Variables expression in the JDBC. problems need help!

Dear Sir,
I am a beginner of using JDBC. I have a problem in the
variable expression.
the fragment code is below
int today = 20041009;
ResultSet rs = statement.executeQuery("select * from
recorder where bookeddate >= 'today' order by
bookeddate");
I want to order the "booked date" and select the date
after today. Unfortunately, it doesn't work. Any
person can tell me how to add variable 'today' in to
the SQL? I used mySQL
If this question is tooo stupid, please forgive me.
what type of field is bookeddate.. a DATE or TIMESTAMP?
the format you have is a TIMESTAMP.
once you have solved the problem of actually including the variable in your query (as the previous poster mentioned) if you still have difficulty make sure you are using the right format for the given field.
you might well want to look at using prepared statements as well because they will handle the formatting for you.. you will need to convert your int variable today into an actual Date object though.

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Still stuck with the same old producer consumer weight problem need help

Hello All,
This is the problem I am stuck with right now.
I have two array lists one producer array list and one consumer array list denoted by a and b
P1 P2 P3 P4 P5
5 6 7 8 9
C1 C2 C3 C4 C5
2 3 4 5 6
Now we find all those producer consumer pairs which satisfy the criteria Pi>=Ci
We have the following sets
(5,2)(6,2)(7,2),(8,2),(9,2)
(5,3)(6,3)(7,3),(8,3),(9,3)
(5,4)(6,4)(7,4),(8,4),(9,4)
(5,5)(6,5)(7,5),(8,5),(9,5)
(6,6)(7,6)(8,6),(9,6)
Let us done each of them with Si
so we have S1,S2,S3,S4,S5
we assign a third parameter called weight to each element in Si which has satisfied the condition Pi>=Ci;
so we we will have
(5,2,ai),(6,2,bi),(7,2,ci)....etc for S1
similarly for S2 and so on.
We need to find in each set Si the the pair which has the smallest weight.
if we have (5,2,3) and (6,2,4) then 5,2,3 should be chosen.We should make sure that there is only one pair in every set which is finally chosen on the basis of weight.
Suppose we get a pair (5,2,3) in S1 and (5,2,3) in S2 we should see that (5,2,3) is not used to compare to compare with any other elements in the same set S2,
Finally we should arrive at the best element pair in each set.They should be non repeating in other sets.
Given a problem
P0 P1 P2 P3 P4
9 5 2 2 8
6 5 4 5 3
C0 C1 C2 C3 C4
we have So as (P0,C0) and (P4,C0)
assuming that the one with the smaller index has lesser weight PO is selected.In the program I have used random weights.from set S1 we select the pair PO,CO
S1 =(P0,C1),(P1,C1) and (P4,C1)
since P0 and P4 are already used in previous set we dont use them for checking in S1 so we have (P1,C1) as best.
S2=(P0,C2),(P1,C2) and (P4,C2) so we dont use P0,C2 and P1 and C2 because PO and P1 are already used in S1.
So we choose P4,C2
in S3 and S4 ae have (P0,C3),(P1,C3),(P4,C3) so we dont choose anything
and same in S4 also.
So answer is
(P0,C0),(P1,C1) and (P4,C2).
My program is trying to assign weights and I am trying to print the weights along with the sets.It doesnt work fine.I need help to write this program to do this.
Thanks.
Regards.
NP
What I have tried till now.
I have one more question could you help me with this.
I have an array list of this form.
package mypackage1;
import java.util.*;
public class DD
private int P;
private int C;
private int weight;
public void set_p(int P1)
P=P1;
public void set_c(int C1)
C=C1;
public void set_weight(int W1)
weight=W1;
public int get_p()
return P;
public int get_c()
return C;
public int get_x()
return weight;
public static void main(String args[])
ArrayList a=new ArrayList();
ArrayList min_weights_int=new ArrayList();
ArrayList rows=new ArrayList();
ArrayList temp=new ArrayList();
Hashtable h=new Hashtable();
String v;
int o=0;
DD[] d=new DD[5];
for(int i=0;i<4;i++)
d=new DD();
for(int i=0;i<4;i++)
d[i].set_p(((int)(StrictMath.random()*10 + 1)));
d[i].set_c((int)(StrictMath.random()*10 + 1));
d[i].set_weight(0);
System.out.println("Producers");
for(int i=0;i<4;i++)
System.out.println(d[i].get_p());
System.out.println("Consumers");
for(int i=0;i<4;i++)
System.out.println(d[i].get_c());
System.out.println("Weights");
for(int i=0;i<4;i++)
System.out.println(d[i].get_x());
for(int i=0;i<4;i++ )
int bi =d[i].get_c();
ArrayList row=new ArrayList();
for(int j=0;j<4;j++)
if( d[j].get_p() >=bi)
d[j].set_weight((int)(StrictMath.random()*10 + 1));
row.add("(" + bi + "," + d[j].get_p() + "," +d[j].get_x() + ")");
else
d[j].set_weight(0);
row.add("null");
rows.add(row);
System.out.println(rows);
int f=0;
for(Iterator p=rows.iterator();p.hasNext();)
temp=(ArrayList)p.next();
String S="S" +f;
h.put(S,temp);
String tt=new String();
for(int j=0;j<4;j++)
if(temp.get(j).toString() !="null")
// System.out.println("In if loop");
//System.out.println(temp.get(j).toString());
String l=temp.get(j).toString();
System.out.println(l);
//System.out.println("Comma matches" + l.lastIndexOf(","));
//System.out.println(min_weights);
f++;
for(Enumeration e=h.keys();e.hasMoreElements();)
//System.out.println("I am here");
int ii=0;
int smallest=0;
String key=(String)e.nextElement();
System.out.println("key=" + key);
temp=(ArrayList)h.get(key);
System.out.println("Array List" + temp);
for( int j=0;j<4;j++)
String l=(temp.get(j).toString());
if(l!="null")
System.out.println("l=" +l);
[\code]

In your example you selected the pair with the greatest
distance from the first set, and the pair with the least
distance from the second. I don't see how the distance
function was used.
Also it's not clear to me that there is always a solution,
and, if there is, whether consistently choosing the
furthest or the closest pairs will always work.
The most obvious approach is to systematically try
all possibilities until the answer is reached, or there
are no possibilities left. This means backtracking whenever
a point is reached where you cannot continue. In this case
backtrack one step and try another possibility at this
step. After all possible choices of the previous step,
backtrack one more step and so on.
This seems rather involved, and it probably is.
Interestingly, if you know Prolog, it is ridiculously
easy because Prolog does all the backtracking for you.
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way as Prolog implementations do it--keep a list of all the
choice points and work through them until success or there
are none left.
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nikhil.gonsalves wrote:
Case 1:
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1. Does the driver jar file and/or driver class need to be changed on the connection pool front?See above.
2. If yes then, do we need to change any of the jdbc coding to ensure compatibility?
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Message was edited by:
Atena

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Message was edited by:
Atena

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Variables expression in the JDBC. problems need help!

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