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oredy have the answer ..
he he he ^.^
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I've got an applet that needs to retrieve a "preview" of sorts of a file from within the applet. I think the best way to do this is to do what Windows Explorer does, and return an image of the file as a preview (or an icon if it can't preview, like returning the Microsoft Word icon for a .doc or .docx file). I have a feeling that this could be done through Shell32.dll. If so, how can I access that? And if not, is there another way to do this without having to save an image for every extension I might use beforehand?
Thanks.
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hi i have written this code to execute a batch file in windows 2003 server R2
but it does not execute anything. it opens the cmd.exe window and does nothing also
the "Exit Value" and "Return Code" is 0.
can anyone help me about this?
is there any windows registry setting to restrict the execution of batch files programatically?
Thank you!!!!
try
{
Runtime runtime = Runtime.getRuntime();
Process process = null;
try
{
System.out.println( "Running: " + fName + ".bat " );
process = runtime.exec( "cmd /c start " + fName + ".bat" );
int returnCode = process.waitFor();
System.out.println("Exit value " + process.exitValue());
System.out.println("Return code value " + returnCode);
System.out.println("Finished running the SQL Loader script's process. " + " Exit Value = " + ( returnCode == 0? "Success" : "Failure" ) );
}
catch ( Exception e )
{
System.out.println(e.getMessage());
e.printStackTrace();
}
}
catch ( Exception e )
{
System.out.println( "Error running the sqlldr script: " + e );
e.printStackTrace();
}
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if your bat script contains spaces or special characters you will need to enclose it in quotes... Dunno if that helps.
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also ensure your path to the bat is readable by javaw -> try putting it in "C:\yourbatscript.bat"
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i am getting the same results. the window title of cmd.exe is the exact path of the bat file but i didn't execute. maybe there is some kind of security setting for java not to invoke the batch files or a setting in windows server...
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No i did this last week. Didn't set anything else.
webtools.java:
private void jButton13ActionActionPerformed(ActionEvent event) {
cmdExec cmd = new cmdExec();
String s = (cmd.run("ping www.google.co.uk"));
jTextPane0.setText(jTextPane0.getText() + s + "\n-----------------------------\n");
}
cmdExec.java:
package webtools;
import java.io.BufferedReader;
import java.io.InputStreamReader;
public class cmdExec {
public String run(String cmdLine) {
String line;
String output = "";
try {
Process p = Runtime.getRuntime().exec(cmdLine);
BufferedReader input = new BufferedReader
(new InputStreamReader(p.getInputStream()));
while ((line = input.readLine()) != null) {
output += (line + '\n');
}
input.close();
}
catch (Exception ex) {
ex.printStackTrace();
}
return output;
}
}
modified on Wednesday, April 22, 2009 10:13 AM
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I'm starting to learn Java and have been testing Visual Editor in Eclipse. I notice there are several different versions & plug-ins available that are all quite different.
Which Plug-in / Version of Visual Editor do you use? What advantages does it have over other versions / products?
I'm on 0.9 (Has a palette - which I couldnt see on the other versions). It can be a bit buggy though - it doesn't like you making mistakes - and goes nuts if you do.
Which is your favourite and why?
(For items with multiple versions or items with the same name please include update links or version numbers, if possible.)
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I am a C# developer by profession, but recently I have been playing around with Java (have been asked to create a Java game for my company). My question is, in C# you can return a value from a class using something like this:
<br />
public string this[int index]<br />
{<br />
return "yadayadayada";<br />
}<br />
my question is, how do i do the same in Java?
Big thanks, sorry for the noob question but I couldnt get a straight answer from Google
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I am assuming the class extends a List of some kind.
For example:
public class TestClass extends ArrayList<String>
Then use something like this:
public String returnStringAtIndex(int index)
{
return this.get(index);
}
hmmm pie
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is it right that Java doesnt support indexers? guess it doesn't really make a difference
Jonathan Harker
preecesoftware.co.uk
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jharker1987 wrote: is it right that Java doesnt support indexers?
