Getting MYSQL data to appear in dynamic drop down menu

[K0108788]

Hi All,

I am trying to get 2 pieces of data (Student ID and their name) to appear in a dropdown box. I have written the following code but keep getting the same error on the first echo statement:

$sql = "SELECT SK_ID, name FROM student"; 
$result = mysql_query($sql, $conn); 
if(mysql_num_rows($result)) { 
while($row = mysql_fetch_assoc($result)) { 
echo ("<option value=\"$row['SK_ID']\">$row['SK_ID'] , $row['name']</option>"); 
} 
} else { 
echo ("<option value=\"\">No students created yet</option>"); 
}

Here is the error message;

Parse error: syntax error, unexpected T_ENCAPSED_AND_WHITESPACE, expecting T_STRING or T_VARIABLE or T_NUM_STRING ...

Any help appreciated
Regards,
K0108788

[thorpe]

echo "<option value=\"{$row['SK_ID']}\">{$row['SK_ID']} , {$row['name']}</option>";

[K0108788]

Thanks for the code, but it doesn't display the information in a drop down box that can be selected.

It just echos IDs and names, one after the other.

Is there anyway of altering this so that the data is displayed in a drop down box?

Any information appreciated.
Rich

[K0108788]

I have tried adding the following but then 2 drop down boxes appear with no data!

echo "<select option value=\"{$row['SK_ID']}\">{$row['SK_ID']} , {$row['name']}</option></select>"; 

[Roger_Ramjet]

$sql = "SELECT SK_ID, name FROM student";
$result = mysql_query($sql, $conn) or die ($sql . '<br />' . mysql_error);
if (!$result) { 
   echo 'Query failed'; 
   exit;
}
if (mysql_num_rows($result)>0) {
   echo '<select name="students">';
   while($row = mysql_fetch_array($result)) {
      echo '<option value="' . $row['SK_ID'] . '">' . $row['SK_ID'] . ', ' . $row['name'] . '</option>';
   }
   echo '</select>';
} else {
   echo 'No students created yet';
} 

[K0108788]

Thanks roger, that has solved it! 🙂