RE: [PHP] Undefined variable? [Solution]
| From: | Alex Hogan | Date: | Wed, 28 Apr 2004 19:40:24 +0000 |
| Subject: | RE: [PHP] Undefined variable? [Solution] | ||
| Groups: | php.general | ||
| Request: | Send a blank email to php-general+get-184898@lists.php.net to get a copy of this message | ||
Nevermind I found it.
At the beginning of each of the loops I have the following;
$row = mssql_fetch_array($result);
And in the condition of the loop I'm running the same call;
While($row = mssql_fetch_array($result)){
I am not getting the first record in the query. Since the where clause of
the query will probably only return one record I'm getting the 'undefined
variable' error.
I can't believe I had missed that for so long.
Pardon the top post...
alex hogan
On Wed, 2004-04-28 at 16:09, Alex Hogan wrote:
> Hi All,
>
> I'm getting an undefined variable on a return value from a function.
> The function is;
> function spanky_select($array1, $array2, $where){
> $query = 'SELECT %s FROM %s WHERE %s';
> $query = sprintf($query, implode(', ',$array1), implode(',
> ',$array2), $where);
> //print $query;
> $result = mssql_query($query);
> $row = mssql_fetch_array($result);
> $j = 0;
> while($row = mssql_fetch_array($result)){
> $lname = $row['lastname'];
> $fname = $row['firstname'];
> $fields = $row['tkt_title'];
> $name[$j] = $lname.", ".$fname;
> $j++;
> }
> return ($name);
> }
>
> When I call this from my page I get the error.
> $array1[0] = 'firstname';
> $array1[1] = 'lastname';
> $array1[2] = 'tkt_title';
> $array2[0] = 'personnel';
> $array2[1] = 'ticket';
> $where = "lastname = \"smith\"";
>
> $row = myselect($array1, $array2, $where);
>
>
> However if I change this line;
>
> $query = 'SELECT %s FROM %s WHERE %s';
>
> to this;
>
> $query = 'SELECT %s FROM %s ORDER BY %s';
>
> and the $where var to;
> $where = "lastname, firstname";
>
> it returns properly.
>
> Can someone explain this to me? Why is it having trouble when I change
> the where var?
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