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Thread: opp help !

  1. #1
    Join Date
    Aug 2011
    Posts
    4

    opp help !

    i have this code how can i call $result in the while looping who's in other function ?

    <?php
    class project0 {

    public function mysql_connection() {
    $host = 'localhost';
    $user = 'root';
    $password = '';
    $db = 'om';
    $dbtbl = 'omtbl';
    $connetion = mysql_connect($host,$user,$password);
    $select = mysql_select_db($db);
    $result = mysql_query("SELECT * FROM $dbtbl");



    }

    public function show_my_users(){
    while ($row = mysql_fetch_array(HOW CAN I CALL $result HERE ?)) {
    print $row['id'];

    }
    }

    }



    ?>

  2. #2
    Join Date
    Aug 2010
    Location
    Ohio
    Posts
    136
    Please use the boards [html [/html] [code [/code] and/or [php [/php] tags where applicable when posting code. Also:

    PHP Code:
    class project0 {

    public function 
    mysql_connection() {
    $host 'localhost';
    $user 'root';
    $password '';
    $db 'om';
    $dbtbl 'omtbl';
    $connetion mysql_connect($host,$user,$password);
    $select mysql_select_db($db);
    $this->result mysql_query("SELECT * FROM $dbtbl");



    }

    public function 
    show_my_users(){
    while (
    $row mysql_fetch_array($this->result)) {
    print 
    $row['id'];

    }
    }



  3. #3
    Join Date
    Aug 2011
    Posts
    4
    i tried this but i have an error
    Warning: mysql_fetch_array() expects parameter 1 to be resource

  4. #4
    Join Date
    Aug 2010
    Location
    Ohio
    Posts
    136
    There's no checking for valid return:
    PHP Code:
    class project0 {

    public function 
    mysql_connection() {
    $host 'localhost';
    $user 'root';
    $password '';
    $db 'om';
    $dbtbl 'omtbl';
    $connetion mysql_connect($host,$user,$password);
    $select mysql_select_db($db);
    $this->result mysql_query("SELECT * FROM $dbtbl");
    if( !
    $this->result ) die('Query failed - ('.mysql_errno().') '.mysql_error());
    if( !
    $this->result ) die('Query returned no results.');

    }

    public function 
    show_my_users(){
    while (
    $row mysql_fetch_array($this->result)) {
    print 
    $row['id'];

    }
    }


    SHould atleast help identify why its not a valid resource ;-)

  5. #5
    Join Date
    Aug 2011
    Posts
    4
    same errrorr ... when every thing is in one function i dont get this error ?

  6. #6
    Join Date
    Aug 2010
    Location
    Ohio
    Posts
    136
    Well this line:
    PHP Code:
    if( !$this->result ) die('Query returned no results.'); 
    shoulda been:
    PHP Code:
    if( !mysql_num_rows($this->result) > ) die('Query returned no results.'); 
    However I think you should have a function to query separate from the connection, then call that function in the display user function. i would rearrange things a bit as such. Tell me if you still get problems:

    PHP Code:
    <?php

    class project {
        private 
    $host 'localhost';
        private 
    $user 'root';
        private 
    $password '';
        private 
    $db 'om';
        private 
    $dbtbl 'ombtbl';
        
        private 
    $conn;
        
        public function 
    __construct() {
            
    $this->conn mysql_connect($this->host,$this->user,$this->password);
            
    mysql_select_db($this->db);
        }
        
        public function 
    get_users() {
            
    $qry "SELECT * FROM ".$this->dbtbl;
            if( !(
    $result mysql_query($qry,$this->conn)) ) die('Query Failed - ('mysql_errno() .') 'mysql_error());
            if( !
    mysql_num_rows($result) > ) return FALSE;
            return 
    $result;
        }
        
        public function 
    show_my_users(){
            if( !(
    $result $this->get_users()) ) trigger_error('No results returned.',E_USER_ERROR);
            while (
    $row mysql_fetch_array($result)) {
                print 
    $row['id'];
            }
        }
    // end class project

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