cannot access database using php class

Question

I created a class to access my database. The simplified class is following (I named it dbaccess.php)

class dbaccess {
function read($db) {
    $con = mysqli_connect($db);
    if (mysqli_connect_errno()){
        echo "Failed to connect to MySQL: " . mysqli_connect_error();
        }
    $result = mysqli_query($con,"SELECT * FROM equipment");
    while($row = mysqli_fetch_array($result)){
        $print = $print . $row['ID'] . " " . $row['name'] . " " . $row['new_price'] . " " . $row['residual_value'] . "<br>";
        }
    echo $print;
    mysqli_close($con);
    }
}

To access the class, I use this code

include './dbaccess.php';
//define db address
$add = '"localhost","myuser","mypassword","mydbname"';
$db = new dbaccess;
$db->read($add);

This code resulting

Failed to connect to MySQL: Unknown MySQL server host '"localhost","myuser","mypassword","mydbname"'(2)

I don't know how to fix it, can anyone here help me?

You need to pass each value in as a separate parameter, not as string of parameters. — Orangepill
– Orangepill, Commented Aug 5, 2013 at 16:01
it is clear but I don't think I can't define all parameter in one string — Hakim
– Hakim, Commented Aug 5, 2013 at 16:04
that is correct ... you cannot. you have to have 4 different arguments for your read method, $host, $user, $password, $db, when you call mysqli_connect pass each of these arguments in separately. Currently you are passing a single string argument in so it's trying to connect to a database on "localhost","myuser","mypassword","mydbname" — Orangepill
– Orangepill, Commented Aug 5, 2013 at 16:06

ale · Accepted Answer · 2013-08-05 16:02:29Z

4

You're passing a single string to mysqli_connect. You need to pass "localhost", "myuser",... as separate variables.

http://php.net/manual/en/function.mysqli-connect.php

answered Aug 5, 2013 at 16:02

ale

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no name · Accepted Answer · 2013-08-05 16:03:03Z

0

I have something similar with yours, download my files from dropbox and have a look inside DB Connect

answered Aug 5, 2013 at 16:03

no name

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Hakim Over a year ago

Great, you did it three years ago. Thank you.

silentw · Accepted Answer · 2013-08-05 16:03:35Z

0

class dbaccess {
    function read($db) {
        $con = mysqli_connect($server, $user, $password, $dbname);
(...)

Then in your code, you should divide the parameters.

include './dbaccess.php';
//define db address
$db = new dbaccess;
$db->read("localhost","myuser","mypassword","mydbname");

REF

answered Aug 5, 2013 at 16:03

silentw

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Comments

user1748315 · Accepted Answer · 2013-08-05 16:06:52Z

0

It's probably a better idea to store your host, username, password and database in separate variables like this:

$host = "localhost";
$user = "myuser";
$pass = "mypassword";
$data = "mydbname";
$db = new dbaccess(); // <-- It's good practice to use parentheses in the constructor statement.
$db->read($host,$user,$pass,$data);

And then the implementation of your dbaccess class could be more like this:

$con = mysqli_connect($host,$user,$pass,$data);

This is because the mysqli_connect function takes the host, username, etc. as separate parameters, not as a single string.

answered Aug 5, 2013 at 16:06

user1748315

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