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;
 });

• aldrin27: Where's the attempted code?
• Ben Kolya Mansley: Ajax will do what you need, have a look at api.jquery.com/jquery.getjson . It gets automatically parsed to a JS array
• ductiletoaster: var location should be defined outside of the getJson function
• user1594938: I edited code still doesn't work, where do I wrong?
• aldrin27: Try to console.log(json)

Answer 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);

• charlietfl: 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