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PHP sql data selection and compare

// I have problem. it does not take tha date and assign it to the  $date_comingfrom_database
what should I do?

$query = "SELECT Email,Date FROM Mytable where Email='$visitoremail'";
$sql=mysql_query($query);
$result = mysql_fetch_array($query); // I have error message Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /
// I have problem here it does not take tha date and assign it to the  $date_comingfrom_database

$date_comingfrom_database = $result['Date'];
$databaseemail =$result['Email'];

$todaysdate = date('Y-m-d H:i:s');

if ( ( strtotime('Y-m-d H:i:s') - strtotime($date_comingfrom_database) >= 50 ) && ($databaseemail = $visitormail))
{
echo "record exist."  . $date_comingfrom_database . " ".$databaseemail. " ".$visitormail    ;
}
else
{
echo "record not exist." . $date_comingfrom_database . " ".$databaseemail. " ".$visitormail    ; // it only writes record not exist. and $visitormail here
}

I'll be back in 25 minutes..
0
Braveheartli
Asked:
Braveheartli
  • 2
1 Solution
 
SnarflesCommented:
change

$result = mysql_fetch_array($query); // I have error message Warning: mysql_fetch_array(): supplied argument

to

$result = mysql_fetch_array($sql); // I have error message Warning: mysql_fetch_array(): supplied argument
0
 
BraveheartliMarketingAuthor Commented:
you are great,
thank you
0
 
SnarflesCommented:
You're welcome :)
0

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