Warning: MySQLi_Select_Db() Expects Exactly 2 Parameters, 1 Given

Warning: mysqli_select_db() expects exactly 2 parameters, 1 given

mysqli_select_db() should have 2 parameters, the connection link and the database name -

mysqli_select_db($con, 'phpcadet') or die(mysqli_error($con));

Using mysqli_error in the die statement will tell you exactly what is wrong as opposed to a generic error message.

mysqli_select_db() expects exactly 2 parameters

do the correction as below:

$conn = mysqli_connect('localhost', 'root', '');

the first thing that you have to pass connection variable in the select_db as first parameter. as below.

mysqli_select_db($conn,'altislife-dev');

also you have to pass connection variable in mysqli_query() as first parameter as given below.

$records=mysqli_query($conn,$sql);

Warning: mysqli_select_db() expects exactly 2 parameters, 1 given *13*

The first parameter of mysqli_select_db() is connection object. This is the syntax :

mysqli_select_db(connection,dbname);

Change your code to:

$con = mysqli_connect("$host", "$username", "$password")or die("cannot connect server "); 
mysqli_select_db($con,$db_name)or die("cannot select DB");

Add the connection object as first parameter in mysqli_query and mysqli_fetch_array functions too. Please refer this link for the syntax.

mysqli_select_db() expects exactly 2 parameters, 1 given

Posting as a community wiki. I don't want rep from this and there shouldn't.

$connexionbase = mysqli_select_db($base)

Just as the error states. You need to pass the db connection as the first argument:

$connexionbase = mysqli_select_db($connexion, $base)

Reference:

• http://php.net/manual/en/mysqli.select-db.php

Example from the manual:

bool mysqli_select_db ( mysqli $link , string $dbname )

Sidenote:

return(($connexion && $connexionbase));

TBH, I've never seen this type of syntax for a return. Far as I know, you can return only once or using an array.

Pulled from this answer https://stackoverflow.com/a/3815243/1415724

You can only return one value. But you can use an array that itself contains the other two values:
return array($uid, $sid);

Instead of going through all that trouble, just use the 4 parameters:

$connexion = mysqli_connect($server,$loginsql,$passsql, $base)

as per the manual:

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

then return with and if it's really needed.

return $connexion;

Plus, why are you intending to use MD5 to store passwords with? That hashing function is no longer considered safe to use.

You're better off using password_hash().

• http://php.net/manual/en/function.password-hash.php

This is the 21st century after all.

and $HTTP_COOKIE_VARS, that's deprecated.

• http://php.net/manual/en/reserved.variables.cookies.php

I've no idea why you're using that code or where you got it from.

PHP session variables not carrying over to my logged in page, but session ID is

I don't see a session_start() in your login script. If you aren't starting the session I don't think php will save any data you place in the $_SESSION array. Also to be safe I'd explicitly place variables into the $_SESSION array instead of just overwriting the whole thing with $_SESSION = mysql_fetch_array($result);.

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