php check and validate form input with mysql database [duplicate] - php

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PHP's white screen of death [duplicate]
What to do with mysqli problems? Errors like mysqli_fetch_array(): Argument #1 must be of type mysqli_result and such
(1 answer)
Closed 3 years ago.
Basically I have this form that asks for email and password.
What I want to do is to compare and check if the inputs match with the data from my table/database.
This is my registration.php (the form)
<form action="Authentication.php" method="post">
<b>Returning Intern Login</b><br/><br/>
Enter your e-mail address: <input type="text" name="email" /><br/><br/>
Enter your password: <input type="password" name="pw2"/><br/><i>(Passwords are case-sensitive and must be 6 characters long)</i><br/><br/>
<input type="reset" value="Reset Login Form" />
<input type="submit" name="submit2" value="Log In" /><hr/><br/>
</form>
and this is the Authentication.php
session_start();
$link = mysqli_connect('localhost','root','');
$database = mysqli_select_db($link,'internship');
$user = $_POST['email'];
$pass = $_POST['pw2'];
// User is logging in
if (isset($_POST["submit2"]))
{
if (empty ($user)) //if username field is empty echo below statement
{
echo "<font color='red'>***You must enter your unique username (email).***</font><br/><br/>";
}
if (empty ($pass)) //if password field is empty echo below statement
{
echo "<font color='red'>***You must enter your unique password.***</font><br/><br/>";
}
}
else
{
$query = "SELECT * FROM Interns WHERE email = '". mysqli_real_escape_string($link,$user) ."' AND password = '". mysqli_real_escape_string($link,$pass) ."'" ;
$result = mysqli_query($link,$query);
if (mysqli_num_rows($result) == 1)
{
echo "pass"; //Pass, do something
}
else
{
echo "fail"; //Fail
}
}
session_write_close();
It works with the empty inputs.
But when I gave an email and password exactly same from the database/table,
It displays white blank page..

Answer 1

You need to write the entire code within an if statement to ensure the field is filled in, like so:
if (isset($_POST["submit2"]))
{
if (empty ($user)) //if username field is empty echo below statement
{
/* Code */
}
if (empty ($pass)) //if password field is empty echo below statement
{
/* Code */
}
$query = "SELECT * FROM Interns WHERE email = '". mysqli_real_escape_string($link,$user) ."' AND password = '". mysqli_real_escape_string($link,$pass) ."'" ;
$result = mysqli_query($link,$query);
if (mysqli_num_rows($result) == 1)
{
echo "pass"; //Pass, do something
}
else
{
echo "fail"; //Fail
}
}
else
{
echo "Empty input submit2"; // empty $_POST["submit2"]
}
Hope this helps.

Answer 2

Mysqli takes 4 parameters hostname,username,password, and dbname:
<?php
session_start();
$link = mysqli_connect('localhost','root','','internship');
// User is logging in
if (isset($_POST["submit2"]))
{
$user = $_POST['email'];
$pass = $_POST['pw2'];
if (empty($user)) //if username field is empty echo below statement
{
echo "<font color='red'>***You must enter your unique username (email).***</font><br/><br/>";
}
else if (empty ($pass)) //if password field is empty echo below statement
{
echo "<font color='red'>***You must enter your unique password.***</font><br/><br/>";
}
else
{
$query = "SELECT * FROM Interns WHERE email = '". $user ."' AND password = '".$pass."'" ;
$result = mysqli_query($link,$query);
if (mysqli_num_rows($result) == 1)
{
echo "pass"; //Pass, do something
}
else
{
echo "fail"; //Fail
}
}
session_write_close();
?>