PHP echo all data from database based on input - php

Question

I want to find out how to output data from database based on a single key,for example my database column are :
kodeDosen(PrimaryKey),namaDosen,email,telepon,password
and my login screen the user can only input kodeDosen and password,and i want to show the other data exept password,this is my register php:
<?php
include 'connectdb.php';
$data = json_decode(file_get_contents('php://input'), true);
$kodeDosen =$data["kodeDosen"];
$namaDosen = $data["namaDosen"];
$email = $data["email"];
$telepon = $data["telepon"];
$password= $data["password"];
$message = array("message"=>"Success");
$failure = array("message"=>"Failure,kodeDosen already used");
$sql = "INSERT INTO tbl_dosen (kodeDosen, namaDosen, email, telepon, password) VALUES ('$kodeDosen', '$namaDosen', '$email', '$telepon','$password')";
if (mysqli_query($conn, $sql)) {
echo json_encode($message);
} else {
echo json_encode($failure) ;
}
?>
and this is my login php:
<?php
include 'connectdb.php';
$data = json_decode(file_get_contents('php://input'), true);
$kodeDosen =$data["kodeDosen"];
$password = $data["password"];
$message = array("message"=>"Data found");
$failure = array("mesage"=>"Data not found");
if ($stmt = mysqli_prepare($conn, "SELECT kodeDosen, namaDosen, email, telepon FROM tbl_dosen WHERE kodeDosen =? and password = ?")) {
/* bind parameters for markers */
mysqli_stmt_bind_param($stmt, "ss", $kodeDosen,$password);
/* execute query */
mysqli_stmt_execute($stmt);
/* store result */
mysqli_stmt_store_result($stmt);
if(mysqli_stmt_num_rows($stmt) > 0) {
echo json_encode($row);
}else {
echo json_encode($failure);
}
}
?>

Answer 1

It's not a good idea to insert a variable directly into an SQL query because of SQL injection.
I would suggest to use prepared statements on both of the queries. To pull the result from the db with prepared statements it's something like:
OOP style:
$stmt = $db->prepare("SELECT kodeDosen, namaDosen, email, telepon FROM tbl_dosen WHERE kodeDosen = ? and password = ?");
$stmt->bind_param('ss', $kodeDosen, $password);
$stmt->execute();
$result = $stmt->get_result();
while ($row = $result->fetch_assoc()) {
//result is in row
var_dump($row);
}
Procedural style:
$stmt = mysqli_prepare($conn, "SELECT kodeDosen, namaDosen, email, telepon FROM tbl_dosen WHERE kodeDosen = ? and password = ?");
mysqli_stmt_bind_param($stmt, 'ss', $kodeDosen, $password);
mysqli_stmt_execute($stmt);
$result = mysqli_stmt_get_result($stmt);
while ($row = $result->fetch_assoc()) {
//result is in row
var_dump($row);
}

Answer 2

You can change in sql SELECT statement in login.php
$sql = "SELECT kodeDosen, namaDosen, email, telepon FROM tbl_dosen WHERE kodeDosen ='$kodeDosen' and password = '$password'";
in SELECT * means return all columns.

Answer 3

I think you want echo json_encode($row); rather than echo json_encode($message);
Try:
<?php
include 'connectdb.php';
$data = json_decode(file_get_contents('php://input'), true);
$kodeDosen =$data["kodeDosen"];
$password = $data["password"];
$message = array("message"=>"Data found");
$failure = array("mesage"=>"Data not found");
if ($stmt = mysqli_prepare($conn, "SELECT kodeDosen, namaDosen, email, telepon FROM tbl_dosen WHERE kodeDosen =? and password = ?")) {
/* bind parameters for markers */
mysqli_stmt_bind_param($stmt, "ss", $kodeDosen,$password);
/* execute query */
mysqli_stmt_execute($stmt);
/* store result */
$result = mysqli_stmt_get_result($stmt);
$row = mysqli_fetch_assoc( $result );
if(mysqli_num_rows($result) > 0) {
echo json_encode($row);
}else {
echo json_encode($failure);
}
}
?>