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related to java.util.hashtable

 
Greenhorn
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here is code :
import java.util.*;
class abc
{
public static void main(String[] args)
{
Hashtable hm=new Hashtable();
for (int i=0;i<5;i++ )
{
hm.put("i","vipin"+String.valueOf(i));
System.out.println("Hello World!"+i+"\t"+hm.get("i"));
}
line-1System.out.println("Hello World!"+hm.get("i"));
line-2System.out.println("Hello World!"+hm.get("4"));
}
}
my question is i am using hashtable put and get methods for storing and retrieving values.i store values using "i" as integer variable in double quotes(b'cos method accept object)but when i retrieve values using variable "i" in line 1 it gives out put using i=4 but when i direct put the value of key i=4 as in line 2 it gives null value ?? why??
 
Ranch Hand
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i in quotes becomes a String "i". This is not perl it will not translate your variable withing quotes. In PERL you could have $i =2
and then "$i" = "2" in Java it does not work.


[This message has been edited by Aleksey Matiychenko (edited July 18, 2001).]
 
Ranch Hand
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Hello,
Please Don't do it this way. You are not putting value of i while you do "i", It becomes a String (that is just 'i')...the correct way is...
hm.put(new Integer(i), "XYZ" + String.valueOf(i));
and to retrive plz use....
hm.get(new Integer(i));


------------------
Amit Agrawal,
New Delhi, India.
 
Ranch Hand
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The key in your hashtable is actually a String object equal to "i". Since the hashtable must have a unique key, it overwrites the previous record, so you end up with one entry, key = "i" and value = 4. Instead of:
hm.put("i","vipin"+String.valueOf(i));
use:
hm.put(new Integer(i), "vipin"+String.valueOf(i));
This will make your key an Integer object with each unique value of i.
 
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