Assign output of PHP function to a variable - php

Question

Here's my current code:
function get_coins() {
global $conn;
$new_sql = "SELECT * FROM marketcaps ORDER BY cap DESC LIMIT 25";
$new_result = $conn->query($new_sql);
while ($row = $new_result->fetch_assoc()) {
$id = $row["id"];
$coin = $row["coin"];
$cap = $row["cap"];
echo $coin . '~USD,';
}
}
$coins = get_coins();
Here's what I can't figure out:
The line $coins = get_coins(); is echoing all of the data from my function and I don't understand why. I know that my function has this line: echo $coin . '~USD,';
But why is it echoing when I add $coins = get_coins();?
Shouldn't I have to add this instead?
$coins = get_coins();
$echo coins;
Basically I don't want to echo anything, I just want to assign the output to a variable. Is there a simple way to do that?

Answer 1

You need to add a return statement to your function, and as you are dealing with multiple rows, collate them into an array (index by ID).
http://php.net/manual/en/function.return.php
function get_coins()
{
global $conn;
$coins = array();
$new_sql = "SELECT * FROM marketcaps ORDER BY cap DESC LIMIT 25";
if( $new_result = $conn->query($new_sql) )
{
while( $row = $new_result->fetch_assoc() )
{
$coins[ $row["id"] ] = $row["coin"] . '~USD,';
}
}
return ( !empty( $coins ) ) ? $coins : null;
}

Answer 2

Here are the principles that I have baked in with my solution which returns the full resultset array:
Don't use global, pass $conn as a parameter to the function.
Don't declare single use variables, unless it dramatically improves readability or assists in the debugging process.
Only query for the specific columns and rows that you intend to process.
Suggested Code:
function get_coins($conn): array {
return $sql
->query("SELECT id,coin,cap
FROM marketcaps
ORDER BY cap DESC
LIMIT 25")
->fetch_all(MYSQLI_ASSOC);
}
foreach (get_coins($conn) as $row) {
// ... reference $row['id'] , $row['coin'] , $row['cap'] during your processes
}