PHP JSON into javascript variable

Question

befora I ask this i tried every solution on google and here on stackoverflow.

So in my js file I have this

    var locations = [
    ['Name','Country','City','lat','lng'],
    ['Name','Country','City','lat','lng'],
];

When i wrote this in file manually my mapp shown locations but I need to generate content of locations variable from mysql in php, is there something that I missing.I tried with ajax,console.log...etc

My PHP file

    $result = mysqli_query($link, "SELECT hospital_name,country,city,lat,lng FROM hospitals");
$to_encode = array();
while($row = mysqli_fetch_row($result)) {
  $to_encode[] = $row;
}
echo json_encode($to_encode);

I tried this but no success

$.getJSON( "process.php", function( json ) {
  var locations = json;
 });

Ajax will do what you need, have a look at api.jquery.com/jquery.getjson . It gets automatically parsed to a JS array — Ben Kolya Mansley
– Ben Kolya Mansley, Commented Sep 22, 2015 at 1:38
var location should be defined outside of the getJson function — ductiletoaster
– ductiletoaster, Commented Sep 22, 2015 at 1:42

Marcus Abrahão · Accepted Answer · 2015-09-22 02:20:38Z

1

I just switched your mysqli_fetch_row to mysqli_fetch_array

    $result = mysqli_query($link, "SELECT hospital_name,country,city,lat,lng FROM hospitals");
    $to_encode = array();
    while($row = mysqli_fetch_array($result)) {
      $to_encode[] = $row;
    }
    echo json_encode($to_encode);

edited Sep 22, 2015 at 2:20

answered Sep 22, 2015 at 2:15

Marcus Abrahão

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1 Comment

charlietfl Over a year ago

Please add explanation not just code. It now becomes a game of find the difference whereas a simple explanation of what you changed goes a long way helping others see it and understand why. Your code solution might be great....but the answer is very low quality without people being able to review it quickly

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