Pull data from MySQL into JSONArray (with PDO!)

Question

I was reading something on the Internet and many people say that it's recommended to switch to PDO from old mysql (and mysqli) extensions.

I'm new with this PDO, I learned something about it. For my problem I tried to search stackoverflow, Google, etc, but it didn't helped me.

Original script with mysqli :

$data = mysqli_query($con, "SELECT x1, x2, x3, x4 FROM Table1 WHERE x1 = $variable1");
$row = mysqli_fetch_row($data);
$result_data = array(
  'data1' => $row[0],
  'data2' => $row[1]
 );

echo json_encode($result_data);

This code outputs something like this :

{"data1", 1, "data2", 2}

I tried to change it with this PDO code :

 $STH = $DBH->prepare("SELECT x1, x2, x3, x4 FROM Table1 WHERE x1 = ?");
 $STH->bindParam(1, $variable1);
 $STH->execute();
 $row = $STH->fetchAll();

 $result_data = array( 
    'data1' => $row[0],
    'data2' => $row[1],
 ); 

  echo json_encode($result_data);

This back me something very strange like

{"data1", 1, 2, 1, "data2", 1, "data2"}, data1 null, data2 null

It should back like with original mysqli script...

I tried with multiple fetch modes like assoc, num, column, what I was found on Internet, but result was very similar and I always get this error :

Notice:Undefined offset: 1 in

What can be the problem, how to fix it?

Your Common Sense · Accepted Answer · 2014-03-24 18:25:25Z

1

 $STH = $DBH->prepare("SELECT x1 data1, x2 data2 FROM Table1 WHERE x1 = ?");
 $STH->execute([$variable1]);
 echo json_encode($STH->fetchAll(PDO::FETCH_ASSOC));

answered Mar 24, 2014 at 18:25

Your Common Sense

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