Java doesn't support properties.[^] So you have to use functions instead.
Is that what you mean?
hmmm pie
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any body here have huffman coding decoding and encoding in JSCRIPT ..
if, have, please share with me ...
or give me a link to download it,..
thx very much ...
regards
novhard
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Are you only looking for the a copy of code to cut and paste or are you looking an explaination of how it works and how to write it? I don't know of any examples in jscript but I have done this in C and to be honest the concept is going to be pretty much the same no matter what language you write it in. If you're interested I'd be glad to explain how it generally works in any language (can't say it would be specific to jscript).
Mike
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thx for answering my question ..
i am looking for the copy of code ,
thanks too if u want to give me explanation about the huffman code ...
just give the link, where i can download your huffman code in c,
i will try to translate it to jscript ...
thank you ver ymuch ...
regards
novhard
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I'll have to find the usb drive I have it on since I did it a little while ago. Hopefully I'll find it by tommorrow and post the code so you can see it.
But here's an explaination of how it works (please note if you don't know what a tree, structure, or linked list is I would suggest reading the wikipedia articles to learn that first since huffman uses them):
Suppose you had a long text file where all the letters were either an A,B,C,or D and you wanted to save them in the minumum space possible. Also suppose you knew that 80% were A, 15% were B, 3% were C, 2% were D. You could give each letter a 2 digit binary code but that would not be as efficient as giving the most commons letter(s) a shorter binary code even if this means the less common ones have to be slightly longer code.
For instance huffman would give them these codes:
A = 0
B = 10
C = 110
D = 111
It creates these codes by creating an unbalanced binary tree. To do this you create a list of nodes that are placed in order according to frequency. Then you remove the 2 least common nodes and create a parent node for them that contains the frequency of both nodes and place the parent node back into the list in order. You keep doing this till you have only 1 node and that node is your root node.
Here's what the tree looks like:
Root...................................(ABCD Freq 100%)
Level 1..........(A Freq 80% Code 0)..................(BCD Freq 20%)
Level 2...............................(B Freq 15% Code 10).........(CD Freq 5%)
Level 3.............................................(C Freq %3 Code 110)..(D Freq 2% Code 111)
The codes are determinded by where the node is on the tree. Left is 0 right is 1. To encode you create the tree and write the binary code for each symbol. To decode you start at the root and go left if the number is 0 and right if it is 1. You continue doing this until you reach a leaf node. Once you reach a leaf node you write the letter and start again at the root.
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thx mates ..
i am oready know about the teory ..
but i feel hard to write it down to a code ..
specially in jscript ..
thank you very much for answer my question
regards
novhard
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Basically this is a project I did in college, it reads a file with 0s and 1s (he didn't encript the punctuation) and writes out the message in plain text. He also gives us a file with the frequencies and we also had to print some other misecleous stuff in the console window if I recall.
ps code project system took out the indents in the code and stripped out my includes for some reason, just so you know.
