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Home/ Questions/Q 7638281
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Editorial Team
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Editorial Team
Asked: May 31, 20262026-05-31T08:08:18+00:00 2026-05-31T08:08:18+00:00

Finally completed the frontend of my website, I am now looking at coding the

  • 0

Finally completed the frontend of my website, I am now looking at coding the backend which needs an overhaul, the coding is very messy and uses far too many SQL Connections and commands, so much so that the host is complaining about it.

One of the main problems I am having is the Site Settings page, problem is this page will soon contain over 10 different options, and I prefer not to have a MySQL Update simply updating the option field to what it already is, so I am wondering if anyone has any ideas?

This is the structure of my options table, nice and simple which I recently changed from an awful layout.

Table structure for options

What would be the best way to edit these options without having to update every single one, and what would be the best way to have them in a function? Currently, the function I have was made back when I was new to coding, and you can see that is is very uneffective(note it uses the old table structure)

function site_upd($site_title, $site_email, $maint_status, $maint_mess, $upload_disable, $site_url, $reg_status, $land_mess)
{
    if( !$site_title ) 
    {       
        echo $this->errorMessage('There was no <b>site title</b> supplied, therefore we can not continue with this request.', 'error');
    } 

    else 
    {
        $this->logQuery('Updated site settings');
        $query = "UPDATE `table`.`settings` SET `site_title` = '".$site_title."', `site_email` = '".$site_email."', `maint_status` = '".$maint_status."', `maint_mess` = '".$maint_mess."', `upload_disable` = '".$upload_disable."', `site_url` = '".$site_url."', `registration_status` = '".$reg_status."', `landing_mess` = '".$land_mess."' WHERE `settings`.`sid` = '1'";
        mysql_query($query) or die(''.mysql_error()); } }

So yeah, there is the awful old structure and my old way of doing things, before I get stuck into coding I want other peoples opinions on what is the best way to do this!

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-31T08:08:20+00:00Added an answer on May 31, 2026 at 8:08 am

    You can work with $_POST Variable to simplify the update method.

    This is what I usually do:

    First I create a non specific function to update:

    function update_db($table, $id,$idvalue,$field,$fieldvalue){
       return mysql_query("update ".$table." set ".$field."=".$fieldvalue." where ".$id."=".$idvalue);
    }
    

    Then i create the specific function to the table (config i.e):

    function update_table_config($name,$value){
       return update_db("config","name",$id,"value",$value);
    }
    

    In the form i call all <input> like its name on the table:

    Site Name: <input type="text" name="site_name">
    ...
    

    and in the “action” page i check $_POST array:

    $data_array=$_POST;
    
    if(check_variables($data_array)){
        foreach ($data_array as $key=>$value){
            update_table_config($key,$value);
        }
    }
    

    the function check_variables has all the checks that you need for all the fields:

    function check_variables($data){
         if($data["site_title"]=="") return false;
         if(!is_numeric($data["landing_mess"])) return false;
         ...
         return true;
    }
    

    with this methods if in the future you have to include more fields in the configuration table you only have to add the <input> field and a check in check_variables function if needed

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