I'm a student having troubles with PHP that i am trying to teach myself after a year without touching it...i've managed to overcome a few problems.. but i'm attempting a query but getting the aforementioned error for...

**Unknown column 'Dresscode' in 'field list' **

Here's the code...

PHP Code:

<?php
        
        include 'trnconnect.php';
        
        $ratingdata = mysqli_query($link, "SELECT * FROM venues order by (vRatingtotal DIV vRatingnumber) desc, vRatingnumber desc limit 10");
        while($ratinginfo = mysqli_fetch_array($ratingdata)) {
            $venueid           = $ratinginfo['venueID'];             
            $vRatingnumber    = $ratinginfo['vRatingnumber'];           
            $vRatingtotal     = $ratinginfo['vRatingtotal'];      
            $vName            = $ratinginfo['vName'];     
            $vAddress        = $ratinginfo['vAddress'];      
            $vPostcode        = $ratinginfo['vPostcode'];     
            $vPhone            = $ratinginfo['vPhone'];
            $vDesc            = $ratinginfo['vDesc'];         
            $vCapacity        = $ratinginfo['vCapacity'];         
            $vPriceID        = $ratinginfo['vPriceID'];
            $vRestrictionID    = $ratinginfo['vRestrictionID'];         
            $vDressID        = $ratinginfo['vDressID'];         
            $vGenreID        = $ratinginfo['vGenreID'];             
            $locationID        = $ratinginfo['locationID'];             
            $regionID        = $ratinginfo['regionID']; 
            
            $q="SELECT Dresscode, Category, Restriction, Genre, lName, rName".
                   "FROM vdress, vprice, vrestriction, vgenre, locations, regions".
                   "WHERE     (vdress.vDressID=$vDressID) AND
                          (vprice.vPriceID=$vPriceID) AND  
                          (vrestriction.vRestrictionID=$vRestrictionID) AND
                          (vgenre.vGenreID=$vGenreID) AND
                          (locations.locationID=$locationID) AND
                          (regions.regionID=$regionID)";
                        
            $venue_data = mysqli_query($link, $q) or die (mysqli_error($link));
            $venue_info = mysqli_fetch_array($venue_data);
            $Dresscode        = $venue_info['Dresscode']; 
            $Category        = $venue_info['Category']; 
            $Restriction    = $venue_info['Restriction']; 
            $Genre            = $venue_info['Genre']; 
            $lName            = $venue_info['lName']; 
            $rName            = $venue_info['rName']; 
            
            
            
                        
            echo "<h2>",$vName,"</h2>\n";
            $output_address = str_replace(",", ",<br/>", "$vAddress");
            echo "<p>",$output_address,"</p>\n";
            echo "<p>",$vPostcode,"</p>\n";
            echo "<p>Tel: ",$vPhone,"</p>\n";
            
            echo "<p>",$vDesc,"</p>\n";
            
            echo "<p>",$vCapacity,"</p>\n";
            echo "<p>",$Dresscode,"</p>\n";
            echo "<p>",$Category,"</p>\n";
            echo "<p>",$Restriction,"</p>\n";
            echo "<p>",$Genre,"</p>\n";
            

        echo "<p>\n";
        if($vRatingnumber >0){
            $vRatingaverage = round(($vRatingtotal/$vRatingnumber), 1);
        }else{
            $vRatingaverage =0;
        }
        echo "Current Venue Rating: ",$vRatingaverage,"/5 - ",$vRatingnumber," votes\n";
        echo "</p>\n";
        
    }
        
        
?>

Any help would be greatly appreciated, thanks Craig.

Is Dresscode in the table that you are trying to access?

AND if your server is running on *nix column and table names are case sensitive. So "Dresscode" and "dresscode" are NOT the same thing.

Hi guys, thank you for the response.. the field is Dresscode and does exist and is in the table in question. The error message has now cleared and been replaced with my new favourite

Warning: mysqli_fetch_array() expects parameter 1 to be mysqli_result, boolean given in ....

As i'm very new to the PHP again i've managed to fix a few different issues, but now this one has me stumped. I've checked many disscussions online, i'm getting the feeling the error is due to a lack of error handling for the array in question, but i'm not sure..

Thanks for the responses with regard to the initial problem. Any opinions on the new error with regard to the same code would be hugely appreciated (',)

Line numbers or better yet highlighting the line causing the error will help immensely!

Originally Posted by Sleeping Troll

Line numbers or better yet highlighting the line causing the error will help immensely!

Looking at the coding above there will only be one line number i dont see another query to constitute multiple line possibilities - scrap that i spotted more i should read more closely


IF you want to you can perform an sql dump of the table only dont bother with the data its unimportant.

post the resulting create table coding here for further inspection. maybe our eyes can see what you might be missing

Warning: mysqli_fetch_array() expects parameter 1 to be mysqli_result, boolean given in ....

Means that the query returned true || false rather than a recordset.

Assuming it is this:

PHP Code:

$venue_info = mysqli_fetch_array($venue_data); 


; echo out the value of $venue_data to see what it contains.

oh and use code delimiters please NOT text formatting.