Re: Drop down box NOT populated

[Date Prev][Date Next][Thread Prev][Thread Next][Date Index][Thread Index]

 



From: <Tristan.Pretty@xxxxxxxxxxxxxxxx>

> Jsut a guess...
> Your row has a capital 'A' in the SQL statement, and a lower case 'a' in
> teh $row[] call..
>
> does that matter?

Yep, that would matter, but not the exact problem. I don't know if this
thread has already been answered or not, so...

The real problem is with this:

> $sql = mysql_query("SELECT distinct(Account) FROM Backlog")or die
> ("Something bad happened here: " . mysql_error()) ;
>
> echo "<select name=\"account\">\n";
> echo "<option>\n";
>
> while ($row = mysql_fetch_array($sql))
>   {
>       echo ' <option
> value="'.$row["account"].'">'.$row["account"]."</option>\n";

because there is no "account" index in $row. You're not selecting "account",
you're selecting "distinct(Account)".

So, you could do it the hard way and use $row['distinct(Account)'] as your
value or change your SQL to:

$sql = mysql_query("SELECT distinct(Account) AS acc FROM Backlog")or die

and use $row['acc']. This is called making an alias. You alias the
distinct(account) column to be called "acc". You can name the alias what
ever you want.

If you developed with your error_reporting() set to E_ALL, you'd have gotten
a notice about "Undefined index 'account' in $row" that may have tipped you
off to all of this.

Hope this helps.

---John Holmes...

-- 
PHP Database Mailing List (http://www.php.net/)
To unsubscribe, visit: http://www.php.net/unsub.php


[Index of Archives]     [PHP Home]     [PHP Users]     [Postgresql Discussion]     [Kernel Newbies]     [Postgresql]     [Yosemite News]

  Powered by Linux