tucker@disney.com]
>> *Sent:* Tuesday, May 15, 2012 3:33 PM
>>
>> *To:* user@hive.apache.org
>> *Subject:* RE: Date format - any easier way
>>
>> What about wrapping it in regexp_replace(…, “[-]”, “”) ? It may not
>> be the cleanest, but I’d recommend passing variables from the shell :)
>>
>>
>>
>> Matt Tucker
>>
>>
>>
>> *From:* Zoltán Tóth-Czifra [mailto:zoltan.tothczifra@softonic.com]
>> *Sent:* Tuesday, May 15, 2012 9:27 AM
>> *To:* user@hive.apache.org
>> *Subject:* RE: Date format - any easier way
>>
>>
>>
>> Nitin,
>>
>>
>>
>> Thank you. As you see below I know and use this function. My problem is
>> that it doesn't give YYYYMMDD format, but YYYY-MM-DD instead, and
>> formatting is not trivial as you can see it too.
>>
>>
>> ------------------------------
>>
>> *From:* Nitin Pawar [nitinpawar432@gmail.com]
>> *Sent:* Tuesday, May 15, 2012 3:24 PM
>> *To:* user@hive.apache.org
>> *Subject:* Re: Date format - any easier way
>>
>> you may want to have a look at this function
>>
>>
>>
>> date_sub(string startdate, int days)
>>
>> Subtract a number of days to startdate: date_sub('2008-12-31', 1) =
>> '2008-12-30'
>>
>>
>>
>> On Tue, May 15, 2012 at 6:41 PM, Zoltán Tóth-Czifra <
>> zoltan.tothczifra@softonic.com> wrote:
>>
>> Hi guys,
>>
>>
>>
>> Thanks you very much in advance for your help.
>>
>>
>>
>> My problem in short is getting the date for yesterday in a YYYYMMDD
>> format. As I use this format for partitions, I need this format in quite
>> some queries.
>>
>>
>>
>> So far I have this:
>>
>>
>>
>> concat(
>>
>> year( date_sub( to_date( from_unixtime( unix_timestamp() ) ), 1 ) ),
>>
>> CASE
>>
>> WHEN month( date_sub( to_date( from_unixtime( unix_timestamp() ) ), 1 ) )
>> < 10
>>
>> THEN concat( '0', month( date_sub( to_date( from_unixtime(
>> unix_timestamp() ) ), 1 ) ) )
>>
>> ELSE trim( month( date_sub( to_date( from_unixtime( unix_timestamp() ) ),
>> 1 ) ) )
>>
>> END,
>>
>> CASE
>>
>> WHEN day( date_sub( to_date( from_unixtime( unix_timestamp() ) ), 1 ) ) <
>> 10
>>
>> THEN concat( '0', day( date_sub( to_date( from_unixtime( unix_timestamp()
>> ) ), 1 ) ) )
>>
>> ELSE trim(day( date_sub( to_date( from_unixtime( unix_timestamp() ) ), 1
>> ) ) )
>>
>> END
>>
>> );
>>
>>
>>
>>
>>
>> ...but it seems to be a bit crazy, especially if you have to repeat it in
>> hundreds of queries. Is there any other (better) way to get this format
>> from yesterday? - there has to be. As I can't use local user variables nor
>> macros whatsoever, I need to repeat myself a lot here. If there is no other
>> way, probably I need to change my partitions.
>>
>>
>>
>> Any ideas are appreciated. Thank you!
>>
>>
>>
>> Zoltan
>>
>>
>>
>>
>>
>> --
>> Nitin Pawar
>>
>
>
>
> --
> Nitin Pawar
>
>
Re: Date format - any easier way
Posted by Nitin Pawar <ni...@gmail.com>.
I will write an UDF for array concatenation and upload on GIT if anyone
does not have it already
On Tue, May 15, 2012 at 7:24 PM, Zoltán Tóth-Czifra <
zoltan.tothczifra@softonic.com> wrote:
> Matt, thanks!
>
> Luckily the order of the parts of the date is correct (reordering them
> would bet he same craziness).
>
> Finally it is:
>
> regexp_replace(
> date_sub(
> to_date(
> from_unixtime(
> unix_timestamp()
> )
> ), 1
> ), "[-]", ""
> )
>
> Nitin, concat apparently doesn't take arrays, and I did not find any
> other way to join arrays in HQL. However, it would be very handy.
>
> Thanks guys!
