PHP database connection class

Question

I am trying to get a users ID from a database within a class, but I have very little to no experience with classes, how could I go about getting the uid from the DB and then return the uid?

so basically something like this,

class hello {
   public function getUid(){
      //connect to the db
      //get all of the users info
      $array = mysql_fetch_array($result);
      $uid = $array['uid'];

      return $uid;
   }
}

Like I said, I am still new to classes, so any advice or help would be greatly appreciated!

Thanx in advance!

Answer 1

First build a MySQL class library... suiting the requirements like in this sample piece:

<?php

include '../config/Dbconfig.php';

class Mysql extends Dbconfig    {

public $connectionString;
public $dataSet;
private $sqlQuery;

    protected $databaseName;
    protected $hostName;
    protected $userName;
    protected $passCode;

function Mysql()    {
    $this -> connectionString = NULL;
    $this -> sqlQuery = NULL;
    $this -> dataSet = NULL;

            $dbPara = new Dbconfig();
            $this -> databaseName = $dbPara -> dbName;
            $this -> hostName = $dbPara -> serverName;
            $this -> userName = $dbPara -> userName;
            $this -> passCode = $dbPara ->passCode;
            $dbPara = NULL;
}

function dbConnect()    {
    $this -> connectionString = mysql_connect($this -> serverName,$this -> userName,$this -> passCode);
    mysql_select_db($this -> databaseName,$this -> connectionString);
    return $this -> connectionString;
}

function dbDisconnect() {
    $this -> connectionString = NULL;
    $this -> sqlQuery = NULL;
    $this -> dataSet = NULL;
            $this -> databaseName = NULL;
            $this -> hostName = NULL;
            $this -> userName = NULL;
            $this -> passCode = NULL;
}

function selectAll($tableName)  {
    $this -> sqlQuery = 'SELECT * FROM '.$this -> databaseName.'.'.$tableName;
    $this -> dataSet = mysql_query($this -> sqlQuery,$this -> connectionString);
            return $this -> dataSet;
}

function selectWhere($tableName,$rowName,$operator,$value,$valueType)   {
    $this -> sqlQuery = 'SELECT * FROM '.$tableName.' WHERE '.$rowName.' '.$operator.' ';
    if($valueType == 'int') {
        $this -> sqlQuery .= $value;
    }
    else if($valueType == 'char')   {
        $this -> sqlQuery .= "'".$value."'";
    }
    $this -> dataSet = mysql_query($this -> sqlQuery,$this -> connectionString);
    $this -> sqlQuery = NULL;
    return $this -> dataSet;
    #return $this -> sqlQuery;
}

function insertInto($tableName,$values) {
    $i = NULL;

    $this -> sqlQuery = 'INSERT INTO '.$tableName.' VALUES (';
    $i = 0;
    while($values[$i]["val"] != NULL && $values[$i]["type"] != NULL)    {
        if($values[$i]["type"] == "char")   {
            $this -> sqlQuery .= "'";
            $this -> sqlQuery .= $values[$i]["val"];
            $this -> sqlQuery .= "'";
        }
        else if($values[$i]["type"] == 'int')   {
            $this -> sqlQuery .= $values[$i]["val"];
        }
        $i++;
        if($values[$i]["val"] != NULL)  {
            $this -> sqlQuery .= ',';
        }
    }
    $this -> sqlQuery .= ')';
            #echo $this -> sqlQuery;
    mysql_query($this -> sqlQuery,$this ->connectionString);
            return $this -> sqlQuery;
    #$this -> sqlQuery = NULL;
}

function selectFreeRun($query)  {
    $this -> dataSet = mysql_query($query,$this -> connectionString);
    return $this -> dataSet;
}

function freeRun($query)    {
    return mysql_query($query,$this -> connectionString);
  }
}
?>

and the configuration file...

<?php
class Dbconfig {
    protected $serverName;
    protected $userName;
    protected $passCode;
    protected $dbName;

    function Dbconfig() {
        $this -> serverName = 'localhost';
        $this -> userName = 'root';
        $this -> passCode = 'pass';
        $this -> dbName = 'dbase';
    }
}
?>

Answer 2

There is a problem with the way your code is written, not with the class. Take a closer look at this line:

$array = mysql_fetch_array($result);

This is the first time variable $result appears on the function. Therefore, it is not possible to communicate with the database.

A possible pseudo-code would be:

• connect to the DB server
• query the database
• fetch the results
• return the uid field.

Have a look at the relevant documentation first:

• http://php.net/manual/en/function.mysql-connect.php
• http://www.php.net/manual/en/function.mysql-query.php
• http://www.php.net/manual/en/function.mysql-fetch-array.php

Answer 3

You can use custom database class.
Code :

<?php 
Class Database
{
    private $user ;
    private $host;
    private $pass ;
    private $db;

    public function __construct()
    {
        $this->user = "root";
        $this->host = "localhost";
        $this->pass = "";
        $this->db = "db_blog";
    }
    public function connect()
    {
        $link = mysql_connect($this->user, $this->host, $this->pass, $this->db);
        return $link;
    }
}
?>

Answer 4

Try following:

ini_set("display_errors", 'off');
    ini_set("error_reporting",E_ALL);   

class myclass {
    function myclass()   {    
        $user = "root";
        $pass = "";
        $server = "localhost";
        $dbase = "";

           $conn = mysql_connect($server,$user,$pass);
           if(!$conn)
        {
            $this->error("Connection attempt failed");
        }
        if(!mysql_select_db($dbase,$conn))
        {
            $this->error("Dbase Select failed");
        }
        $this->CONN = $conn;
        return true;
    }
    function close()   {   
        $conn = $this->CONN ;
        $close = mysql_close($conn);
        if(!$close)
        {
            $this->error("Connection close failed");
        }
        return true;
    }       function sql_query($sql="")   {    
        if(empty($sql))
        {
            return false;
        }
        if(empty($this->CONN))
        {
            return false;
        }
        $conn = $this->CONN;
        $results = mysql_query($sql,$conn) or die("Query Failed..<hr>" . mysql_error());
        if(!$results)
        {   
            $message = "Bad Query !";
            $this->error($message);
            return false;
        }
        if(!(eregi("^select",$sql) || eregi("^show",$sql)))
        {
            return true;
        }
        else
        {
            $count = 0;
            $data = array();
            while($row = mysql_fetch_array($results))
            {
                $data[$count] = $row;
                $count++;
            }
            mysql_free_result($results);
            return $data;
         }
    }      
} 
$obj = new myclass();
$obj->sql_query("");

Answer 5

Alright, one piece of advice:

Do everything for a reason. Don't use things you don't know. Go and learn them instead.

One may give you an answer for this specific question, but until you don't know what Object Oriented means and why there are classes at all, you shouldn't use them.

Answer 6

• Create two classes. One for working with database, second one for managing User or Auth data.
• For SQL class create methods connect(), query(), fetch() etc
• For User class create methods get($id) etc