I'm having

[sahm]

Hi every one

I having problem using count() function
when I use count() the result came multiply 2

and this is my code :

////////////////////////////////////

$sql2 = "select * from user_previlag WHERE User_email= '$userEmail'";
$data2 = mysql_query($sql2);
$info = mysql_fetch_array($data2);

echo count($info);

\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\

[=?iso-8859-1?Q?=C1lvaro?= G. Vicario]

*** sahm escribió/wrote (Sat, 20 Sep 2008 01:57:28 -0700 (PDT)):

I having problem using count() function
when I use count() the result came multiply 2

and this is my code :

////////////////////////////////////

$sql2 = "select * from user_previlag WHERE User_email= '$userEmail'";
$data2 = mysql_query($sql2);
$info = mysql_fetch_array($data2);

echo count($info);

From http://php.net/mysql_fetch_array:

array mysql_fetch_array ( resource $result [, int $result_type ] )

result_type: The type of array that is to be fetched. It's a constant and
can take the following values: MYSQL_ASSOC, MYSQL_NUM, and the default
value of MYSQL_BOTH.

Return Values: [...] The type of returned array depends on how result_type
is defined. By using MYSQL_BOTH (default), you'll get an array with both
associative and number indices.

So your problem has nothing to do with count(). Also, have a look at
http://php.net/print_r
--
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[Kari Hanski]

sahm <sa*****@gmail.comkirjoitti 20.09.2008:

>
I having problem using count() function
when I use count() the result came multiply 2

and this is my code :

////////////////////////////////////

$sql2 = "select * from user_previlag WHERE User_email= '$userEmail'";
$data2 = mysql_query($sql2);
$info = mysql_fetch_array($data2);

echo count($info);

\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\

You should be using mysql_num_rows() function.

If you do print_r($info), you'll see why count() does that.