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

[shimabuku]

Hello all! New here and new to PHP. I have an old script that was built for PHP 5.X and need some help getting it to work with PHP 7.X. According to the PHP manual I need to pass the db connection as the first argument. I do get the display error "SQL Error: Can't select DB" and in the logs show "PHP Warning:  mysqli_select_db() expects exactly 2 parameters, 1 given".

<?php
extract($_REQUEST);

$DB_HOST="localhost";                           //Host
$DB_DATABASE="babyyoda";                        //database name
$DB_USER="mando";                                //database username
$DB_PASSWORD="password";                              //databse password



$ADMIN_EMAIL='admin@example.com';
$TEMPLATES='./templates/';

$URL_BASE='http://example.com/banners/';

$PATH='/var/www/vhosts/example.com/httpdocs/banners/';  
$BANNERS_PATH=$PATH.'banners/';  
$BANNERS_URL=$URL_BASE.'banners/';  


$DB_TABLE_CATEGORIES='ban_categories';
$DB_TABLE_SUBCATEGORIES='ban_subcategories';
$DB_TABLE_SUBCATEGORIES_BANNERS='ban_subcategories_banners';
$DB_TABLE_BANNERS='ban_banners';
$DB_TABLE_USERS='ban_users';
$DB_TABLE_SETTINGS='ban_settings';

$IMAGE_MAGICK_PATH='/usr/bin/';
//$IMAGE_MAGICK_PATH='/usr/local/bin/';  	// on some servers it may be like this


$TEMPLATE_EXECUTE_PHP=1; // set to 1 to allow use of php inside of templates


$CRYPT_SALT='ef&dF%HGRTSvsw35';


$TYPES[1]='Jpeg';
$TYPES[2]='Gif';
$TYPES[3]='Flash'; // dont change flash should be "3"


//$LINKED_SCRIPTS['Galleries Manager']='/galleries/admin/';
//$LINKED_SCRIPTS['Flash Manager']='/flash/admin/';


$NATS=0;
//$PASS_VARIABLES=array('campaign','program');



define('CONFIG_INCLUDED',1);

mysqli_connect($DB_HOST, $DB_USER, $DB_PASSWORD) or die("SQL ERROR: Can't Connect to DB");
mysqli_select_db($DB_DATABASE) or die("SQL ERROR: Can't select DB");


error_reporting(0);
ini_set('session.gc_maxlifetime',3600);

?>

 

I came up with the below but I don't know where to insert these two statements and what to remove.

$link = mysqli_connect($DB_HOST, $DB_USER, $DB_PASSWORD);
mysqli_select_db($link, $DB_DATABASE);

 

[requinix]

Were you able to find the parts in your existing where you call those two functions?

[ginerjm]

Needing 2 parms means that your connect didn't provide the proper result for the select_db call.  Figure out first why your connection fails.  Usually a good thing to check whenever you open a db connection.

if ($mysqli->connect_errno)
{
	echo "Connect failed, error is: " . $mysqli->connect_error);
	exit();
}

 

[shimabuku]

The original parts were;

mysql_connect($DB_HOST, $DB_USER, $DB_PASSWORD) or die("SQL ERROR: Can't Connect to DB");
mysql_select_db($DB_DATABASE) or die("SQL ERROR: Can't select DB");

So i figured I'd just change it to mysqli_connect and mysqli_select_db since it's PHP7.X.

[requinix]

1 hour ago, shimabuku said:

The original parts were;

mysql_connect($DB_HOST, $DB_USER, $DB_PASSWORD) or die("SQL ERROR: Can't Connect to DB");
mysql_select_db($DB_DATABASE) or die("SQL ERROR: Can't select DB");

So i figured I'd just change it to mysqli_connect and mysqli_select_db since it's PHP7.X.

Okay... I know you figured out the parameters because you posted what it should be, and now I know you have seen where those functions (which already have the "i" added) are in the code.

So what's the problem?

[shimabuku]

On 1/20/2020 at 10:15 AM, requinix said:

Okay... I know you figured out the parameters because you posted what it should be, and now I know you have seen where those functions (which already have the "i" added) are in the code.

So what's the problem?

When I change it to below. I just get an internal 500 error. Nothing in the logs.

$link = mysqli_connect($DB_HOST, $DB_USER, $DB_PASSWORD);
mysqli_select_db($link, $DB_DATABASE);

 

[mac_gyver]

On 1/19/2020 at 5:51 PM, shimabuku said:

error_reporting(0);

the code is intentionally disabling error reporting via the above line. comment out that line and see what you get.

[shimabuku]

33 minutes ago, mac_gyver said:

the code is intentionally disabling error reporting via the above line. comment out that line and see what you get.

Roger that. I'll try that when I get home. Thanks!