Select Option Menu

[bjblackmore]

Hi,

I'm trying to build a select option field, from 2 tables, so after a user logs in, a select field will be build based on their user_id. Table1 has user info, (username/password/user_id), Table2 has the options for each user, (user_id/option_value/option_description).
So far I have...
<SELECT NAME="room_type">';
mysql_select_db("dbname");
$sql="SELECT option_value, option_description FROM table2, hotels WHERE user_id = user_id";
{
while ($row=mysql_fetch_array($result,MYSQL_ASSOC))
{
echo "<OPTION VALUE=".$row['code_id'].">".$row['room_type']."</option>";
};
}
else
{
echo mysql_error();
};
'</select>

But this gives a parse error can anyone see something wrong?

[fleymingmasc]

SELECT option_value, option_description
FROM table2, hotels
WHERE table2.user_id = hotels.user_id

Any time you have fields in 2 joined tables that have the same name, you have to declare which table you mean.

[[deleted]]

Either you have done a poor job at copy/paste, or you have completely forgotten to use the <? and ?> tags to define where your PHP starts and stops. :-)

[bjblackmore]

its a php script, inside a html form, inside a php page. I thought the '; after <SELECT NAME="name">'; told it to stop html, and the ' before '</select> told it to start html again!

[[deleted]]

hm... I gues you are very new to PHP? :-)

all PHP block must be enclosed by <? and ?>. Allways. Even if it's an includefile read by another PHP script. PHP code is only ever executed if it's between <? and ?> tags.

<?
echo "this is PHP";
?>
<h2>And this is plain HTML</h2>
<SELECT NAME="room_type">
<? // start of PHP
echo "And this is PHP again";

mysql_select_db("dbname");
$sql="SELECT option_value, option_description FROM table2, hotels WHERE user_id = user_id";
{
while ($row=mysql_fetch_array($result,MYSQL_ASSOC))
{
echo "<OPTION VALUE=".$row['code_id'].">".$row['room_type']."</option>";
};
}
else
{
echo mysql_error();
};

// End of PHP
?>
</select>

[bjblackmore]

Actually I tried that, and I get..
Parse error: parse error, expecting ','' or';'' in /users/user1/www/php/script.php on line 38

[[deleted]]

And what's the code of/around line 38?

[bjblackmore]

Erm its is echo
"<OPTION VALUE=".$row['code_id'].">".$row['room_type']."</option>"; But I'm sure this is right, cause I copied it from another script that works in the same kind of way!
Thanks for your help by the way :-)

[[deleted]]

It doesn;t have to be this line that causesthe error, you could have left out a ; or ) somewhere before this line...

[Anon]

You do not need the semi-colon at the closing brace on the while statement.

while ($row=mysql_fetch_array($result,MYSQL_ASSOC))
{
echo "<OPTION VALUE=".$row['code_id'].">".$row['room_type']."</option>";
};
^
Get rid of this semi-colon.

[[deleted]]

what's with all these people and semicolons?
What did that semicolon ever do to you? :-)

Semicolons help you make your code more readable
You put a semicolon after all your other commands don't you? so why not after closing brackets?