modified on Monday, April 20, 2009 12:44 PM
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#include <stdio.h>
#include <stdlib.h>
#include <ctype.h>
#include <string.h>
#define TRUE 1
#define FALSE 0
#define MAXBITS 26
void prq_create(struct prqueue*);
int prq_empty(struct prqueue*);
void prq_insert(struct prqueue*, struct info_node*);
void prq_remove(struct prqueue*, struct info_node*);
void merge_nodes(struct info_node*, struct info_node*, struct info_node*);
void init_node(struct info_node*);
void print_level_order(FILE*, struct info_node*);
void print_code(FILE*, struct info_node*);
int isleft(struct info_node*);
int build_numcode(struct info_node*, int []);
void print_code(FILE*,struct info_node* pleaf_location[]);
void build_pque (FILE*, struct prqueue*, struct info_node* []);
struct info_node* build_tree (struct prqueue*);
void print_inorder(FILE*, struct info_node*);
void decode (FILE* ,struct info_node* , FILE*);
#define TRUE 1
#define FALSE 0
struct info_node
{
char code[MAXBITS] ;
float weight ;
int level ;
struct info_node* father ;
struct info_node* left ;
struct info_node* right ;
struct info_node* next ;
} ;
struct prqueue
{
struct info_node * prqtop ;
};
struct lo_queue
{
struct info_node* front ;
struct info_node* rear ;
};
int main()
{
struct prqueue pque ;
struct info_node* pleaf_location[MAXBITS] ;
struct info_node* tree_root ;
FILE* infile; FILE* outfile; FILE* dec_result; FILE* bitfile;
int count = 0;
infile = fopen("symb_freq.txt", "r");
outfile = fopen("code_tree.txt", "w");
bitfile = fopen("bit_file.txt", "r");
dec_result = fopen("decoded_file.txt", "w");
//builds the priority queue
pque.prqtop = NULL;
build_pque(infile, &pque, pleaf_location);
fclose(infile);
////builds the tree
tree_root = build_tree(&pque);
//does the inorder traversal
print_inorder(outfile, tree_root);
fprintf(outfile, "\n\n");
//prints the level order traversal
print_level_order(outfile, tree_root);
//prints the code for each letter
print_code(outfile, pleaf_location);
//decodes the zeros and ones into a message
decode(bitfile, tree_root, dec_result);
fclose(outfile);
fclose(bitfile);
fclose(dec_result);
return( 0 ) ;
} // end of function main()
void build_pque (FILE* infile, struct prqueue* pque, struct info_node* pleaf_location[])
{
int n = 0; struct info_node* node;
//creates the priority queue and loads pleaf_location array
while(!feof(infile))
{
node = (info_node*) malloc(sizeof(struct info_node));
init_node(node);
fscanf(infile, "%s", &node->code[0]);
fscanf(infile, "%f", &node->weight);
prq_insert(pque, node);
pleaf_location[n] = node;
n++;
}
}
struct info_node* build_tree (struct prqueue* pque)
{
struct info_node* p1; struct info_node* p2;
struct info_node* rnode; struct info_node* tree_root;
//builds the tree
while(pque->prqtop->next != NULL)
{
p1 = pque->prqtop;
p2 = p1->next;
pque->prqtop = p2->next; //moves top to the next spot
rnode = (info_node*) malloc(sizeof(struct info_node));
init_node(rnode);
merge_nodes(p1, p2, rnode);
prq_insert(pque, rnode);
}
tree_root = pque->prqtop;
pque->prqtop = NULL;
return tree_root;
}
void print_code(FILE* outfile, struct info_node* node)
{
int n = 0;
while(node->code[n] != '\0')
{
fprintf(outfile, "%c",node->code[n]);
n++;
}
fprintf(outfile, " ");
}
//makes a node and inititializes it
void init_node (struct info_node* node)
{
node->father = NULL;
node->left = NULL;
node->right = NULL;
}
//test if it is empty
int prq_empty(struct prqueue* prq)
{
if (prq->prqtop == NULL)
return TRUE;
else
return FALSE;
}
//inserts node into priority queue
void prq_insert(struct prqueue* prq, struct info_node* node)
{
struct info_node* curnode; struct info_node* lastnode = NULL;
curnode = prq->prqtop;
if ((curnode == NULL) || (node->weight <= curnode->weight)) //inserts at the start of the queue
{
node->next = curnode;
prq->prqtop = node;
}
else //moves through the list and inserts at the right spot
{
//lastnode = curnode;
//curnode = lastnode->next;
while ((curnode != NULL) && (node->weight > curnode->weight))
{
lastnode = curnode;
curnode = lastnode->next;
}
lastnode->next = node;
node->next = curnode;
