SQL results to PHP array - php

Question

I've got the following sql query:
$sql = "SELECT lat, lang
FROM users";
I then use the following code to put the results of the array into two arrays, one for lat and one for lang.
$i = 0;
foreach($results as $row) {
$latArray = array();
$langArray = array();
$latArray[$i] = $row['lat'];
$langArray[$i] = $row['lang'];
$i = ($i + 1);
}
However, it seems that only the last value that is passed to the array is stored. When I echo out each value of the array I get the following error: Undefined offset: 0 which I believe means theres nothing at latArray[0].
I'm sure I've missed something obvious here but why aren't all the values copied to the new array?

Answer 1

$i = 0;
$latArray = array(); //Declare once, do not redeclare in the loop
$langArray = array();
foreach($results as $row) {
$latArray[$i] = $row['lat'];
$langArray[$i] = $row['lang'];
$i = ($i + 1);
}

Answer 2

Declare your array before the loop
$latArray = array();
$langArray = array();
foreach($results as $row) {
$latArray[$i] = $row['lat'];
$langArray[$i] = $row['lang'];
$i = ($i + 1);
}

Answer 3

You should put
$latArray = array();
$langArray = array();
Before foreach cycle (like you do with your counter $i), 'cause with every new value from $result it resets thous values..
so your code will look like:
$i = 0;
$latArray = array();
$langArray = array();
foreach($results as $row) {
$latArray[$i] = $row['lat'];
$langArray[$i] = $row['lang'];
$i = ($i + 1);
}

Answer 4

use fetch_assoc instead:
$latArray = array();
$langArray = array();
while($row = mysql_fetch_assoc($your_query)){
$latArray[$i] = $row['lat'];
$langArray[$i] = $row['lang'];
$i++;
}
if u are using msqli us mysqli_fetch_assoc