Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean online 30

Question

I am newly to ajax and php .so cant understand actually what happened .Here i am using ajax with jquery for show the result in dropdownlist in php.Here i am retrieving the data from database and shows in dropdownlist but my code not working.it shows warning error.what is the problem.

    <?php
 include "connection.php";
  $value;
$valueAcomoType;
$valueAcomoTypename;
$valueSuperArea;
$value=$_GET['projectid'];
$valueAcomoType=$_GET['accomo_scale_id'];
$valueAcomoTypename=$_GET['accomo_type'];
//$valueSuperArea=$_GET['Super_area_sqrft'];
if(isset($_GET['Super_area_sqrft'])){
$valueSuperArea=$_GET['Super_area_sqrft'];
 }

  $scaleID=$_GET['accomo_scale_id'];
 $Accomodation=$_GET['accomo_scale'];
 $AccomoType=$_GET['accomo_type'];
//$SuperArea = $_GET['Super_area_sqrft'];
 $sql_super_area= "";
  echo $sql_super_area="SELECT * from antheia.1bhk_accomodation_scale where    accomo_scale_id='".$scaleID."'AND accomo_scale='".$Accomodation."'  AND  accomo_type='".$AccomoType."' AND project_id='".$value."'";


 $resultArea = mysql_query($sql_super_area);
 $SupArea = "area";
 $zero = 0;
echo "<input name ='superarea' id = 'superarea'>";
echo "<input value ='".$zero."'> ". $SupArea."";
while ($row_super_area = mysql_fetch_array($resultArea)) {
$row_super_area['accomo_scale_id'];
$row_super_area['accomo_type'];
 $row_super_area['Super_area_sqrft'];                 
 echo "<input type = 'text' name = value=".$row_super_area['Super_area_sqrft']."/>";

}

 ?>

jquery file is here

        <script type="text/javascript">
       $(document).ready(function(){  
      $('#dropType').change(function(){
      var value2 = $('#dropType').val();
      alert(value2);
      var projectId = $('#dropProjectList').val();
      var element = value2.split("-"); 

      var scaleID1=element[0];
     var Accomodation=element[1];
     var AccomoType=element[1];
  var SuperArea;

   var chk=$('#superarea').val();
   $.ajax({url:"findSuperArea.php?accomo_scale_id="+scaleID+"&  accomo_scale="+Accomodation+"&accomo_type="+AccomoType+"&Super_area_sqrft="+SuperArea+"&  projectid="+projectId,success:function(result){
  res2=result;
 alert(res2);
  $('#super_area').html(res2);
 }});

    });
   });
 </script>

`$resultArea = mysql_query($sql_super_area) or die(mysql_error());` t check that what error is given — Awlad Liton, Mar 11 '14 at 07:09
I presume you haven't looked up the error message yet? Or looked at one of the many relevant questions with exactly the same titles? — John Dvorak, Mar 11 '14 at 07:09
(and if someone disagrees this is a duplicate, you can still closevote as "typo") — John Dvorak, Mar 11 '14 at 07:10

score 2 · Answer 1 · answered Mar 11 '14 at 07:10

2

You are missing a space before the first AND in your query

echo $sql_super_area="SELECT * from antheia.1bhk_accomodation_scale where    accomo_scale_id='".$scaleID."' AND accomo_scale='".$Accomodation."'  AND  accomo_type='".$AccomoType."' AND project_id='".$value."'";

answered Mar 11 '14 at 07:10

Nambi

11,766
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same error dear..after put a space before AND in query. – user1330693 Mar 11 '14 at 07:14

score 0 · Answer 2 · answered Mar 11 '14 at 07:09

Need a space in your query

$sql_super_area="SELECT * from antheia.1bhk_accomodation_scale where    accomo_scale_id='".$scaleID."' AND accomo_scale='".$Accomodation."'  AND  accomo_type='".$AccomoType."' AND project_id='".$value."'";
                                                                               // here                ^

score 0 · Accepted Answer · answered Mar 11 '14 at 07:11

You need to use mysql_real_escape_string() for each $_GET variables.

You have error in here:

accomo_scale_id='".$scaleID."'AND

you should have space between AND and your php variable.

Try this:

$sql_super_area="SELECT * from antheia.1bhk_accomodation_scale where    accomo_scale_id='".$scaleID."' AND accomo_scale='".$Accomodation."'  AND  accomo_type='".$AccomoType."' AND project_id='".$value."'";

You should switch to mysqli_* or PDO since mysql_* functions are deprecated.

score 0 · Answer 4 · answered Mar 11 '14 at 07:15

I think you should be use inner join and also quotes in query. I have add quotes in your query kindly try

$sql_super_area="SELECT * from `antheia`.`1bhk_accomodation_scale` where    `accomo_scale_id`='".$scaleID."'AND `accomo_scale`='".$Accomodation."'  AND  `accomo_type`='".$AccomoType."' AND `project_id`='".$value."'";

Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean online 30

4 Answers4