I am trying to overwrite the current data in MySQL to be able to update everything.
I am new to this I don't see any errors with the below code:
PHP code:
<?php
// see if the form has been completed
if (isset($_POST['submit'])){
$firstname = $_POST['firstname'];
$surname = $_POST['surname'];
if($firstname && $surname){
// connect to the server
include_once("php_includes/db_conx.php");
// check if that user exist
$exists = mysql_query ("SELECT * FROM users WHERE firstname='$firstname'") or die ("the query could not be connected");
if (mysql_num_rows ($exists) != 0) {
// update the description in the database
mysql_query("UPDATE firstname SET surname='$surname' WHERE firstname='$firstname'") or die ("update could not be applied");
echo "successful";
} else echo "the name does not edist";
} else echo "you need to enter both of the fields try again:";
}
?>
The error I get is
The query could not be connected
but I tried the query and it is fine.
HTML:
<html>
<head>
<title>update MySql form</title>
</head>
<body>
<div id="pageMiddle">
<form action="user1.php" method="POST">
<div>
<p>First Name: <input type="text" name="firstname" id="firstname" ></p>
<p>Surname: <input type="text" name="surname" id="surname"></p>
<p><input type="submit" name="submit" id="submit" value="Update Description"></p>
</div>
</form>
</body>
</html>
Your table is called 'users' but you are running UPDATE on firstname. Change it to:
UPDATE users SET surname...
Then do
$exists = mysql_query ("SELECT * FROM users WHERE firstname='" . $firstname . "'")
in order to isolate whether you have value for that variable. I would recommend printing the query string for testing purposes.
Related
Following is the Code for LOGIN page used with html & php.
The problem I am facing is that , even after submitting correct information Login is failed .
Is there any problem with the query I used?
<html>
<head>
<title>login</title>
<link rel="stylesheet" href="css/insert.css" />
</head>
<body>
<div class="maindiv">
<!--HTML form -->
<div class="form_div">
<form action="<?php echo $_SERVER['PHP_SELF']; ?>" method="post"> <!-- method can be set POST for hiding values in URL-->
<h2>Login Form</h2>
<label>Email:</label><br />
<input class="input" type="email" name="mail" />
<br />
<label>Password:</label><br />
<input class="input" type="text" name="pass" />
<br />
<input class="submit" type="submit" name="submit" value="Login" />
PHP
//Selecting Database from Server
$db = mysql_select_db("tanni", $connection);
if(isset($_POST['submit'])){
//Fetching variables of the form which travels in URL
$mail = $_POST['mail'];
$pass = $_POST['pass'];
if($mail!=''&&$pass!=''){
$query=mysql_query("SELECT* FROM user WHERE mail='".$mail."' and pass='".$pass."'") or die(mysql_error());
$res=mysql_fetch_row($query);
if($res){
$_SESSION['mail']=$mail;
}else {
echo'You entered username or password is incorrect';
}
}else{
echo'Enter both username and password';
}
}
//Closing Connection with Server
mysql_close($connection);
?>
</form>
</div>
<div class="formget"><a href=http://www.formget.com/app><img src="formget.jpg" alt="Online Form Builder"/></a>
</div>
</div>
</body>
</html>
What is the problem in the code?
Need space between select and * at SELECT* FROM
Your query would be
SELECT * FROM user WHERE...
Use mysql_num_rows() to check number of rows return from your query instead mysql_fetch_row
mysql is deprecated instead use mysqli or PDO
You need to start session at the top of your page
session_start();
Don't store plain password into database use password hashing technique
http://php.net/manual/en/function.password-hash.php
http://php.net/manual/en/faq.passwords.php
Your code is open for sql injection read
How can I prevent SQL injection in PHP?
Your whole code would be
<?php
session_start();
//Establishing Connection with Server
$connection = mysql_connect("localhost", "root", "");
//Selecting Database from Server
$db = mysql_select_db("tanni", $connection);
if (isset($_POST['submit'])) {
//Fetching variables of the form which travels in URL
$mail = $_POST['mail'];
$pass = $_POST['pass'];
if ($mail != '' && $pass != '') {
$query = mysql_query("SELECT * FROM user WHERE mail='" . $mail . "' and pass='" . $pass . "'") or die(mysql_error());
$res = mysql_num_rows_row($query);
if ($res == 1) {
$_SESSION['mail'] = $mail;
} else {
echo'You entered username or password is incorrect';
}
} else {
echo'Enter both username and password';
}
}
//Closing Connection with Server
mysql_close($connection);
?>
This seems like it should be simple, but I've tried it every which way and can't seem to get it to work.