>
> ------------------------------
> *From:* Tucker, Matt [Matt.Tucker@disney.com]
> *Sent:* Tuesday, May 15, 2012 3:33 PM
>
> *To:* user@hive.apache.org
> *Subject:* RE: Date format - any easier way
>
> What about wrapping it in regexp_replace(…, “[-]”, “”) ? It may not be
> the cleanest, but I’d recommend passing variables from the shell :)
>
>
>
> Matt Tucker
>
>
>
> *From:* Zoltán Tóth-Czifra [mailto:zoltan.tothczifra@softonic.com]
> *Sent:* Tuesday, May 15, 2012 9:27 AM
> *To:* user@hive.apache.org
> *Subject:* RE: Date format - any easier way
>
>
>
> Nitin,
>
>
>
> Thank you. As you see below I know and use this function. My problem is
> that it doesn't give YYYYMMDD format, but YYYY-MM-DD instead, and
> formatting is not trivial as you can see it too.
>
>
> ------------------------------
>
> *From:* Nitin Pawar [nitinpawar432@gmail.com]
> *Sent:* Tuesday, May 15, 2012 3:24 PM
> *To:* user@hive.apache.org
> *Subject:* Re: Date format - any easier way
>
> you may want to have a look at this function
>
>
>
> date_sub(string startdate, int days)
>
> Subtract a number of days to startdate: date_sub('2008-12-31', 1) =
> '2008-12-30'
>
>
>
> On Tue, May 15, 2012 at 6:41 PM, Zoltán Tóth-Czifra <
> zoltan.tothczifra@softonic.com> wrote:
>
> Hi guys,
>
>
>
> Thanks you very much in advance for your help.
>
>
>
> My problem in short is getting the date for yesterday in a YYYYMMDD
> format. As I use this format for partitions, I need this format in quite
> some queries.
>
>
>
> So far I have this:
>
>
>
> concat(
>
> year( date_sub( to_date( from_unixtime( unix_timestamp() ) ), 1 ) ),
>
> CASE
>
> WHEN month( date_sub( to_date( from_unixtime( unix_timestamp() ) ), 1 ) )
> < 10
>
> THEN concat( '0', month( date_sub( to_date( from_unixtime(
> unix_timestamp() ) ), 1 ) ) )
>
> ELSE trim( month( date_sub( to_date( from_unixtime( unix_timestamp() ) ),
> 1 ) ) )
>
> END,
>
> CASE
>
> WHEN day( date_sub( to_date( from_unixtime( unix_timestamp() ) ), 1 ) ) <
> 10
>
> THEN concat( '0', day( date_sub( to_date( from_unixtime( unix_timestamp()
> ) ), 1 ) ) )
>
> ELSE trim(day( date_sub( to_date( from_unixtime( unix_timestamp() ) ), 1 )
> ) )
>
> END
>
> );
>
>
>
>
>
> ...but it seems to be a bit crazy, especially if you have to repeat it in
> hundreds of queries. Is there any other (better) way to get this format
> from yesterday? - there has to be. As I can't use local user variables nor
> macros whatsoever, I need to repeat myself a lot here. If there is no other
> way, probably I need to change my partitions.
>
>
>
> Any ideas are appreciated. Thank you!
>
>
>
> Zoltan
>
>
>
>
>
> --
> Nitin Pawar
>
--
Nitin Pawar
RE: Date format - any easier way
Posted by Zoltán Tóth-Czifra <zo...@softonic.com>.
Matt, thanks!
Luckily the order of the parts of the date is correct (reordering them would bet he same craziness).
Finally it is:
regexp_replace(
date_sub(
to_date(
from_unixtime(
unix_timestamp()
)
), 1
), "[-]", ""
)
Nitin, concat apparently doesn't take arrays, and I did not find any other way to join arrays in HQL. However, it would be very handy.
Thanks guys!
________________________________
From: Tucker, Matt [Matt.Tucker@disney.com]
Sent: Tuesday, May 15, 2012 3:33 PM
To: user@hive.apache.org
Subject: RE: Date format - any easier way
What about wrapping it in regexp_replace(…, “[-]”, “”) ? It may not be the cleanest, but I’d recommend passing variables from the shell :)
Matt Tucker
From: Zoltán Tóth-Czifra [mailto:zoltan.tothczifra@softonic.com]
Sent: Tuesday, May 15, 2012 9:27 AM
To: user@hive.apache.org
Subject: RE: Date format - any easier way
Nitin,
Thank you. As you see below I know and use this function. My problem is that it doesn't give YYYYMMDD format, but YYYY-MM-DD instead, and formatting is not trivial as you can see it too.
