Re: [PHP] contents of fetch array into a string

[Steve Cayford]

On Wednesday, March 6, 2002, at 12:13  PM, Kris Vose wrote:

> I have a problem with reading the contents of $r[1] into a string 
> called = $mail.  Does anyone have any suggestions?  Any help is greatly 
> appreciated.
>
> Kris Vose
>
>
> if ($czero != "")
> {
>
> $t = mysql_query("SELECT * FROM AddExisting");

$t is now a resource variable, not an array. If you want to know the 
number of rows returned
use mysql_num_rows($t) not count($t)

>
> $number_of_customers = count($t);
>
> while($r = mysql_fetch_array($t))

Note: you'll loop through this section $number_of_customer times, 
pulling one table row into $r
each time through.

> {
> extract($r);

Why are you using extract here?

>   for ($i = 0; $i<$number_of_customers; $i++)

You're looping through $number_of_customers columns on each row in the 
table. Is this what you want?  Is your table really $number_of_customers 
wide?

>   {
> echo $r[1].", ";
Do you mean $r[$i] ?
>   }
>
> }
>
Now you've dropped out of both loops. $r will be holding the last row 
you retrieved from the database table.

> $mail = ' ';
> foreach($r[1] as $mailline)

$r[1] is not an array here. What are you trying to do?

> {
> $mail.=$mailline;
> }
>
> }
>


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[PHP] contents of fetch array into a string

[Kris Vose]

I have a problem with reading the contents of $r[1] into a string called = $mail.  
Does anyone have any suggestions?  Any help is greatly appreciated.
 
Kris Vose
 
 
if ($czero != "")
{
 
$t = mysql_query("SELECT * FROM AddExisting");
 
$number_of_customers = count($t);
 
while($r = mysql_fetch_array($t))
{
extract($r);
 
  for ($i = 0; $i<$number_of_customers; $i++)
  {
echo $r[1].", ";
  }
 
}
 
$mail = ' ';
foreach($r[1] as $mailline)
{
$mail.=$mailline;
}
 
}