How to output JSON from recordset in Dreamweaver with minimal manual code? [duplicate] - sql

Question

I have a basic MySQL database table People with 3 columns (ID, Name, Distance)
I am trying to output this with PHP as a JSON so my web app can pick it up. I tried using the solution from another answer:
$sth = mysql_query($con,"SELECT * FROM People");
$rows = array();
while($r = mysql_fetch_assoc($sth)) {
$rows[] = $r;
}
print json_encode($rows);
However I am just returning a blank array of [].
Update: Changed to mysqli and added dumps:
$sth = mysqli_query("SELECT * FROM Events",$con);
$rows = array();
var_dump($sth);
while($r = mysqli_fetch_assoc($sth)) {
var_dump($rows);
$rows[] = $r;
}
print json_encode($rows);
Returns:
NULL []

Answer 1

Swap your query and connection in the mysql_query-statement:
$sth = mysql_query("SELECT * FROM People", $con);
Besides that, the mysql-library is deprecated. If you're writing new code consider using mysqli_ or PDO.

Answer 2

Why no mysql_connect ?
var_dump($sth); // check the return of mysql_query()
var_dump($rows); // check the array of mysql_fetch_assoc result

Answer 3

I would check if your query is returning any rows first.
<?php
$rows = array();
array_push($rows, array("name" => "John"));
array_push($rows, array("name" => "Jack"));
array_push($rows, array("name" => "Peter"));
print json_encode($rows);
?>
Produces this output.
[{"name":"John"},{"name":"Jack"},{"name":"Peter"}]

Answer 4

Solved using a combination of answer to get:
$sth = mysqli_query($con,"SELECT * FROM Events");
$datarray = array();
while($row = mysqli_fetch_assoc($sth)) {
$datarray[] = $row;
}
print_r(json_encode($datarray));