"Wrong datatype for second argument" while using in_array()

[LOSTBOY]

hi 

  am an PHP Beginner when i try this code it throws

 if (!in_array($qno,$tempArray))

 

Warning: in_array() [function.in-array]: Wrong datatype for second argument

 

what can i do to overcome this warning!

 

 

Thanks in Advance

[rajivgonsalves]

well it just states that the second argument is not an array

[rajivgonsalves]

as for overcoming it this code should work, you just stop the error not the functionality

 

if (!@in_array($qno,$tempArray))

[LOSTBOY]

thanks for ur valuable reply ............

[rajivgonsalves]

No problem at all, but I would suggest to you never use the @ operator to overide an error it becomes very hard to debug, find out why $tempArray is not an array and try to overcome it before the if statement (something like if its not an array, initialize it to an array), it would be a good programming practice

[LOSTBOY]

here is my code.....actually i want to generate Randam Quiz's .....as single quiz per page

here $_session['$max'] is the maximum questions that user select from our db

 

if (!isset ($_SESSION['$max'])){
$_SESSION['$max']=$_POST['sel_ques'];
$tempArray= array();
} 
  

do
{   
  $qno=rand(1, $_SESSION['max']);   
  if (!@in_array($qno,$tempArray)) { 
    $tempArray[$r] = $qno;
    $query="SELECT * FROM question WHERE qid=$qno" ;
    $result=mysql_query($query);
    $id=mysql_result($result,$i,"qid");
    $question=mysql_result($result,$i,"question");
    mysql_result($result,$i,"question");
    $query1="SELECT * from answer where qid=$id";
    $result1=mysql_query($query1);
$r++;
  }

 

 

But it not working properly .......plz give ur suggestions

[rajivgonsalves]

try this code

 

if (!isset ($_SESSION['$max'])){
$_SESSION['$max']=$_POST['sel_ques'];
}

if (!isset($_SESSION['tempArray']))
{
   $_SESSION['tempArray'] = array();
} 

$query = "SELECT * FROM question limit 1";
if (!empty($_SESSION['tempArray']))
{	     
    $query="SELECT * FROM question WHERE qid not in (".implode(",",$_SESSION['tempArray']).") limit 1" ;
}
    $result=mysql_query($query);
    $id=mysql_result($result,1,"qid");
    $_SESSION['tempArray'][] = $id;
    $question=mysql_result($result,1,"question");
    mysql_result($result,$i,"question");
    $query1="SELECT * from answer where qid=$id";
    $result1=mysql_query($query1);

 

and please backup your old code incase I am wrong 

[LOSTBOY]

ur code is not wrong ...but doesnt bring any data from data base....................................................