Re: [PHP] Cannot use a scalar value as an array

[Steve Cayford]

On Tuesday, December 4, 2001, at 06:04  AM, Xavier Antoviaque wrote:

>
> Hello all,
>
> I have a problem with a php script, which I want to use to fetch two 
> data
> (integer) from a MySQL table, divide the first by the second, and store 
> the
> value resulting in an double array. That seems not very difficult, but I
> always have the error 'Cannot use a scalar value as an array' when I 
> use my
> script.
>
> Here it is :
>
>   $result = mysql_query("SELECT timestamp_c,connectes,num FROM
> stats_serveurs WHERE timestamp_c LIKE '".$date."%' "AND serveur='".
> $serveur."'", $link);
>
>   $max = 0;
>   $reps = mysql_num_rows($result);
>   for($i = 1; $i<=$reps; $i++){
> $row = mysql_fetch_row($result);
> $serveur[$i]['heure'] = $row[0];
> $serveur[$i]['connectes'] = (int) ($row[1] / $row[2]); // Error!
> $max = max($max, $serveur[$i]['connectes']);
>   }
>

How does the operator precedence work in the statement ($row[1] / 
$row[2]) ? You might try (($row[1]) / ($row[2])) instead, or pull those 
into scalar variables before trying the division.

-Steve

> Any help is welcome ! :-)
>
> --
> Xavier.
>
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[PHP] Cannot use a scalar value as an array

[Xavier Antoviaque]

Hello all,

I have a problem with a php script, which I want to use to fetch two data 
(integer) from a MySQL table, divide the first by the second, and store the 
value resulting in an double array. That seems not very difficult, but I 
always have the error 'Cannot use a scalar value as an array' when I use my 
script.

Here it is :

  $result = mysql_query("SELECT timestamp_c,connectes,num FROM 
stats_serveurs WHERE timestamp_c LIKE '".$date."%' "AND serveur='".
$serveur."'", $link);

  $max = 0;
  $reps = mysql_num_rows($result);
  for($i = 1; $i<=$reps; $i++){
$row = mysql_fetch_row($result);
$serveur[$i]['heure'] = $row[0];
$serveur[$i]['connectes'] = (int) ($row[1] / $row[2]); // Error!
$max = max($max, $serveur[$i]['connectes']);
  }

Any help is welcome ! :-)

-- 
Xavier.

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