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Error when connecting to DB

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Apr 12.2015

Hi guys, i am basically new to php.. I have recently tried creating a web app using mysql and php.. However, i got a error that states:

Warning: mysqli_select_db() expects exactly 2 parameters, 1 given in C:wampwwwmain.php on line 8
Call Stack

# Time Memory Function Location

1 0.0000 241960 {main}( ) ..main.php:0
2 0.0010 251336 mysqli_select_db ( ) ..main.php:8
Could not find database (line 8)

whenever i tried to open the page via web browser.. the following are my connection codes:

[code=php]$hostname = “localhost”;
$username = “DBuser1”;
$dbname= “Newdb”;
$password = “newcomer”;
$conn = mysqli_connect($hostname,$username,$password);
mysqli_select_db($conn) or die(“Could not find database”);[/code]

May i know how do i fix this problem? It is really getting on my nerves… Please help!!! thank you!!!

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2 Comments(s) ↴

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@iBeZiApr 12.2015 — You need to tell it which database you want to select.

[code=php]
mysqli_select_db($conn,$dbname) or die("Could not find database");

[/code]
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@rootApr 13.2015 — 
  • 1. you should avoid creation of variables for the sake of it. Instead of having $hostname="localhost"; simply use the term "localhost" where it is needed.

  • 2. You can use a 4th parameter to select the database you wish to use in the opening command

  • 3. Keeping your login details separate from your main script is a good way of ensuring some kind of security should your main page become susceptible to a code injection attack, for example.

  • 4. Using an OO approach will reduce the need to keep a separate variable for the connection handle.


    • [code=php]include("database.php"); // database, username and password info :)
    $db = new mysqli("localhost",$username,$password,$databaseName); [/code]

    then any use of the $db object like a query requires only $result = $db->query("SELECT * FROM somedatabase WHERE wotsits='prawncoctail'"); and then $results contains your needed information based on your query.
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