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after executing this code in browser i am getting warning message 'un defined variable query result ". I think the query is not executing correctly. However if the same query is run in phpadmin it retrieves the values . what is the problem with select statement in php

<?php



$con=mysqli_connect("localhost","root","neel","sitams");
// Check connection
if (mysqli_connect_errno())
  {
  echo "Failed to connect to MySQL: " . mysqli_connect_error();
  }


$sql="SELECT call_no,author_id,edition,title,publisher,vendor,cost_i,cost_f,plo_pub  from lib_acc  where access_no='6'";

if ($result=mysqli_query($con,$sql))
  {
  while ($obj=mysqli_fetch_object($result))
    {
     $queryResult[] = $obj->call_no;
     $queryResult[] = $obj->author_id;
     $queryResult[] = $obj->edition;
     $queryResult[] = $obj->title;
     $queryResult[] = $obj->publisher;
     $queryResult[] = $obj->vendor;
     $queryResult[] = $obj->cost_i;
     $queryResult[] = $obj->cost_f;
     $queryResult[] = $obj->plo_pub;


    }
}

$textboxValue1 = $queryResult[0];
$textboxValue2 = $queryResult[1];
$textboxValue3 = $queryResult[2];
$textboxValue4 = $queryResult[3];
$textboxValue5 = $queryResult[4];
$textboxValue6 = $queryResult[5];
$textboxValue7=  $queryResult[6];
$textboxValue8 = $queryResult[7];
$textboxValue9 = $queryResult[8];




echo json_encode(array('first'=>$textboxValue1,'second'=>$textboxValue2,'third'=>$textboxValue3,'four'=>$textboxValue4,'five'=>$textboxValue5,'six'=>$textboxValue6,'seven'=>$textboxValue7,'eight'=>$textboxValue8,'nine'=>$textboxValue9));
?>
Puligundla Neelakantan
  • 19
  • 1
  • 4
  • Before you start assigning values to it, declare it `$queryResult = array();`. Also, likely `if ($result=mysqli_query($con,$sql))` fails. Debug it. – DeDee Sep 16 '15 at 18:00
  • define the `$queryResult` var before you use it as an array. – Danijel Sep 16 '15 at 18:00

0 Answers0