How do I connect phpmyadmin database to my website using PHP

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I have created a database in phpmyadmin but when i am making a connection to that database through php, i am getting a connection error. I have rechecked the code and i don't think there is any problem with the code in php.

What I have tried:

<?php 
define('DB_HOST', 'localhost'); 
define('DB_NAME', 'alabdouli_users'); 
define('DB_USER','safeer123'); 
define('DB_PASSWORD','******'); 
$con=mysqli_connect(DB_HOST,DB_USER,DB_PASSWORD) or die("Failed to connect to MySQL: " . mysqli_error()); 
$db=mysqli_select_db(DB_NAME,$con) or die("Failed to connect to database: " . mysqli_error()); 
function NewUser() 
{ 
$fullname = $_POST['name']; 
$userName = $_POST['user']; 
$email = $_POST['email'];
$password = $_POST['pass']; 
$query = "INSERT INTO websiteusers (fullname,userName,email,pass) VALUES ('$fullname','$userName','$email','$password')"; 
$data = mysqli_query ($query)or die(mysqli_error()); 
if($data) 
{ 
echo "YOUR REGISTRATION IS COMPLETED..."; 
} } 
function SignUp() 
{ 
if(!empty($_POST['user'])) //checking the 'user' name which is from Sign-Up.html, is it empty or have some text 
{ 
$query = mysqli_query("SELECT * FROM websiteusers WHERE userName = '$_POST[user]' AND pass = '$_POST[pass]'") or die(mysqli_error()); 
if(!$row = mysqil_fetch_array($query) or die(mysqli_error())) { newuser(); 
} 
else 
{ 
echo "SORRY...YOU ARE ALREADY REGISTERED USER..."; 
} } } 
if(isset($_POST['submit'])) 
{ 
SignUp(); 
} 
?>

Posted 24-Jan-18 2:23am

lookwhosback

Updated 24-Jan-18 2:34am

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Comments

[no name] 24-Jan-18 8:24am

Can you please post what is the error you are getting?

lookwhosback 24-Jan-18 8:27am

failed to connect to the database

Jochen Arndt 24-Jan-18 8:30am

Where does it fail and what is the error message?

[no name] 24-Jan-18 8:36am

lookwhosback,Please don't post issues/problems as solution

[no name] 24-Jan-18 8:37am

Can you please provide the stack trace.

2 solutions

Solution 2

<div id="Sign-Up"> 
<fieldset style="width:30%">
<legend>Registration Form</legend> 
<table border="0"> 
<tr> 
<form method="POST" action="signup.php"> 
<td>Name</td>
<td> <input type="text" name="name"></td> 
</tr> 
<tr> <td>Email</td>
<td> <input type="text" name="email"></td> 
</tr> 
<tr> 
<td>UserName</td>
<td> <input type="text" name="user"></td> 
</tr> 
<tr> <td>Password</td>
<td> <input type="password" name="pass"></td> 
</tr> 
<tr> 
<td>Confirm Password </td>
<td><input type="password" name="cpass"></td> 
</tr> 
<tr> 
<td><input id="button" type="submit" name="submit" value="Sign-Up"></td> 
</tr> 
</form> 
</table> 
</fieldset> 
</div>

this is my html code in which when i press the signup button and it should execute the signup.php, but i recieve an error from my php code saying "Failed to connect to the database"

Posted 24-Jan-18 2:34am

lookwhosback

Comments

[no name] 24-Jan-18 8:36am

Please don't post your problems as solution.

Add a Solution

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Jochen Arndt · Accepted Answer · 2018-01-24T02:33:00

Solution 1

See PHP: mysqli::select_db - Manual[^]. The first argument is the connection and the second is the name but you are passing them reversed.

It must be

PHP

$db=mysqli_select_db($con,DB_NAME) or die("Failed to connect to database: " . mysqli_error());

[EDIT]
Because you are using the procedural style, you have to use it too for the error function. Note also that mysqli_error() can't be used when connecting fails. You have to use mysqli_connect_error() instead:

PHP

$con=mysqli_connect(DB_HOST,DB_USER,DB_PASSWORD) or die("Failed to connect to MySQL: " . mysqli_connect_error()); 
$db=mysqli_select_db($con,DB_NAME) or die("Failed to connect to database: " . mysqli_error($con));

[/EDIT]

[EDIT2]
And similar for all other SQL operations where you did not pass the connection/link $con.
[/EDIT2]

Posted 24-Jan-18 2:33am

Jochen Arndt

Updated 24-Jan-18 2:52am

v4

Comments

lookwhosback 24-Jan-18 8:44am

Thankyou so much for that, this feels so good first time someone solved my problem on the internet :D i really didn't know that was the problem, that fixed the database connectivity but now i don't know why i am not getting any message for adding the user to the database.