________________________________
From: Nitin Pawar [nitinpawar432@gmail.com]
Sent: Tuesday, May 15, 2012 3:24 PM
To: user@hive.apache.org<ma...@hive.apache.org>
Subject: Re: Date format - any easier way
you may want to have a look at this function
date_sub(string startdate, int days)
Subtract a number of days to startdate: date_sub('2008-12-31', 1) = '2008-12-30'
On Tue, May 15, 2012 at 6:41 PM, Zoltán Tóth-Czifra <zo...@softonic.com>> wrote:
Hi guys,
Thanks you very much in advance for your help.
My problem in short is getting the date for yesterday in a YYYYMMDD format. As I use this format for partitions, I need this format in quite some queries.
So far I have this:
concat(
year( date_sub( to_date( from_unixtime( unix_timestamp() ) ), 1 ) ),
CASE
WHEN month( date_sub( to_date( from_unixtime( unix_timestamp() ) ), 1 ) ) < 10
THEN concat( '0', month( date_sub( to_date( from_unixtime( unix_timestamp() ) ), 1 ) ) )
ELSE trim( month( date_sub( to_date( from_unixtime( unix_timestamp() ) ), 1 ) ) )
END,
CASE
WHEN day( date_sub( to_date( from_unixtime( unix_timestamp() ) ), 1 ) ) < 10
THEN concat( '0', day( date_sub( to_date( from_unixtime( unix_timestamp() ) ), 1 ) ) )
ELSE trim(day( date_sub( to_date( from_unixtime( unix_timestamp() ) ), 1 ) ) )
END
);
...but it seems to be a bit crazy, especially if you have to repeat it in hundreds of queries. Is there any other (better) way to get this format from yesterday? - there has to be. As I can't use local user variables nor macros whatsoever, I need to repeat myself a lot here. If there is no other way, probably I need to change my partitions.
Any ideas are appreciated. Thank you!
Zoltan
--
Nitin Pawar
RE: Date format - any easier way
Posted by "Tucker, Matt" <Ma...@disney.com>.
What about wrapping it in regexp_replace(..., "[-]", "") ? It may not be the cleanest, but I'd recommend passing variables from the shell :)
Matt Tucker
From: Zoltán Tóth-Czifra [mailto:zoltan.tothczifra@softonic.com]
Sent: Tuesday, May 15, 2012 9:27 AM
To: user@hive.apache.org
Subject: RE: Date format - any easier way
Nitin,
Thank you. As you see below I know and use this function. My problem is that it doesn't give YYYYMMDD format, but YYYY-MM-DD instead, and formatting is not trivial as you can see it too.
________________________________
From: Nitin Pawar [nitinpawar432@gmail.com]
Sent: Tuesday, May 15, 2012 3:24 PM
To: user@hive.apache.org<ma...@hive.apache.org>
Subject: Re: Date format - any easier way
you may want to have a look at this function
date_sub(string startdate, int days)
Subtract a number of days to startdate: date_sub('2008-12-31', 1) = '2008-12-30'
On Tue, May 15, 2012 at 6:41 PM, Zoltán Tóth-Czifra <zo...@softonic.com>> wrote:
Hi guys,
Thanks you very much in advance for your help.
My problem in short is getting the date for yesterday in a YYYYMMDD format. As I use this format for partitions, I need this format in quite some queries.
So far I have this:
concat(
year( date_sub( to_date( from_unixtime( unix_timestamp() ) ), 1 ) ),
CASE
WHEN month( date_sub( to_date( from_unixtime( unix_timestamp() ) ), 1 ) ) < 10
THEN concat( '0', month( date_sub( to_date( from_unixtime( unix_timestamp() ) ), 1 ) ) )
ELSE trim( month( date_sub( to_date( from_unixtime( unix_timestamp() ) ), 1 ) ) )
END,
CASE
WHEN day( date_sub( to_date( from_unixtime( unix_timestamp() ) ), 1 ) ) < 10
THEN concat( '0', day( date_sub( to_date( from_unixtime( unix_timestamp() ) ), 1 ) ) )
ELSE trim(day( date_sub( to_date( from_unixtime( unix_timestamp() ) ), 1 ) ) )
END
);
...but it seems to be a bit crazy, especially if you have to repeat it in hundreds of queries. Is there any other (better) way to get this format from yesterday? - there has to be. As I can't use local user variables nor macros whatsoever, I need to repeat myself a lot here. If there is no other way, probably I need to change my partitions.