I have this login script which I adapted from an online tutorial. What I'd like to do is have users sign in with a username and password, and if these are correct, have it go after their lab results in another table (same database) and display them. I can get it to sign in, but that's it. Here's the login code:
<?php //Start the Session
session_start();
require('connect.php');
//3. If the form is submitted or not.
//3.1 If the form is submitted
if (isset($_POST['username']) and isset($_POST['password'])){
//3.1.1 Assigning posted values to variables.
$username = $_POST['username'];
$password = $_POST['password'];
//3.1.2 Checking the values are existing in the database or not
$query = "SELECT * FROM `user` WHERE username='$username' and password='$password'";
$result = mysql_query($query) or die(mysql_error());
$count = mysql_num_rows($result);
//3.1.2 If the posted values are equal to the database values, then session will be created for the user.
if ($count == 1){
$_SESSION['username'] = $username;
}else{
//3.1.3 If the login credentials doesn't match, he will be shown with an error message.
echo "Invalid Login Credentials.";
}
}
//3.1.4 if the user is logged in Greets the user with message
if (isset($_SESSION['username'])){
$username = $_SESSION['username'];
echo "Hi " . $username . "! ";
echo "This is the results of your inquiry.<br><br>";
/*This is where I'm assuming the new query needs to go.
Query a different table named "data" and pick out information according to
$username which was put in earlier */
echo "<br><a href='logout.php'>Logout</a>";
}else{
//3.2 When the user visits the page first time, simple login form will be displayed.
?>
<!DOCTYPE html>
<html>
<head>
<title>Lab Sign In Page</title>
<link rel="stylesheet" type="text/css" href="style.css">
</head>
<body>
<!-- Form for logging in the users -->
<div class="register-form">
<?php
if(isset($msg) & !empty($msg)){
echo $msg;
}
?>
<h1>Login</h1>
<form action="" method="post">
<p><label>User Name :</label> <input id="username" type="text" name="username" placeholder=
"username"></p>
<p><label>Password :</label> <input id="password" type="password" name="password"
placeholder="password"></p><a class="btn" href="register.php">Signup</a> <input class=
"btn register" type="submit" name="submit" value="Login">
</form>
</div><?php }
?>
</body>
</html>
A "Join" is what I get when I google it but that doesn't seem right. Could someone help?
You can just use a different query for making it easy:
$sql = mysql_query("SELECT * FROM data WHERE user = '" . mysql_real_escape_string($username) . "'");
Then you can process with this.
Please note that you should not use MySQL driver as it is deprecated, use MySQLi(mproved) instead. And you should escape the POSTed values, this is very important!
Your last else statement is trying to echo the HTML.
You should be using mysqli or PDO since mysql is deprecated.
<?php //Start the Session
session_start();
// require('connect.php');
// Establish connection with database
$con=mysqli_connect("localhost","root","","test");
// Check connection
if (mysqli_connect_errno()){ echo "Failed to connect to MySQL: " . mysqli_connect_error(); }
//3. If the form is submitted or not.
//3.1 If the form is submitted
if (isset($_POST['username']) and isset($_POST['password'])){
//3.1.1 Assigning posted values to variables.
$username = $_POST['username'];
$password = $_POST['password'];
//3.1.2 Checking the values are existing in the database or not
$query = "SELECT * FROM `people` WHERE username='$username' and password='$password'";
$result = mysqli_query($con,$query) or die(mysqli_error());
$count = mysqli_num_rows($result);
//3.1.2 If the posted values are equal to the database values, then session will be created for the user.
if ($count == 1){
$_SESSION['username'] = $username;
echo "Valid";
}else{
//3.1.3 If the login credentials doesn't match, he will be shown with an error message.
echo "Invalid Login Credentials.";
}
}
//3.1.4 if the user is logged in Greets the user with message
if (isset($_SESSION['username'])){
$username = $_SESSION['username'];
echo "Hi " . $username . "! ";
echo "This is the results of your inquiry.<br><br>";
echo "Username: $username";
echo "Session Username:".$_SESSION['username'];
// This is where I'm assuming the new query needs to go.
// Query a different table named "data" and pick out information according to $username which was put in earlier
echo "
Logout";
}else{
//3.2 When the user visits the page first time, simple login form will be displayed.
