CODEWITHSUNDEEP
php
user3004356
user3004356

Reputation: 880

Script returns null on fetching results

I'm trying to query a database using PHP script and pass JSON string. I'm using this script from jQuery Ajax.

<?php  

$con = mysqli_connect("localhost","root","password","test") or die("Some error occurred during connection " . mysqli_error($con));          
$strSQL = "SELECT name,id from build";        
$query = mysqli_query($con, $strSQL);
$builder_json=array();
$row_array=array();    
while($result = mysqli_fetch_array($query))
{
  // $builder_json[]=$result['name'].":".$result['id'];           
    $row_array['name'] = $result['name'];
    $row_array['id'] = $result['id'];
    array_push($builder_json[],$row_array);
}
echo json_encode($builder_json);
mysqli_close($con);

?>

Now if the data fetched is like :

Name - A,B

Id - 1,2

I want JSON string to be like - {"A":"1","B":"2"}; With the above code I get NULL . What is the mistake I have done

Upvotes: 0

Views: 35

Answers (1)

Cl&#233;ment Malet
Cl&#233;ment Malet

Reputation: 5090

You don't need the [] after your variable in your array_push call

//array_push($builder_json[],$row_array); 
array_push($builder_json,$row_array);

Otherwise it gives the following error :

Warning: array_push() expects parameter 1 to be array, null given on line

Upvotes: 1

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