April 20, 2012
Dynamic Select list using sql data
Question by Baruch
I am trying to create a select drop-down list using php. Every time i try, i get an error.
Here is my code:
The function:
function dropDown(){
$options="<select>";
$connect = mysql_connect('localhost','id','pass') or die ("couldn't connect!").mysql_error;
mysql_select_db('db') or die('could not connect to database!');
$sql="SELECT * FROM DESC";
$result=mysql_query($sql);
while ($row=mysql_fetch_array($result)) { ****this is line 60
$name=$row["name"];
$options.="<option value="$name">".$name."</option>";
}
$options.= "</SELECT>";
return "$options";
}
and then i just call it in my code
<?php
include ('includes/functions.php');
....
$list = dropDown();
echo "$list";
....
>
the error i get is:
Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/garagenj/public_html/dispatch/includes/functions.php on line 60
Answer by JT Smith
If your table name is DESC then that’s your problem. It’s all about namespacing. You can’t even have a field named desc, I’ve tried. It errors everytime
Answer by Starx
SELECT * FROM DESC ??????
- From what table
- Or, if the table name is DESC. escape them with ticks
And You dont have start a new connection, every time you call the function. Put your connection part somewhere else
$connect = mysql_connect('localhost','id','pass') or die ("couldn't connect!").mysql_error;
mysql_select_db('db') or die('could not connect to database!');
function dropDown(){
$options="<select>";
$sql="SELECT * FROM `DESC`";
$result=mysql_query($sql);
while ($row=mysql_fetch_array($result)) {
$name=$row["name"];
$options.="<option value="$name">".$name."</option>";
}
$options.= "</SELECT>";
return $options;
}
