dropdown selection, post id

itsme85

Hello,

I want to use the ID from the selected item and insert it into my join table.

This are my tables:

product
product_id
product_name

category
category_id
category_name

product_category
product_category_id
product_id
category_id

I use this to insert the data and make a selection.

<html>
<head>
<title></title>

</head>

<?php
// Connect database.
include("connectdb.php");

$categoryquery = "SELECT DISTINCT category_name, category_id FROM category";

$categoryresult = mysql_query($categoryquery) or die(mysql_error());
?>
<form name="input" action="post.php" method="post">

<p>Product:</p>
<input size="30" type="text" name="product_name">

<p>Category:
<select name="category_name">
<!-- Drop down -->
<?php
while($row = mysql_fetch_array($categoryresult)){
echo '<option value="' .$row['category_id']. '">'. $row['category_name']. '</option>' ;
}
?>
<option value="category_name">(category_name)</option>
</select>

<p><input type="submit" value="Submit" name="submit"><p>

</form>

</body>
</html>

and I use this to post the data into the database:

<html>
<head>
<title>Untitled Document</title>
<meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1">
</head>

<body>

<?php
// Connect database.
include("connectdb.php");

// sample new data to insert into our db
// -------------------------------------
$product_name=$_POST['product_name'];
$category_name=$_POST['category_name'];
$category_id=$_POST['category_id'];
// -------------------------------------

// Inserting the data into product
// -----------------------------
mysql_query("insert into product(product_name) values('$product_name')");

# get the product id
$product_id = mysql_insert_id();

# and insert the product_category
mysql_query("insert into product_category(product_id, category_id) values ('$product_id','$category_id')");

?>
</body>
</html>

The problem is, it will post only the product_id into table product_category.

Could anyone help me to post the category_id as well?

Thank you in advance.

Stryker250

Your $category_name variable has the value of your category id. And the $category_id variable is empty because there is nothing posted by that name.
If you change

$category_name = $_POST['category_name'];
$category_id = $_POST['category_id'];

$category_id = $_POST['category_name'];

Then it will work.

itsme85

you're right, it works

itsme85

thanks for helping

Stryker250

No Problem 😉

itsme85

I tried to put the dropdownbox between my other code like this:

echo $EXTRA_INFO;
echo "<form action='$PHP_SELF' method='post'>
       <input type='hidden' value='$view' name='view'>
	   <p>Name:
        <input type='$type' name='product_name' value='$product_name'><br>
		<p>Category: <select name='category_name'>
            <!-- Drop down -->';

        while($row = mysql_fetch_array($categoryresult)){
   echo "<option value=".$row['category_id'].">".$row['category_name']."</option>";    

}
    echo '<option value='category_name'>(category_name)</option>
            </select>
        <!-- and so on for the rest -->";
echo $INPUT_BUTTON;
echo "</form>";

When I put the

{
   echo "<option value=".$row['category_id'].">".$row['category_name']."</option>";    

}

it doesn't display the form anymore. There is something wrong with my echo's I think. Do you have any idea?

Stryker250

This section should be closed off with a double quote and not a single quote.

echo "<form action='$PHP_SELF' method='post'>
       <input type='hidden' value='$view' name='view'>
       <p>Name:
        <input type='$type' name='product_name' value='$product_name'><br>
        <p>Category: <select name='category_name'>
            <!-- Drop down -->';

And this section should be opened with a double quote and not a single quote.

    echo '<option value='category_name'>(category_name)</option>
            </select>
        <!-- and so on for the rest -->";

So it should look like

echo $EXTRA_INFO;
echo "<form action='$PHP_SELF' method='post'>
       <input type='hidden' value='$view' name='view'>
       <p>Name:
        <input type='$type' name='product_name' value='$product_name'><br>
        <p>Category: <select name='category_name'>
            <!-- Drop down -->";

        while($row = mysql_fetch_array($categoryresult)){
   echo "<option value=".$row['category_id'].">".$row['category_name']."</option>";    

}
    echo "<option value='category_name'>(category_name)</option>
            </select>
        <!-- and so on for the rest -->";
echo $INPUT_BUTTON;
echo "</form>";

itsme85

That works fine, thanks.

Now I am trying to post the category_id and the category_name in different posts.

I tried this:

echo $EXTRA_INFO;
echo "<form action='$PHP_SELF' method='post'>
      <span name='input_form' style='display: $type'>
       <input type='hidden' value='$view' name='view'>
       <p>Name:
       <input type='text' name='product_name' value='$product_name'><br>
        <p>Category: <select name='category_name'>
            <!-- Drop down -->";

$categoryquery = "SELECT DISTINCT category_name, category_id FROM category"; 

$categoryresult = mysql_query($categoryquery) or die(mysql_error());  

while($row = mysql_fetch_array($categoryresult)){
echo "<option value=".$row['category_id']."".$row['category_name'].">".$row['category_name']."</option>";    

}
echo "<option value='category_name'>(category_name)</option>
            </select>
        <!-- and so on for the rest -->
        </span>";
echo $INPUT_BUTTON;
echo "</form>";

But now it posts the category_name and category_id in the same post. I think I have to change something in the "select name". But I have no idea how to do that.