Hello
Can someone help me?
I encountered a failed database connection.
Please correct my mistake
Try mysqli_* instead
<?php
$server = "localhost";
$username = "id2538939_syafeimohammed";
$password = "database";
$database = "id2538939_db_kelurahan";
$tanggal=date("Y-m-d H:i:s");
$entries=3;
$koneksi = mysqli_connect($server,$username,$password) or die ('Koneksi gagal');
if($koneksi){
mysqli_select_db($database) or die ('Database belum dibuat');
}
?>
Thankyou for the response.
I have done that but the information that appears “database belum dibuat”
what’s wrong?
Please replace
mysqli_select_db($database) or die ('Database belum dibuat');
By
mysqli_select_db($koneksi,$database) or die ('Database belum dibuat');
Can not use “mysql”?
Migration “mysql” to “mysqli” is very troublesome
Is there an easier alternative way?
I tried to change it but it did not work
Mysql
<?php
// Perintah untuk menampilkan data //menampikan SEMUA data dari tabel berita
$query = mysql_query("select * from berita limit 3");
while ($data = mysql_fetch_array($query)){
// perintah untuk membaca dan mengambil data dalam bentuk array
$id = $data['id_berita'];
?>
Mysqli
<?php
// Perintah untuk menampilkan data dari tabel berita
$result = mysqli_query($link, "SELECT * FROM berita limit 3");
while ($data=mysqli_fetch_array($result,MYSQLI_ASSOC)){
// perintah untuk membaca dan mengambil data dalam bentuk array
$id = $data['id_berita'];
?>
Try this one
<?php
$server = "localhost";
$username = "id2538939_fei";
$password = "root10";
$database = "id2538939_kel";
$tanggal=date("Y-m-d H:i:s");
$entries=3;
$koneksi = mysqli_connect($server,$username,$password) or die ('Koneksi gagal');
if($koneksi){
mysqli_select_db($koneksi,$database) or die ('Database belum dibuat');
}
?>