Jochen Arndt 24-Jan-18 8:49am

Thank you for the feedback and accepting my solution.

See my updated answer. It explains why you did not get the PHP error messages and also why the queries did not work: Same reason; you have to use the procedural style passing the link/connection.

lookwhosback 24-Jan-18 8:55am

Sir Jochen, I got the previous problem solved, that is linking the database with the php file with your help. Still the queries are not executing and the user is not being added to the database

Jochen Arndt 24-Jan-18 9:03am

See update 2.
You have to use the procedural style everywhere because you decided initially to use that style.
So it must be
$data = mysqli_query($con, $query) or die(mysqli_error($con));
and similar ($con as first/only parameter) for all mysqli function calls.

If then a call fails you will get at least a meaningful error message that helps finding the problem like passing wrong parameters or typos in table or field names.

lookwhosback 24-Jan-18 9:26am

I have replaced as you said, still the record is not being added and i am not receiving any message. this is how my code looks like after changing it to the procedural style.

<?php 
define('DB_HOST', 'localhost'); 
define('DB_NAME', 'alabdouli_users'); 
define('DB_USER','safeer123'); 
define('DB_PASSWORD','3333333'); 
$con=mysqli_connect(DB_HOST,DB_USER,DB_PASSWORD) or die("Failed to connect to MySQL: " . mysqli_connect_error()); 
$db=mysqli_select_db($con, DB_NAME) or die("Failed to connect to database: " . mysqli_error($con)); 
function NewUser() 
{ 
$fullname = $_POST['name']; 
$userName = $_POST['user']; 
$email = $_POST['email'];
$password = $_POST['pass']; 
$query = "INSERT INTO websiteusers (fullname,userName,email,pass) VALUES ('$fullname','$userName','$email','$password')"; 
$data = mysqli_query ($con, $query)or die(mysqli_error($con)); 
if($data) 
{ 
echo "YOUR REGISTRATION IS COMPLETED..."; 
} } 
function SignUp() 
{ 
if(!empty($_POST['user'])) //checking the 'user' name which is from Sign-Up.html, is it empty or have some text 
{ 
$query = mysqli_query($con,"SELECT * FROM websiteusers WHERE userName = '$_POST[user]' AND pass = '$_POST[pass]'") or die(mysqli_error($con)); 
if(!$row = mysqil_fetch_array($con, $query) or die(mysqli_error($con))) 
{ 
NewUser(); 
} 
else 
{ 
echo "SORRY...YOU ARE ALREADY REGISTERED USER..."; 
} } } 
if(isset($_POST['submit'])) 
{ 
SignUp(); 
} 
?>

Jochen Arndt 24-Jan-18 9:43am

See http://php.net/manual/en/language.variables.scope.php

You have functions that did not "see" the global $con variable but use a local (empty) instance.

lookwhosback 24-Jan-18 10:00am

i didn't get you, can you please elaborate?

Jochen Arndt 24-Jan-18 10:25am

The $con variable used to connect to the database is not visible inside the functions. That use a another (local) variable with the same name but different content (it is an empty variable). You have to declare it as global as described in the link or passed as parameter to the function.

lookwhosback 24-Jan-18 10:46am

i know im asking for too much, but can you please give me an example from my code please?

<? php
function SignUp($con) 
{ 
if(!empty($_POST['user'])) //checking the 'user' name which is from Sign-Up.html, is it empty or have some text 
{ 
$query = mysqli_query("SELECT * FROM websiteusers WHERE userName = '$_POST[user]' AND pass = '$_POST[pass]'") or die(mysqli_error($con)); 
if(!$row = mysqil_fetch_array($con, $query) or die(mysqli_error($con))) 
{ 
NewUser($con); 
} 
else 
{ 
echo "SORRY...YOU ARE ALREADY REGISTERED USER..."; 
} } } 
if(isset($_POST['submit'])) 
{ 
SignUp(); 
} 
?>

Jochen Arndt 24-Jan-18 10:58am

Check it step by step. First put the code below your setup code outside any function. If that works, put it inside a function.
This line is suspicious:
if(!$row = mysqil_fetch_array($con, $query) or die(mysqli_error($con)))
That is an if and an or. Which is checked first? Is the if block ever executed?
I don't know the answer because I would never use such a construct.
Use an if else instead (success - error).
There is also a typo: mysqil.