Re: [PHP] a generic getsql() function

[Don Read]

On 25-Jun-2003 Thomas Hochstetter wrote:
> Hi guys,
>  
> I wrote a generic getsql() function for my project's class. However, I
> only manage to retrieve a single row. What it should really do is:
> Read all rows into the array (with multiple results).
> The code is below:
>  
> function getsql($sql,$conn,$dbase,&$arr)
> {
> if($conn)
> {
> mysql_select_db($dbase,$conn);
> $result = mysql_query($sql,$conn);
> 
> while ($arr = mysql_fetch_assoc($result))
> {
> return $arr;
> // Something has to happen here!!!

  Nothing will happen here!!!
  You've already returned from the function.

> }


Regards,
-- 
Don Read [EMAIL PROTECTED]
-- It's always darkest before the dawn. So if you are going to 
   steal the neighbor's newspaper, that's the time to do it.

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RE: [PHP] a generic getsql() function

[French, Alastair]

Thomas Hochstetter wrote:
> Hi guys,
> 
> I wrote a generic getsql() function for my project's class. However, I
> only manage to retrieve a single row. What it should really do is:
> Read all rows into the array (with multiple results). The code is
> below: 
> 
> function getsql($sql,$conn,$dbase,&$arr)
> {
> if($conn)
> {
> mysql_select_db($dbase,$conn);
> $result = mysql_query($sql,$conn);
> 
> while ($arr = mysql_fetch_assoc($result))
> { return $arr;
> // Something has to happen here!!!
> }
> }

I think your problem is that you are returning after the first row has been
read, what you need to do is add the row ($arr) to another array using
$new_array[] = $arr or similar. and then move the return $arr to after the
while loop, but then return $new_array not $arr.

Hope this makes sense.

Alastair


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[PHP] a generic getsql() function

[Thomas Hochstetter]

Hi guys,
 
I wrote a generic getsql() function for my project's class. However, I
only manage to retrieve a single row. What it should really do is:
Read all rows into the array (with multiple results).
The code is below:
 
function getsql($sql,$conn,$dbase,&$arr)
{
if($conn)
{
mysql_select_db($dbase,$conn);
$result = mysql_query($sql,$conn);

while ($arr = mysql_fetch_assoc($result))
{
return $arr;
// Something has to happen here!!!
}
}
else
{
$arr["Error"] = "No connection";
return $arr;
}   
}
 
Thanks.
 
Thomas