}
?>
<!DOCTYPE html>
<html>
<head>
<title>Lab Sign In Page</title>
<link rel="stylesheet" type="text/css" href="style.css">
</head>
<body>
<!-- Form for logging in the users -->
<div class="register-form">
<?php
if(isset($msg) & !empty($msg)){
echo $msg;
}
?>
<h1>Login</h1>
<form action="test.php" method="post">
<p><label>User Name :</label> <input id="username" type="text" name="username" placeholder=
"username"></p>
<p><label>Password :</label> <input id="password" type="password" name="password"
placeholder="password"></p><a class="btn" href="register.php">Signup</a> <input class=
"btn register" type="submit" name="submit" value="Login">
</form>
</div>
</body>
</html>
I'm trying to create a set of webpages that work together to allow users to view, delete, and edit rows of a MS Access database using PHP.
Membership.php shows a list of the names of members in the Access database. Their names are also hyperlinks that, when clicked, take users to another page EditRecord.php where all of information on the member whose name was clicked on Membership.php is displayed in text boxes with the option to completely delete the record, or just update certain fields.
Membership.php and EditRecord.php are displayed below. The error code is for line 91 of my source for EditRecord.php, but I cut some things out of this post for privacy. Instead, the line has been marked like so:
//--------This is the error line----------
code
[Membership.php]
<!DOCTYPE html>
<html lang="en">
<head>
<meta charset="utf-8">
<link rel="stylesheet" href="Accounts.css">
<style type="text/javascript" src="Validate.js"></style>
<style type="text/javascript" scr="Redirect.js"></style>
<style type="text/javascript" src="Utilities.js"></style>
<title>Member Information Input</title>
</head>
<body>
<div id="content">
<?php
//Establish data connection using external file
require("connection.php");
//Issue SQL SELECT Statement
$sql = "SELECT * FROM Membership";
//Stores any results that match the search term.
$rs = odbc_exec($conn, $sql);
//Set counter for search results to zero
$results = 0;
//Iterates through search results and prints information on records that match
while($row = odbc_fetch_array($rs))
{
$results += 1;
echo '<p>' . $row['FirstName'] . " " . $row['LastName'] . "</p>";
}
?>
</div>
</body>
</html>
[EditRecord.php]
<?php
//Retrieve ID value - if the page is loading for the first time, use $_GET[]. If the
//delete or edit button has been clicked, use $_POST[]
if (isset($_GET['ID'])) {
$userID = $_GET['ID'];
}
else {
$userID=$_POST['ID'];
}
//Establish data connection
require("connection.php");
//If the Delete Button is clicked
if (isset($_POST['DelBtn'])) {
//Issue SQL Statement to Delete Selected Record
$sqlDelete = "DELETE FROM Membership WHERE ID = $userID";
//Execute the SQL Delete Query
$rsDelete = odbc_exec($conn,$sqlDelete);
if(odbc_num_rows($rsDelete) == 1) {
echo "Record successfully deleted!";
}
}
//If the Edit Button is clicked
else if (isset($_POST['EditBtn'])) {
//Collect form field values in scalar variables
$FirstName = $_POST['FirstName'];
$LastName = $_POST['LastName'];
$Address = $_POST['Address'];
$City = $_POST['City'];
$State = $_POST['State'];
$Email = $_POST['Email'];
$Gender = $_POST['Gender'];
$Comments = $_POST['Comments'];
//Issue SQL Statement to Update Selected Record
$sqlUpdate = "UPDATE Membership SET FirstName = '$FirstName', LastName = '$LastName', Address = '$Address', City = '$City', State = '$State'" .
"Email='$Email', Gender = '$Gender', Comments = '$Comments' WHERE ID = $userID";
//Execute the SQL UPDATE Query
$rsEdit = odbc_exec($conn,$sqlUpdate);
if(odbc_num_rows($rsEdit) == 1) {
echo "Record successfully updated!";
}
}
//Issue SQL SELECT Statement to Select Record to Edit or Delete
$sql = "SELECT * FROM Membership WHERE ID = $userID";
//Execute the SQL Query
$rs = odbc_exec($conn, $sql);
odbc_close($conn);
?>
<!DOCTYPE html>
<html lang="en">
<head>
<meta charset="utf-8">
<link rel="stylesheet" href="Accounts.css">
<style type="text/javascript" src="Validate.js"></style>
<style type="text/javascript" src="Utilities.js"></style>
<title>Member Information Input</title>
</head>
<body>
<div id="content">
<form method="post" action="EditMember.php" name="EditForm">
<?php
// Loop through and display the recordset returned by SELECT statement. Display the record values in HTML Text Boxes
**//--------This is the error line----------
while ($row = odbc_fetch_array($rs)) {
?>**
First Name: <input type="text" name="FirstName" value="<?php echo $row['FirstName']?>"><br>
Last Name: <input type="text" name="LastName" value="<?php echo $row['LastName']?>"><br>
Address: <input type="text" name="Address" value="<?php echo $row['Address']?>"><br>
City: <input type="text" name="Telephone" value="<?php echo $row['City']?>"><br>
State: <input type="text" name="Telephone" value="<?php echo $row['State']?>"><br>
Email: <input type="text" name="Email" value="<?php echo $row['Email']?>"><br>
Gender: <input type="text" name="Telephone" value="<?php echo $row['Gender']?>"><br>
Comments: <input type="text" name="Comments" value="<?php echo $row['Comments']?>"><br><br>
<input type="hidden" name="ID" value="<?php echo $row['ID']?>" >
<?php
}
?>
<input type="submit" name="EditBtn" value="Edit Record"> <input type="submit" name="DelBtn" value="Delete Record">
</form>
</div>
<div id="footer">
<?php require("Footer.php"); ?>
</div>
</body>
</html>
I also find this strange, because there are five records in my database, not four. Is that because it starts counting at zero?
