[Resolved] What is wrong?

[jkarr]

Can someone please tell me what I am doing wrong. I'm trying to create a drop down lost from a database.

Here is my code:

<?
//Choose products
?>
<table border="1">
<th colspan="2">Please choose the products below and the quantity with which you wish to create a purchase order</th>
<tr>
<td><strong>Item Name </strong></td>
<td><strong>Quantity to Order</strong></td>
</tr>
<?
$vendors = "_vendors";
$company2 = $company;
$company2 = eregi_replace(' ', "_", $company2);
$company_vendors = $company2 . $vendors;
//Get the vendor information
$vendortag = $id;
$vendorinfo = mysql_query("SELECT * FROM $company_vendors WHERE vendorid=$id") or die (mysql_error());
$vendor_name = $vendorinfo['name'];
//Display the items in a drop down box.
?>
<tr>
<td>
<select name="item1">
<option value="none" selected> N/A
<?
$item_list = mysql_query("SELECT * FROM $company WHERE vendor=$vendor_name") or die (mysql_error());
while($row1 = mysql_fetch_row( $item_list )) {
$item_name = $row1['item_name'];
echo "<option value=$item_name>$item_name</option>";
}
?>
</select>
<td>
<input size="5" name="item1qty" id="item1qty">
</td>
</tr>
</table>
<?
 } ?>

The error is inside the drop list, I'll copy straight from View Source:

<select name="item1">
<option value="none" selected> N/A
You have an error in your SQL syntax. Check the manual that corresponds to your MySQL server version for the right syntax to use near '' at line 1

Thanks in advance,

Johnie Karr
Church of Christ
www.godsbible.org

[devinemke]

you probably need quotes around the vendor name. echo the query to browser to see what it looks like:

$sql = "SELECT * FROM $company WHERE vendor = '$vendor_name'";
echo $sql;

[jkarr]

Thanks devinemke.

Putting the single quotes in there fixed the error, but it doesn't display any results. I think that there needs to be no single quotes because I'm pulling a dynamic value.

Please goto http://www.godsbible.org/inventory and create an account and then click purchase order to see what I'm talking about.

Thanks,
Johnie Karr
Church of Christ
www.godsbible.org

[jkarr]

Ok,

I think the problem now is that I am not pulling data correctly from the vendor database first. I added the line echo $vendor_name; and it did not print the vendor name.

Thanks,
Johnie Karr
Church of Christ
www.godsbible.org

[jkarr]

Ok,

For anyone needing to resolve this same issue:

Take my original code and make the following changes to make it work.

First we will add

$row = mysql_fetch_array( $vendorinfo );

and change

$vendor_name = $vendorinfo['name']; 

to

$vendor_name = $row['name'];

We also put single quotes around the values for both of our WHERE commands in both the mysql_query

Thanks and I hope this helps someone,
Johnie Karr
Church of Christ
www.godsbible.org