PHP json_encode() in while loop

Question

I am trying to use json_encode() in a while loop while getting database results. Here is my code:

<?

$database = sqlite_open("thenew.db", 0999, $error);
if(!$database) die($error);

$query = "SELECT * FROM users";
$results = sqlite_query($database, $query);
if(!$results) die("Canot execute query");

while($row = sqlite_fetch_array($results)) {
  $data = $row['uid'] . " " . $row['username'] . " " . $row['xPos'] . " " . $row['yPos'];
}
echo json_encode(array("response"=>$data));

sqlite_close($database);

?>

The output of this is

{"response":"lastUserID lastUser lastXPos lastYPos"}

I want it to be...

{"response":["1 Alex 10 12", "2 Fred 27 59", "3 Tom 47 19"]}

etc.

So I want the json_encode() function to put ALL users into the array rather than the last one. How would I do this? Thanks

Nazin · Accepted Answer · 2011-08-05 22:31:50Z

4

Try:

<?

$database = sqlite_open("thenew.db", 0999, $error);
if(!$database) die($error);

$query = "SELECT * FROM users";
$results = sqlite_query($database, $query);
if(!$results) die("Canot execute query");

$data = array();

while($row = sqlite_fetch_array($results)) {
  $data[] = $row['uid'] . " " . $row['username'] . " " . $row['xPos'] . " " . $row['yPos'];
}
echo json_encode(array("response"=>$data));

sqlite_close($database);

?>

answered Aug 5, 2011 at 22:31

Nazin

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Comments

Pheonix · Accepted Answer · 2011-08-05 22:28:40Z

3

Change this

while($row = sqlite_fetch_array($results)) {
  $data = $row['uid'] . " " . $row['username'] . " " . $row['xPos'] . " " . $row['yPos'];
}

to

$data = array();
while($row = sqlite_fetch_array($results)) {
  $data[] = $row['uid'] . " " . $row['username'] . " " . $row['xPos'] . " " . $row['yPos'];
}

answered Aug 5, 2011 at 22:28

Pheonix

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Comments

Baversjo · Accepted Answer · 2011-08-05 22:30:03Z

3

Push each user to an array:

$data = array();
while($row = sqlite_fetch_array($results)) {
  $data[] = $row['uid'] . " " . $row['username'] . " " . $row['xPos'] . " " . $row['yPos'];
}
echo json_encode(array("response"=>$data));

answered Aug 5, 2011 at 22:30

Baversjo

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