Querying a database within a PHP function [duplicate] - php

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mysqli/mysql query inside function not working [duplicate]
(2 answers)
Closed 9 years ago.
I'm trying to load some data from a database using PHP, but for some reason it doesn't work when I put it inside a function. If I try the code without a function, it works fine:
//$dbc connection
$call1 = 0;
$output = '';
$query = "select * from artists order by lname limit $call1, 15";
$result = mysqli_query($dbc, $query);
while($row = mysqli_fetch_array($result)){
$output .= "<ul>";
$output .= "<li>" . $row['name'] . "</li>";
$output .= "</ul>";
}
However, when I change the code to be inside a function, I don't get anything from the database (or at least it won't return anything):
//$dbc connection
$call1 = 0;
$output = '';
function loadArtists($call){
$query = "select * from artists order by lname limit $call, 15";
$result = mysqli_query($dbc, $query);
while($row = mysqli_fetch_array($result)){
$output .= "<ul>";
$output .= "<li>" . $row['name'] . "</li>";
$output .= "</ul>";
}
}
loadArtists($call1);
What am I doing wrong here?

Answer 1

You cannot use $dbc in your function, because it is a global variable.
You can use either
function loadArtists($call){
global $dbc;
...
}
to make $dbc known to loadArtists() or pass it as a second parameter
function loadArtists($dbc, $call){
...
}
and call it as
loadArtists($dbc, $call1);

Answer 2

As I mentioned in one of my comments, using global to fix the scope of your connection is poor practice. The correct way to pass your connection is like so:
$dbc = mysqli_connect("localhost", "my_user", "my_password", "world");
$call1 = 0;
$output = '';
function loadArtists($call, $dbc){
$query = "select * from artists order by lname limit $call, 15";
$result = mysqli_query($dbc, $query);
while($row = mysqli_fetch_array($result)){
$output .= "<ul>";
$output .= "<li>" . $row['name'] . "</li>";
$output .= "</ul>";
}
}
loadArtists($call1, $dbc);

Answer 3

Declaring your username and password on the same page as the code you're executing every time you want to make a DB connection is bad practice because:
You could potentially have more than one page to edit if you move to a different host or a different environment than the dev environment.
If you declare it outside of the root you can limit access to the DB password from FTP accounts.
I like to use a function for the connection so this way if the connection closes you can reopen it at will (reducing server overhead). Also you don't have to set it as a global var inside the function (not a good idea because of several reasons).
So for those reasons, this connection should be outside (below) the root.
/../safe/connection.php
function openSQL() {
$conn = mysqli('localhost', 'my_user', 'my_password', 'my_db');
return $conn;
}
functions.php
require_once($_SERVER['DOCUMENT_ROOT'].'/../safe/connection.php');
function loadArtists($call){
$dbc = openSQL();
$query = "select * from artists order by lname limit $call, 15";
$result = mysqli_query($dbc, $query);
while($row = mysqli_fetch_array($result)){
$output .= "<ul>";
$output .= "<li>" . $row['name'] . "</li>";
$output .= "</ul>";
}
mysqli_close($dbc);
return $output;
}
$myOutput = loadArtists(4);

Answer 4

The problem is variable scope. That variable does not exist within your function.
There are three ways to deal with that:
Make it global, that means an outside variable is readible from within a function. (Note that using global variables is often considered a security concern.)
global $dbc;
You can pass that variable into the function as an argument
function loadArtists($connection, $call) { ... }
you can make a class and that class variable will now be useable inside of class functions:
class Artists {
public $dbc;
public function __construct() {
$this->dbc = open_the_db_connection(); //etc?
}
public function loadArtists($call) {
$query = "select * from artists order by lname limit $call, 15";
$result = mysqli_query($this->dbc, $query);
while($row = mysqli_fetch_array($result)){
$output .= "<ul>";
$output .= "<li>" . $row['name'] . "</li>";
$output .= "</ul>";
}
return $output;
}
}

Answer 5

It looks like a scoping issue to me. The $output you reference in the function isn't the same as the $output you defined outside the function.
You should change your function to the following:
function loadArtists($call){
$output = "";
$query = "select * from artists order by lname limit $call, 15";
$result = mysqli_query($dbc, $query);
while($row = mysqli_fetch_array($result)){
$output .= "<ul>";
$output .= "<li>" . $row['name'] . "</li>";
$output .= "</ul>";
}
return $output;
}
$output = loadArtists($call1);