https://devcodepro.comdevcodepro

PHP 8 join(): Argument #1 ($pieces) must be of type array, string given

$id = join(',',$val);

##replace with

if (!empty($val)) {
 $id = join(',', $val);
}else{
 $id = join(',', array_filter(array(@$val)));
}

AD · www.interserver.net/r/240055

Interserver | Standard & VPS Cloud Hosting | $2.50 /Month

Flexible VPS hosting platform to deploy your online projects. Economical and balanced between processor cores, memory and storage

rated 1 times (1) (0)

comments: 1 / hits: 2567 / 2 years ago, sun, may 2, 21, 03:34:47

Comments

I have the same problem, Hoe someone giving help for it

[#61] Thursday, June 9, 2022, 4:22:53

commented 1 year ago

Ferdinand Juko Randrianirimiadana

Only authorized users can post. Please sign in first, or register a free account

Posted

go

Member since May 1, 2021

Total Points: 50

Total Code Snippets: 2

Total Comments: 0

Location: n/a

go snippets

ADODB check if MySQL connected

2 years ago, sat, may 1, 21, 5:53:49

<?php include ('classes/adodb.inc.php'); $dbdriver = 'mysqli'; $server = 'localhost'; $user = 'root'; $password = 'testpass'; $database = 'dbname'; $conn = ADONewConnection($dbdriver); $conn->Connect($server, $user, $password, $database); if (!$conn->isConnected()): echo 'No Connected'; die(); else: echo 'Connected'; endif; ?>

comments: 0 / hits: 565

PHP