What a same code dun work the same

Anon

i develop a small database application, during the test on my computer, it worked completely fine, but when i upload the code to my web host, it always show me the error
"Warning:Supplied argument is not a valid MySQL result resource in ......"

The warning is at a function

1 function get_value($query){
2
3 $result=mysql_query($query);
4 $i=0;
5 $column_count=mysql_num_fields($result);
6 if ($column_count==1){
7 while($row=mysql_fetch_array($result)){
8 $bucket[$i]=$row[0];
9 $i++;
10 }
11 }
12 else { //multiple column
13 while($row=mysql_fetch_array($result)){
14 $bucket[$i]=$row;
15 $i++;
16 }
17 }
18 reset($bucket);
19 return $bucket;
20 }

line 5, 13 and 18

what could have cause this

thanks

zhabala

It seems that the connection to the mysql database server is not working. Are you properly connected to DB ? Are the connection parameters the same both at your computer and at your provider ?

Zdenek

Anon

You do not create a connection within your function, so you either need to pass it in as a parameter, or declare it as global within the function.

Anon

hi,
I did not have any error regarding connectin to database, i have checked, and it is ok.

the funny thing is when i test the program at home, it worked completely fine, when i upload it, it is where the problem comes.

thank you

eric

Anon

hi,
the connection parameter is declared as a global variable, so there is no problem there. But it works fine on my computer. only the net version give me problem.

216.74.109.245/~checkpoin/ntu/

take a look?

Anon

if i declare the connection to the db before this function, then there should be no problem right?