}
return;
}
//removes one node and passes it back by reference
void prq_remove(struct prqueue* prq, struct info_node* node)
{
node = prq->prqtop;
prq->prqtop = node->next;
node->next = NULL;
}
//this creates a new node above the 2 existing nodes and calculated values for the new node
void merge_nodes(struct info_node* p1, struct info_node* p2, struct info_node* rnode)
{
//links nodes
rnode->left = p1;
rnode->right = p2;
p1->father = rnode;
p2->father = rnode;
//calculates value for the father
rnode->weight = p1->weight + p2->weight;
strcpy(rnode->code , p1->code);
strcat(rnode->code, p2->code);
//clears old links
p1->next = NULL;
p2->next = NULL;
}
//builds the code in left to right order. //this will be reversed when it is read
int build_numcode(struct info_node* node, int numcode [])
{
int count = 0;
while (node->father != NULL)
{
if (isleft(node))
{
numcode[count] = 0;
count += 1;
}
else
{
numcode[count] = 1;
count += 1;
}
node = node->father;
}
return count;
}
//test if the node is on the left side of the father node
int isleft(struct info_node* node)
{
if (node == node->father->left)
return TRUE;
else
return FALSE;
}
///////////////////////////////////////////////////////////////////////////////
// PRINTS THE CODE FOR EACH LETTER
void print_code(FILE* outfile, struct info_node* pleaf_location[])
{
int n; int a; int count; int numcode[10];
//prints the code for each letter
fprintf(outfile, "\n\n");
for (n = 0; n < MAXBITS; n++)
{
count = build_numcode(pleaf_location[n], numcode);
fprintf(outfile, "Symbol: %c Huffman Code: ", pleaf_location[n]->code[0]);
for (a = (count - 1); a >= 0; a--) //prints code in reverse order
fprintf(outfile, "%d", numcode[a]);
fprintf(outfile, "\n");
}
}
////////////////////////////////////////////////////////////////////////////////
// IN ORDER TRAVERAL CODE
void print_inorder(FILE* outfile, struct info_node* node)
{
if (node != NULL)
{
print_inorder(outfile, node->left);
fprintf(outfile, "FREQUENCY: %f SYMBOL: ", node->weight);
print_code(outfile, node); //prints out all the letters
fprintf(outfile, "\n");
print_inorder(outfile, node->right);
}
}
////////////////////////////////////////////////////////////////////////////////////////
//LEVEL ORDER TRAVERSAL CODE
//prints the current layer and gets nodes of the next layer
void print_level_order(FILE* outfile, struct info_node* tree_root)
{
struct info_node* left; struct info_node* right; int level = 0;
struct lo_queue* queue;
queue = (lo_queue*) malloc(sizeof(struct lo_queue));
//prints out the levels
tree_root->level = 0;
queue->rear = tree_root;
queue->front = tree_root;
fprintf(outfile, "level 0: ");
//print_levels(outfile, &real_queue);
while(queue->rear != NULL)
{
if (queue->rear->level != level) //if level changed
{
fprintf(outfile, "\n");
fprintf(outfile, "level %d: ", queue->rear->level);
}
print_code(outfile, queue->rear);
level = queue->rear->level;
//puts atoms in the queue
left = queue->rear->left;
if (left != NULL)
{
left->level = level + 1;
queue->front->next = left;
queue->front = left;
}
right = queue->rear->right;
if (right != NULL)
{
right->level = level + 1;
queue->front->next = right;
queue->front = right;
}
queue->rear = queue->rear->next;
}
}
////////////////////////////////////////////////////////////////////////////////////////
//Decoding
void decode (FILE* infile,struct info_node* tree_root, FILE* outfile)
{
char num; int n = 0; struct info_node* node;
node = tree_root;
while(!(feof(infile)))
{
fscanf(infile, "%c", &num);
if (num == '0')
{
node = node->left;
if (node->left == NULL) //is a leaf node
{
fprintf(outfile, "%c" , node->code[0]);
node = tree_root;
}
}
else if (num == '1')
{
node = node->right;
if (node->right == NULL) //is a leaf node
{
fprintf(outfile, "%c" , node->code[0]);
node = tree_root;
}
}
else
{
fprintf(outfile, "%c", num);
}
n++;
}
}
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00110100 011011011111011111 00001101000111100 00001010100000001100111 0001101
110000100 01111001111000101-1000010001011111111001010000, 00001101110000
11101111011110 001111001010 11100101100 010010101100111000010011111
