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Register Code / Sign Up Code ERROR - PHP ver. 7.2.5 Xampp ver. 3.2.2 [2015]
Halo agan-agan / sista-sista Salam kenal, ane baru belajar coding otodidak 3 bulan yang lalu, jadi skrg belajar PHP, ada problem yang ga bisa ane pecahin dr kemarin-kemarin, mohon pencerahannya...
Problem : Error aja, ane ga tau mesti evaluasi problem nya dimana
<?php
$host = "localhost";
$user = "root";
$pass = "";
$database = "test";
$db = mysqli_connect($host, $user, $pass);
$first_name = $_POST['first_name'];
$last_name = $_POST['last_name'];
$password = $_POST['password'];
$re_password = $_POST['re_password'];
$tanggal = $_POST['tanggal'];
$bulan = $_POST['bulan'];
$tahun = $_POST['tahun'];
$email = $_POST['email'];
if($_POST['password']==$_POST['re_password'])
{
$_POST['password'];
}
else
{
echo "Password not match !";
}
$sql = ("INSERT INTO user ('first_name, last_name, password, tanggal, bulan, tahun, email') VALUES ('$first_name', '$last_name', '$password', '$tanggal', '$bulan', '$tahun', '$email')");
$result = mysqli_query($db, $sql);
if($result)
{
echo "Data saved";
}
else
{
echo "Error !";
}
?>
18 Jawaban:
Ane coba ilangin sintak echo error nya, blank putih aja Data yang di input ga masuk apapun ke database
"INSERT INTO user ('first_name, last_name, password, tanggal, bulan, tahun, email') VALUES ('$first_name', '$last_name', '$password', '$tanggal', '$bulan', '$tahun', '$email')"
coba ilangin tanda petik yang di sebelum firstname dan email gan jadi
"INSERT INTO user (first_name, last_name, password, tanggal, bulan, tahun, email) VALUES ('$first_name', '$last_name', '$password', '$tanggal', '$bulan', '$tahun', '$email')"
sama aja gan, ga mau jg malah minta parameter nya 1 lagi di mysqli_query
$input = ("INSERT INTO users (first_name,last_name,password,tanggal,bulan,tahun,email) VALUES ('$first_name','$last_name','$password','$tanggal','$bulan','$tahun','$email')");
$result = mysqli_query($input, $connect);
// HASIL //
Database Exist
Warning: mysqli_query() expects parameter 1 to be mysqli, string given in C:\xampp\htdocs\mofu\input.php on line 41
Data save failed
liat di codingan yang di postingan pertanyaan kayak nya belum pilih database nya deh
pastikan juga koneksi nya benar.. dan terkoneksi
// UPDATE //
<?php
$host = "localhost";
$user = "root";
$pass = "";
$database = "mofu";
$table = "users";
$connect = mysqli_connect($host, $user, $pass);
if($connect)
{
echo "Database Exist";
}
else
{
echo "Database is not exist";
}
$check = mysqli_select_db($connect, $database);
// ========================================================== //
$first_name = $_POST['first_name'];
$last_name = $_POST['last_name'];
$password = $_POST['password'];
$re_password = $_POST['re_password'];
$tanggal = $_POST['tanggal'];
$bulan = $_POST['bulan'];
$tahun = $_POST['tahun'];
$email = $_POST['email'];
if ($_POST['password']==$_POST['re_password'])
{
$_POST['password'];
}
else
{
echo "<script>alert('Password not match')</script>";
}
$input = ("INSERT INTO 'users' ('first_name','last_name','password','tanggal','bulan','tahun','email') VALUES ([$first_name],[$last_name],[$password],[$tanggal],[$bulan],[$tahun],[$email])");
$result = mysqli_query($connect, $input);
if ($result)
{
echo "Data saved";
}
else
{
echo "Data save failed";
}
?>
// HASIL //
Database Exist Data save failed
ga mau jg gan, heran ane ntah dimana lah salah nya atau emg bug dari PHP nya kali ya ?
coba ini gan
<?php
$host = "localhost";
$user = "root";
$pass = "";
$database = "mofu";
$table = "users";
$connect = mysqli_connect($host, $user, $pass);
if($connect)
{
echo "Database Exist";
}
else
{
echo "Database is not exist";
}
$check = mysqli_select_db($connect, $database);
// ========================================================== //
$first_name = $_POST['first_name'];
$last_name = $_POST['last_name'];
$password = $_POST['password'];
$re_password = $_POST['re_password'];
$tanggal = $_POST['tanggal'];
$bulan = $_POST['bulan'];
$tahun = $_POST['tahun'];
$email = $_POST['email'];
if ($_POST['password']==$_POST['re_password'])
{
$_POST['password'];
}
else
{
echo "<script>alert('Password not match')</script>";
}
$input = "INSERT INTO users (first_name,last_name,password,tanggal,bulan,tahun,email) VALUES ('$first_name','$last_name','$password','$tanggal','$bulan','$tahun','$email')";
$result = mysqli_query($check, $input);
if ($result)
{
echo "Data saved";
}
else
{
echo "Data save failed";
}
?>
beda nya dmn gan ? explain please
spy ane bisa lebih mengerti
liat query insert into nya beda.. coba pelajari lagi cara penulisan query
mksd nya gmn gan @Sukmaw ?
Bukan nya sql nya betul ya ? INSERT INTO [nama_table] (1,2,3,4,5,6,~) VALUES ($value1, $value2, $value3,~)
bukan nya begitu ya ?
@stefanuspn
// HASIL // Database exist Warning: mysqli_query() expects parameter 1 to be mysqli, boolean given Data save failed
ga bisa gan, mysqli_query() nya ini yg ane kurang paham
Jawaban Terpilih
gini aja lebih simple
<?php
$host = "localhost";
$user = "root";
$pass = "";
$database = "mofu";
$connect = mysqli_connect($host, $user, $pass,$database);
if($connect)
{
echo "Database Exist";
}
else
{
echo "Database is not exist";
}
// ========================================================== //
$first_name = $_POST['first_name'];
$last_name = $_POST['last_name'];
$password = $_POST['password'];
$re_password = $_POST['re_password'];
$tanggal = $_POST['tanggal'];
$bulan = $_POST['bulan'];
$tahun = $_POST['tahun'];
$email = $_POST['email'];
$input = "INSERT INTO users (first_name,last_name,password,tanggal,bulan,tahun,email) VALUES ('$first_name','$last_name','$password','$tanggal','$bulan','$tahun','$email')";
$result = mysqli_query($connect, $input);
if ($result)
{
echo "Data saved";
}
else
{
echo "Data save failed";
}
?>
untuk password biasakan di md5 atau password_hash biar aman
contoh :
$password = md5($_POST['password']); // penggunaan md5;
// Penggunaan Password_hash;
$pass = $_POST['password'];
$password = password_hash($password, PASSWORD_DEFAULT);
dah bisa gan @stefanuspn
thanks, buat yang udah bantu @sukmaw, @stefanuspn
btw, kalau pakai password_hard itu, muncul nya di database gmn ya ?
contoh kalau pakai md5, kan di enkripsi jadi kaya campuran huruf n angka
nah, kalau password_hard itu gmn gan ? bisa di jelasin ?
php 7.2.5 kok aneh ya ? atau perasaan ane aja ?
$connect = mysqli_connect($host, $user, $pass,$database);
itu value nya 4, biasa nya kan 3,
:O
can someone explain to me ?