See anything wrong with this?

Anon

<HEAD>
<TITLE> Add New subcategory </TITLE>
</HEAD>
<BODY>
<?php

if ($submit): // A new author has been entered
// using the form below.

include("articles.inc");
mysql_select_db("petesmc_articles");

$sql = "INSERT INTO subcategory SET " .
"SubCat='$subcat' ," .
"CID='$selectcat' ," .
"Description='$description' ";
if (mysql_query($sql)) {
echo("<P>subcategory added</P>");
} else {
echo("<P>Error adding new subcategory: " .
mysql_error() . "</P>");
}

$select = "SELECT ID, Cat FROM category " .
"ID='$selectid' ," .
"Cat='$selectcat' ";
if (!$select) {
echo("<P>Error retrieving cateogry from database!<BR>".
"Error: " . mysql_error());
exit();

<P><A HREF="<?php echo($PHP_SELF); ?>">Add another subcategory</A></P>
<P><A HREF="subcat.php">Return to subcategory list</A></P>

<?php
else:
?>

<FORM ACTION="<?php echo($PHP_SELF); ?>" METHOD=POST>
<P>Enter the new category:<BR>
SubCategory: <INPUT TYPE=TEXT NAME="subcat" SIZE=25 MAXLENGTH=100 value="<?php echo($subcat); ?>"><BR>
<textarea ROWS=15 COLS=45 WRAP><?php echo($description); ?></textarea><br>
<select>
<option><?php echo($selectcat); ?></option>
</select>
<INPUT TYPE=SUBMIT NAME="submit" VALUE="SUBMIT"></P>
</FORM>

<?php endif; ?>

</BODY>

Anon

Well, I don't have time to take a good look at your code, but this line:
$sql = "INSERT INTO subcategory SET " .

DON'T use SET there... no need for that...

And if (arg):
Please use if (arg) {
: is the old way (Just style, doesn't have anything to do with speed... i think 🙂

Anyway, that's what I think of it...

Anon

I am trying to select from category table, display in a select box, then whichever is selected, gets entered into CID in subcategory table on sumbit.

pnoeric

Peter... can you tell us exactly what goes wrong when the code is run? Do you get any error messages? Do you know if the data is going into the database on the first mysql call but not being retrieved the second time?

The only thing that jumps out at me, when I glance at it quickly, is:

$select = "SELECT ID, Cat FROM category " .
"ID='$selectid' ," .
"Cat='$selectcat' ";

I would throw in the SQL keyword "WHERE" after the table name "category"... it makes it more readable and may be required. :-)

Eric

Anon

Here is the page it is being used on:

http://216.74.101.27/admin/newsubcat.php

As you can see, the select box doesn't have the categories listed in them. I haven;t tried to submit yet without all the data.

pnoeric

So if bug we're dealing with right now is the fact that the 'category' pop-up doesn't show anything, that's somewhat straightforward...

I'm making some assumptions based on what you've posted but here's me "thinking out loud" as I look at your program: your code right now is divided into two parts. The first part handles putting information into the database (the INSERT statement), then retrieves some information (the SELECT) statement. The first part of the code then ends at the else: statement.

The second part of the code displays the form and echos some variables to set the defaults... however, those variables have never been set in the code before this point.

It looks like you actually want to have the first part of the code simply store information in the db... and the second part should retrieve information (like main category names) then display the form, with that information already there (in the pop-up menu, and as defaults in the text fields).

Does that sound right? If so there should be enough hints there to get you moving in the right direction :-) If not, write back and ask some detailed questions and I'll respond.

best,
Eric Mueller

Anon

I want the first part to retrieve and put into a select box, second stores into a different table.

I really have no idea, why this doesn't work. It looks like it should work to me

pnoeric

Peter... your program logic looks like this:

if (we have variables from a submit button)
{
-- put some data in the database
-- get some other data in the database
} else {
-- draw a form on the screen
}

Can you see where the part after the "else" needs data from the database to draw the form, but you don't give it any?

Here's what I think you should do before you go any further: separate the two parts of the program two php files. I'll bet you'll find them easier to work on and easier to debug, too. The first php file should be the form the action= should be the second php file, which will actually do something with the data from the form.

best,
Eric Mueller