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up vote 0 down vote favorite

I have the following code in my addproduct.php which connects to doaddproduct.php after clicking a submit button:

<tr>
<td>Category Name</td>
<td>:</td>
<td>

<select name="categoryname" id="">
<option value="">Select</option>
<?php

while($row=mysql_fetch_array($result)) {
?>
<option value="<?php echo $row['CategoryID']; ?>">

<?php echo $row['CategoryName']; ?></option>
<?php } ?>
</select>

</td>
</tr>

So, the dropdown menu (addproduct.php) gives options from the database (table: category, column: categoryname). For example the categoryname for categoryID '1' is food, the categoryname for categoryid '2' is drink and so on.

Now, I want to insert the data from selected dropdown option to my database (table: product, column: categoryID) by changing the selected categoryname into categoryID. For example, I select 'drink' from dropdown menu and when I click the submit button, the program inserts the data into table product:

|ProductID | CategoryID | ProductName |

| 123231 | 2 | Coca-cola |

Here's my following code in my doaddproduct.php:

<?php
    include("connect.php");

    $productname = $_POST['productname'];
    $categoryname = $_POST['categoryname'];
    $stock = $_POST['stock'];
    $price = $_POST['price'];

    if($productname == NULL || $productname == ""){
        header("location:../addproduct.php?err=You must fill product name");
    }else if($categoryname== "none"){
        header("location:../addproduct.php?err=You must choose category name");
    }else if($stock == NULL || $stock == ""){
        header("location:../addproduct.php?err=You must fill stock");
    }else if($price == NULL || $price == ""){
        header("location:../addproduct.php?err=You must fill price");
    }else if($_FILES["file"]["type"] != "image/jpeg" && $_FILES["file"]["type"] != "image/png" && $_FILES["file"]["type"] != "image/jpg"){
        header("location:../addproduct.php?err=Extention of your photo must be jpg/jpeg/png");
    }else{
        $query = $username . "-" .$fullname . "-" . $phone ."-".$email."-".$password."-".$rpassword."-".$address."-".$gender;
        $qcek = "select * from product where productname = '$productname'";
        $result = mysql_query($qcek);
        if(mysql_num_rows($result) > 0){
            header("location:../addproduct.php?err=Product name already used");
        }else{
            //for upload
            $ext = substr($_FILES["file"]["name"], strrpos($_FILES["file"]["name"], '.'));
            move_uploaded_file($_FILES["file"]["tmp_name"],"../photos/" . $username . $ext);
            $pho = $username . $ext;

            $password = md5($password);
            //still confused here
                    $query = "insert into product values('', '', '$productname', '$pho', '$stock', '$price')";

            mysql_query($query);
            header("location:../addproduct.php?err=Success");
        }

    }

?>

What should I do?

share|improve this question

closed as not a real question by CBroe, hjpotter92, jszumski, Jocelyn, Niels Keurentjes May 17 '13 at 1:28

It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question.
    
what is the problem ? any error you getting? –  Rajeev Ranjan May 16 '13 at 12:41
    
@RajeevRanjan the problem is I can't change the selected dropdown option (categoryname) into categoryid so I can insert it into my new table (which is table product) –  noobprogrammer May 16 '13 at 12:43
    
your <select> name is categoryname but it option it have it has categoryId in its value ,when you select any option you will get categoryId in $_POST['categoryname'] . –  Rajeev Ranjan May 16 '13 at 12:51

2 Answers 2

up vote 1 down vote

If I understand the problem correctly, "by changing the selected categoryname into categoryID.", it appears that you've done most of the legwork already. The drop-down box named "categoryname' is simply the variable name that holds the id val. In the query use {$_POST['categoryname']}, the brackets are required.

share|improve this answer
up vote 0 down vote

You forget category name, change your insert query like this:

 $insert_query = "INSERT INTO `product ` (ProductID,CategoryID,ProductName) VALUES 
                                       ($product_id,$categoryname,'$productname')";
share|improve this answer
    
I have tried it before. And the data couldn't be inserted into the database. –  noobprogrammer May 16 '13 at 12:51

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