db search, product listing

[viveleroi0]

I have written the following code to search a table for a product ID, find the result and fill in a form with all of that products information.
My prob is that the variable SubCat 's value is displayed as the result of all of the other fields. What is wrong? Here is the code:

elseif($URI6=="editprodview"){
$SearchList=mysql_query("SELECT SubCat,ID,ProdName,ProdPrice,ProdDesc,ProdImage,ProdDate,ProdMisc FROM Vintage WHERE 1 AND ID = \"$SearchID\" LIMIT 0, 1");
$SubCat=mysql_result($SearchList,"SubCat");
$ID=mysql_result($SearchList,"ID");
$ProdName=mysql_result($SearchList,"ProdName");
$ProdPrice=mysql_result($SearchList,"ProdPrice");
$ProdDesc=mysql_result($SearchList,"ProdDesc");
$ProdImage=mysql_result($SearchList,"ProdImage");
$ProdDate=mysql_result($SearchList,"ProdDate");
$ProdMisc=mysql_result($SearchList,"ProdMisc");

[maxsslade]

There are many ways to do things in mysql, you are using a form of code that is unstable and even if the code did work without errors it may not run correctly every time online.

To get round this problem you can write your code this way:

<?

$query1 = "SELECT SubCat, ID, ProdName, ProdPrice,
ProdDesc, ProdImage, ProdDate, ProdMisc,
FROM Vintage WHERE ID LIKE '$SearchID'";

$result = mysql_query($query1);

$query_data = mysql_fetch_row($result);

/
The data has the values in an array called $query_data
you can view the data in order. e.g. $query_data[0] is
the value for the 'SubCat' field and $query_data[3] is
the value for the 'ProdDesc' value.
/

if($query_data[0] = '') {
echo"No string similar to the ID you typed";
} else {

display code...

}

?>

I hope this works

• Max