10000100100101000111011110, 00001101110000 0000110100110010
11100101100
10010101111111010011010011111 110000110010 000011010010110101
010010101100111000011010101 0011010110000110 010011000101000111010111010
01000101011101111010111001010111011000011110100 0101101111001101100000111,
00001101110000 11100011111011010110011 00001101000111100 11100101100
11110101100101100, 1111010111100001000101000110010 1110101011111 0000110100
1100010100001010111010001011000 110100101
011011101100011100011011101010001110111. 00001101110000 0001101
011110001001010000101100 00001101000111100 0101101111001101100000111,
1100111101001000100010110100011110010100000111
11100101100
10111010011100010110000100000010011111 11100011111011010110011 001111001010,
111111000101101100100010111010110011 000011010010110101 001001011010000111000
11000111010011010001010111 001010101110100111 0000110100
010011101101001111001010000 110100101 0000110100
11001111010010001000101101010011111. 00001101110000
00110011010010101000010001000101 11101010110010 001011101010100111 110100101
110011110100100010001011010001111001010000 110000100010011101001111000111
111111000111000010101000010010001011001000100 0001101 00001101000111100
1001010111110111, 1011000 10110111 0000110100 010110111100110110000 110100101
0000110100 110001100110111000111110100 0001101 1110111100001000101 11010101
0001101 1110110000110111110101101110110 1011000, 1110101011111 0001101
101110100111000101100001000000100 101010000110
110011110100100010001011010001111001010000, 11110111011001010111010110011
10110000111 001011101010001010111111110000101111011010 11011010
011101000010010110 110001010110111010010011011110001111101000111 1110101011111
1101010111001111101010101100100100010111010110011 10110000111
11000111010011010001010111 10111010
011101000010010110 001011101010100111,
11100111 0001101 000011010000111 0111011011101111011110 011110010000111
0011111010111000 111101011001001110011110110010 0001101
100001010010110001001000 000011010010110101 0111111000101100000110010
1110101011111 011011101100011100011011101010001110111.
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a 8.167
b 1.492
c 2.782
d 4.253
e 12.702
f 2.228
g 2.015
h 6.094
i 6.966
j 0.153
k 0.772
l 4.025
m 2.406
n 6.749
o 7.507
p 1.929
q 0.095
r 5.987
s 6.327
t 9.056
u 2.758
v 0.978
w 2.360
x 0.150
y 1.974
z 0.074
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thx MikeMarq
this code really help me ..
regards
novhard
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I'm trying to display text in the shape of a cross on a JFrame using 4 JTextAreas but when I do all of the JTextAreas display correctly except the last one which ends up in the upper left corner. It is always the last item that ends up in the wrong spot. For instance I tried creating the west one first and it was then fine but then south became wrong. Here's the display code thanks in advance for your help.
Mike
JFrame frame = new JFrame();
jtextnorth = new JTextArea ("North" + '\n' + party[0] + '\n');
jtextnorth.setLocation(150, 0);
jtextnorth.setSize(100,100);
frame.add(jtextnorth);
jtexteast = new JTextArea ("East" + '\n' + party[1] + '\n');
jtexteast.setLocation(300, 150);
jtexteast.setSize(100,100);
frame.add(jtexteast);
jtextsouth = new JTextArea ("South" + '\n' + party[2] + '\n');
jtextsouth.setLocation(150, 300);
jtextsouth.setSize(100,100);
frame.add(jtextsouth);
jtextwest = new JTextArea ("West" + '\n' + party[3] + '\n');
jtextwest.setLocation(25, 150);
jtextwest.setSize(100,100);
frame.add(jtextwest);
frame.setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE);
frame.setSize(400,400);
frame.setVisible(true);
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Hi All,
I am working on an online assessment tool for that i want to show the time to user for the test he/she would be taking so can anyone tell something about how to add time in it.
I am using struts 2.0 framework.
Regards
Kaushal Kishore Sharma
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Can any body tells me how to make a video player in java? 
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