PHP to mySQL array problem

mounters · 02-16-2002, 04:32 PM

The users selects the position they want from the array populated by the following:

PHP Code:



$query = "SELECT posID, position FROM position ORDER BY posID"; 
$result = mysql_query($query); 
while($pos = mysql_fetch_array($result)) 
    { 
    print("<option value=\"$pos[0]\">$pos[1]\n"); 
    } 
print("</select>

My problem is that when this is submitted to the mySQL database then the value "Array" is entered into the database rather the option selected from the drop down? Am I doing something blatantly wrong or is this never going to work this way?

Any help greatfully received.
Mounters

crabboy · 02-16-2002, 09:41 PM

Your program looks correct. I added a table to my db, added some data, and ran your exact program. My results were correct. It has to be the way you are grabbing the data from the form and inserting. it.

Code:

use test;

DROP TABLE IF EXISTS position;
create table position
(
   posID        varchar(35),
   position     int(5)
);

mysql> insert into position values ( 'five', 5 );
mysql> insert into position values ( 'ten', 10 );
mysql> insert into position values ( 'one', 1 );

PHP Code:



<?php 
mysql_connect( "localhost", "test", "test" ) 
           or die ("Unable to connect"); 
mysql_select_db( "test" ) 
   or die( "Unable to select database test" ); 
 
Header("Content-type: text/html"); 
print "<select size=7 name=\"TestMe\">\n"; 
 
$query = "SELECT posID, position FROM position ORDER BY posID"; 
$result = mysql_query($query); 
while($pos = mysql_fetch_array($result)) 
    { 
    print("<option value=\"$pos[0]\">$pos[1]\n"); 
    } 
print("</select>\n"); 
 
?>

Output: (View source from browser)

Code:

<select size=7 name="TestMe">
<option value="five">5
<option value="one">1
<option value="ten">10
</select>

mounters · 02-17-2002, 03:36 AM

Sorry must have not made myself clear about what the porblem was, the drop down displays fine, the problem is when the form data is posted to my script which for this code adds a player to a database. In the column 'position' the value Array is entered rather than the actual position selected from the drop down.

PHP Code:



$query = "INSERT INTO members (student_no, first_name, last_name, email_address, mobile, ncl_phone, other_phone, position) 
      VALUES ('$student_no', '$first_name', '$last_name', '$email', '$mobile', '$ncl', '$other', '$pos')"; 
   $result = mysql_query($query) 
   or die ("Couldn't add player to members table. Check the information and try again.");

The value pos should, I assume, contain the actual data from the array. Although this is not actually happening!
Any further ideas?
Cheers
Mounters

mounters · 02-17-2002, 04:20 PM

Got it sorted, I had put the worng variable to be added to the database, so many similar ones, I was calling the variable for the actual select values rather than the one for the select statement itself.
Cheers for the help anyhow.
Mounters