How to display SQL result separately - php

Question

I'm currently making a webpage which is meant to show all it's content from the database. So I made an SQL command which selects the data needed for only 1 particular field on the webpage.
Is it possible to make the SQL command so that it get's all the content for the page at once and that ill still be able to display it separately?
If so, how? Thanks
function dbGet() {
global $conn;
global $return;
$sql = SELECT * FROM testTable;
$result = $conn->query($sql);
if ($result->num_rows > 0) {
while($row = $result->fetch_assoc()) {
$return = $row["text"];
return $return;
}
}
else {
echo "0 results";
}
// $conn->close();
}

Answer 1

You can use in this way through which you can identify records has been there or not.
function dbGet() {
global $conn;
// I am not sure what is the datatype of $return here.
// if it's having a predefined format,
// please push the records into $return instead of creating as an array,
// which will be taken care by framework if any you are using.
// global $return;
$return = array('status' => false, records => []);
$sql = "SELECT text FROM testTable";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
$return['status'] = true;
while($row = $result->fetch_assoc()) {
$return['records'][] = $row["text"];
}
}
// $conn->close();
return json_encode($return);
}

Answer 2

Just echo the results inside while loop instead of returning in whatever format you wish
If you are using php, you can try something like
echo $row["text"];
Hope it helps
Let me know if you require any further help

Answer 3

First you should fix your return code as #AbraCadaver mentioned :
$return[] = $row["text"];
Then you can use foreach in your html :
<?php
foreach($return as $r){
echo $r['text'];
}
?>

Answer 4

I think a Json encoding : json_encode will work well.
Try this: http://php.net/manual/pt_BR/function.json-encode.php