php problem - retrieving info from a database:

Discussion in 'HTML & Website Design' started by webbhost, Feb 18, 2007.

0
  1. #1
    Hi, having a slight problem which i cannot resolve.

    I am trying to retrieve a value from my database, database is as follows:


    websitesettings

    settingid settingname settingvalue
    1 tocversion 1
    2 regversion 1
    3 publicenabled 0


    there are 2 other files, admindblink.php and adminpanel.php


    Admin dblink.php includes the following...


    <?php
    session_start();
    $Host = "localhost";
    $user = "battles_gb75723n";
    $Password = "********";
    $DBName = "battles_battlesnails";
    $link = mysql_connect ($Host, $user, $Password);

    if (isset($_SESSION['username'])){
    if (isset($_SESSION['accesslevel'])){

    // USERNAME IS LOGGED IN. NOW NEED TO CHECK THEIR ACCESS LEVEL (ENSURE THEY ARE NOT FROZEN) AND CHECK WEBSITE MAINTINENCE STATUS

    if ($_SESSION['accesslevel'] == "0"){
    header ("Location: accountfrozen.php");
    }elseif ($_SESSION['accesslevel'] == "2"){

    }else{
    header ("snailcastle.php");

    }
    }else{
    $_SESSION['username'] = "";
    header ("Location: index.php");
    }
    }else{
    $_SESSION['accesslevel'] = "";
    header ("Location: index.php");
    }



    and the other file beginning is as follows...

    <?php
    require ("admindblink.php");


    //*******************************
    //* find current website status *
    //*******************************

    $Query2 = "SELECT * FROM websitesettings WHERE 'settingid' = '3'";
    $Result2 = mysql_db_query($DBName, $Query2, $link);
    while ($logininfo2 = mysql_fetch_array($Result2)){
    $websitestatus = $logininfo2['settingvalue'];
    }

    if ($websitestatus == "0"){
    $button = "enable site";
    $status = "<font color='#ffffff'>Site disabled</font>";
    $bgcolor = "red";
    }else{
    $button = "disable site";
    $status = "<font color='#ffffff'>Site enabled</font>";
    $bgcolor = "green";
    }
    ?>

    <input type="submit" name="changestatus" value="<?php echo $button; ?>">
    <?php echo $status; ?> - status = <?php echo $websitestatus; ?>. whiled = <?php echo $whiled; ?>. query statement = <?php echo $Query2; ?>, and result statement = <?php echo $Result2; ?></font>
    </form>






    Now the problem that i am getting is that the while statement is not activated (but no error messages occur). On page results are as follwos
    Site enabled - status = . whiled = . query statement = SELECT * FROM websitesettings WHERE 'settingid' = '3', and result statement = Resource id #4





    If anyone can help me debug this and find the cause of the error i would appreciate it alot!

    ?>
     
    webbhost, Feb 18, 2007 IP
  2. sketch

    sketch Well-Known Member

    Messages:
    898
    Likes Received:
    26
    Best Answers:
    0
    Trophy Points:
    148
    #2
    Actually, according to the output you give, the WHILE is triggered, how else would you get "Site enabled" in your output?

    I skimmed your code a couple of times, it looks like it's doing what it's supposed to... what kind of output are you expecting?

    Also, since you're only fetching one value with $logininfo2, I recommend you use mysql_result instead of the while loop, then there's no need to loop and reassign variables and you'll get the actual value instead of "Resource id #4".
     
    sketch, Feb 18, 2007 IP
  3. webbhost

    webbhost Peon

    Messages:
    2
    Likes Received:
    0
    Best Answers:
    0
    Trophy Points:
    0
    #3
    siteenabled = 0 which means website was disabled (so it shouldnt say it was enabled) - i found the error.

    read again:



    if ($websitestatus == "0"){
    $button = "enable site";
    $status = "<font color='#ffffff'>Site disabled</font>";
    $bgcolor = "red";
    }else{
    $button = "disable site";
    $status = "<font color='#ffffff'>Site enabled</font>";
    $bgcolor = "green";
    }



    It couldn't find the "websitestatus" variable, so displays status as site enabled.

    Figured it out in the end, if you are interested, this was the cause of the error..



    $Query2 = "SELECT * FROM websitesettings WHERE 'settingid' = '3'";

    There were 2 ' around "settingid" which should have not been there. That was invalidating the query, so it didn't want to work.
     
    webbhost, Feb 18, 2007 IP