Any ideas are appreciated. Thank you!
Zoltan
--
Nitin Pawar
RE: Date format - any easier way
Posted by Zoltán Tóth-Czifra <zo...@softonic.com>.
Nitin,
Thank you. As you see below I know and use this function. My problem is that it doesn't give YYYYMMDD format, but YYYY-MM-DD instead, and formatting is not trivial as you can see it too.
________________________________
From: Nitin Pawar [nitinpawar432@gmail.com]
Sent: Tuesday, May 15, 2012 3:24 PM
To: user@hive.apache.org
Subject: Re: Date format - any easier way
you may want to have a look at this function
date_sub(string startdate, int days) Subtract a number of days to startdate: date_sub('2008-12-31', 1) = '2008-12-30'
On Tue, May 15, 2012 at 6:41 PM, Zoltán Tóth-Czifra <zo...@softonic.com>> wrote:
Hi guys,
Thanks you very much in advance for your help.
My problem in short is getting the date for yesterday in a YYYYMMDD format. As I use this format for partitions, I need this format in quite some queries.
So far I have this:
concat(
year( date_sub( to_date( from_unixtime( unix_timestamp() ) ), 1 ) ),
CASE
WHEN month( date_sub( to_date( from_unixtime( unix_timestamp() ) ), 1 ) ) < 10
THEN concat( '0', month( date_sub( to_date( from_unixtime( unix_timestamp() ) ), 1 ) ) )
ELSE trim( month( date_sub( to_date( from_unixtime( unix_timestamp() ) ), 1 ) ) )
END,
CASE
WHEN day( date_sub( to_date( from_unixtime( unix_timestamp() ) ), 1 ) ) < 10
THEN concat( '0', day( date_sub( to_date( from_unixtime( unix_timestamp() ) ), 1 ) ) )
ELSE trim(day( date_sub( to_date( from_unixtime( unix_timestamp() ) ), 1 ) ) )
END
);
...but it seems to be a bit crazy, especially if you have to repeat it in hundreds of queries. Is there any other (better) way to get this format from yesterday? - there has to be. As I can't use local user variables nor macros whatsoever, I need to repeat myself a lot here. If there is no other way, probably I need to change my partitions.
Any ideas are appreciated. Thank you!
Zoltan
--
Nitin Pawar
Re: Date format - any easier way
Posted by Nitin Pawar <ni...@gmail.com>.
you may want to have a look at this function
date_sub(string startdate, int days)Subtract a number of days to startdate:
date_sub('2008-12-31', 1) = '2008-12-30'
On Tue, May 15, 2012 at 6:41 PM, Zoltán Tóth-Czifra <
zoltan.tothczifra@softonic.com> wrote:
> Hi guys,
>
> Thanks you very much in advance for your help.
>
> My problem in short is getting the date for yesterday in a YYYYMMDD
> format. As I use this format for partitions, I need this format in quite
> some queries.
>
> So far I have this:
>
> concat(
> year( date_sub( to_date( from_unixtime( unix_timestamp() ) ), 1 ) ),
> CASE
> WHEN month( date_sub( to_date( from_unixtime( unix_timestamp() ) ), 1 ) )
> < 10
> THEN concat( '0', month( date_sub( to_date( from_unixtime(
> unix_timestamp() ) ), 1 ) ) )
> ELSE trim( month( date_sub( to_date( from_unixtime( unix_timestamp() ) ),
> 1 ) ) )
> END,
> CASE
> WHEN day( date_sub( to_date( from_unixtime( unix_timestamp() ) ), 1 ) ) <
> 10
> THEN concat( '0', day( date_sub( to_date( from_unixtime( unix_timestamp()
> ) ), 1 ) ) )
> ELSE trim(day( date_sub( to_date( from_unixtime( unix_timestamp() ) ), 1 )
> ) )
> END
> );
>
>
> ...but it seems to be a bit crazy, especially if you have to repeat it
> in hundreds of queries. Is there any other (better) way to get this format
> from yesterday? - there has to be. As I can't use local user variables nor
> macros whatsoever, I need to repeat myself a lot here. If there is no other
> way, probably I need to change my partitions.
>
> Any ideas are appreciated. Thank you!
>
> Zoltan
>
--
Nitin Pawar