DB connections

Have you tried:

function function_name()
{
global $conn;

Hi,
Here is my way:

1) put your mysql_connect( or mysql_pconnect) and mysql_select_db in an include file, for example "db.inc".
   also you could put your function in the same or another include file, for example "functions.inc".

2) use include or require statement in files you want to get database connection and use your functions. for example:
#inlcude("db.inc");
#include("functions.inc");

my_function();

hope it helps
regards
huweiqi

Basically you don't need to worry about link identifier which is returned by the mysql_connect.

Why ?

Cause when you use function mysql_query and don't give link identifier it would AUTOMATICALLY identify current connection and execute query using that. So no need to use global and stuff, just go ahead in using the mysql_query function.

JD

JD is right, if no link is provided PHP will use the last opened link.

Ok here is the main call to the db, the first one not in any function just in the script.

     @$conn = mysql_connect("localhost",$dbUserName, $dbPassword) or die("An error occured whilst connecting to the database.\n");

     @mysql_select_db($dbName) or die("Could not select database on host");

     //check to see if page is in database

     $sql="select id from pages where name = \"$name\";";
     $result = mysql_query($sql) or die("Query failed");

then i do some other stuff then i have this function that is called form the above script

function buildNavigation()
{
     $username = $GLOBALS["username"];
     
     //get user details      
    $sql="select id from user where user_name = \"$username\";";
    $result = mysql_query($sql) or die("Query failed");    
     
     return "User name " . $username;
}

here the query failes, why is this, i thought that if i provided no more conection details php would simply use the last obtained connection, i think this is what JD is saying??

First make sure it's the connection that is failing:

$result = mysql_query($sql) or die(mysql_error());

The mysql_error function will return a error message from the db if any.

If it's the connection try:

$result = mysql_query($sql,$GLOBALS["conn"]) or die(mysql_error());  

here database object that I created

<?
class DatabaseObj {

  var $dblink;
     var $dbname;
     var $lasterror;
     var $lastresult;
     
     function DatabaseObj($DBName, $DBHost, $DBUser, $DBPass) {

          $this->dbname = 'upgrader'; // Change this back to = $DBName after correcting source code
         
    $this->dblink = @mysql_connect("$DBHost", "$DBUser", "$DBPass")
     or die( "<p><b>Could not connect to database server: ($DBHost)</b></p>" );

    @mysql_select_db ($this->dbname , $this->dblink)
     or die (  "<p><b>Could not connect to database ($DBName)</b></p>" );


     }
     
     function SQLFetchAssoc() {
       return mysql_fetch_assoc($this->lastresult);
     }
     
     function SQLSelect ($sql){
       $this->lastresult = mysql_query($sql, $this->dblink);
       return $this->lastresult;
     }
     
     function SQLInsert($table, $values) {
       $fieldnames = implode(", ",array_keys($values));
          $fieldconts = implode(", ",array_values($values));

          $sql = "INSERT INTO $table ($fieldnames) VALUES ($fieldconts)";

          $result  = mysql_query($sql, $this->dblink)
     or $this->lasterror = mysql_error();

          return $result;
     }

     function SQLUpdate($table, $values, $criteria) {

          $sql = "UPDATE $table SET ";

    foreach ($values as $field => $value)
               $sql .= "$field = $value, ";
               
          $sql = substr ($sql, 0, -2); // strip last comma and space  

          $sql .= " WHERE $criteria";

          $result = mysql_query($sql, $this->dblink) or $this->lasterror = mysql_error();

          return $result;
     }

     function SQLDelete($table, $field, $values) {
          $range = implode(", ",array_values($values));

          $sql = "DELETE FROM $table WHERE $field IN ($range)"; echo $sql;
          $result  = mysql_query($sql, $this->dblink)
     or $this->lasterror = mysql_error();
          if ($result)
                return mysql_affected_rows($this->dblink);
          else
                return $result;
     }
}
?>

use as below

$db = new DatabaseObj('faults', 'localhost', '', '');

if ( $result = $db->SQLSelect("SELECT uid, user, email FROM users"))
{$row = $db->SQLFetchAssoc();}

if you wish to use in a function either pass $db by reference or use global $db

Ok so now using $result = mysql_query($sql) or die(mysql_error()); thanks andriv, i can see that the problem is that no db is selected, as you can see this is not the case as i have selected on earler in the mane script. so what to do.  So i tried adding this line.      mysql_select_db("kplonk") or die(mysql_error());     but it failes with the error
"Access denied for user: '@localhost' to database 'kplonk'"

Try this:

function buildNavigation()
{
    global $name,$conn,$dbname;
   
    //get user details      
   $sql="select id from user where user_name = \"$username\";";
   $result = mysql_db_query("$dbname,$sql,$conn) or die("Query failed");    
   
    return "User name " . $username;
}


The mysql_db_query will allow you to select the database in the query.  The global will make the variables global.

Also, what version of PHP do you have?

Exactly.

So your problem is not about being your connection variable global. But your connection itself is not going through.

Seems like you have permission issue here. Check with MySQL admin about your permissions.

JD

4, still having problems, this is what the function looks like

function buildNavigation()
{
     //get user details
        global $conn,$dbName,$username;
     
       $sql="select id from user where user_name = \"$username\";";
       $result = mysql_db_query($dbName,$sql,$conn) or die("Query failed");    
     
     return "User name " . $username;
}

and the error is now "Warning: Supplied argument is not a valid MySQL-Link resource in /home/sites/site56/web/test/index.php on line 105
"

any ideas

See what happens if you reconnect to the server:

function buildNavigation()
{
    //get user details
       global $dbName,$username;

     @$conn = mysql_connect("localhost",$dbUserName, $dbPassword) or die("An error occured whilst connecting
to the database.\n");

      $sql="select id from user where user_name = \"$username\";";
      $result = mysql_db_query($dbName,$sql,$conn) or die("Query failed");    
   
    return "User name " . $username;
}

One more thing, if you are not using version 3 remove the '@' symble from the $conn, this can be causing the problem.

Odd thing the code now looks like this,

function buildNavigation()
{
     //get user details
        global $user_access_level,$username;
   
     $title_sql="select title,id from nav order by orderby;";
    $title_result = mysql_query($title_sql) or die("Query failed");

and works dont know what changed, appart form removing the reconect, that said i had already tried that must have miss typed or somthing.. my new proble is hwo to give the points to??

Ok thanks well as all in this thread helped me with this problem i feel i will split the points equaly, thanks for the input all of you it reall helped me out.

Glad we can help

Ok have been told to add questions form the rest of you to collect some points have a look