Any insight or advice would be greatly appreciated.
Your problem is that you are calling odbc_close() and closing the connection before your loop calls odbc_fetch_array(). You need to leave the connection open until after you've fetched all of the rows.
Also, the "4" in the error message does not refer to a number of rows or anything like that; it's just the numeric representation of result identifier for the resource created by the odbc_exec() call.
Overview:
I am crating a dummy website for learning purposes therefore its functionalists are basic and security in not on the agenda atm.
Actual Problem:
OK so my application loges in a users who has an option of editing his/hers account if desired. So i have gone ahead and created PHP script that soopose to deal with this BUT IT DOES NOT. When I click edit account button no errors pop up but at the same time when i check MySQL database no changes occurred.
EditAccountForm.php file:
<?php
include('connect_mysql.php');
if(isset($_POST['editAccount'])){
$Newusername = $_GET['username'];
$Newpassword = $_POST['password'];
$Newfirstname = $_POST['first_name'];
$Newlastname = $_POST['last_name'];
$Newemail = $_POST['email'];
if($Newusername != $username)
{
$q1 = ("UPDATE users SET username=$Newusername WHERE username=$username");
}
else if(!mysql_query($q1)){
echo "MySQL ERROR: " . mysql_error() . "" . $sql;
}
///////////////////////////////////////////////////////////////
if($Newpassword != $password)
{
$q2 = ("UPDATE users SET password=$Newpassword WHERE password=$password");
}
else if(!mysql_query($q2)){
echo "MySQL ERROR: " . mysql_error() . "" . $sq2;
}
///////////////////////////////////////////////////////////
if($Newfirstname != $firstname)
{
$q3 = ("UPDATE users SET first_name=$Newfirstname WHERE first_name=$firstname");
}
else if(!mysql_query($q3)){
echo "MySQL ERROR: " . mysql_error() . "" . $sq3;
}
///////////////////////////////////////////////////////////////
if($Newlastname != $lastname)
{
$q4 = ("UPDATE users SET last_name=$Newlastname WHERE last_name=$lastname");
}
else if(!mysql_query($q4)){
echo "MySQL ERROR: " . mysql_error() . "" . $sq4;
}
///////////////////////////////////////////////////////////////
if($Newemail != $email)
{
$q5 = ("UPDATE users SET username=$Newemail WHERE email=$email");
}
else if(!mysql_query($q5)){
echo "MySQL ERROR: " . mysql_error() . "" . $sq5;
}
}
?>
userEditAccount.php:
<html>
<head>
<title>Edit Account</title>
<meta http-equiv="content-type" content="text/html; charset=iso-8859-1" />
<link href="style.css" rel="stylesheet" type="text/css" />
</head>
<body>
<div id="wrapper">
<header><h1>E-Shop</h1></header>
<article>
<h1>Welcome</h1>
<h1>Edit Account</h1>
<div id="login">
<ul id="login">
<form method="post" name="editAccount" action="userEditAccount.php" >
<fieldset>
<legend>Fill in the form</legend>
<label>Select Username : <input type="text" name="username" /></label>
<label>Password : <input type="password" name="password" /></label>
<label>Enter First Name : <input type="text" name="first_name" /></label>
<label>Enter Last Name : <input type="text" name="last_name" /></label>
<label>Enter E-mail Address: <input type="text" name="email" /></label>
</fieldset>
<br />
<input name="Editsubmited" type="submit" submit="submit" value="Edit Account" class="button">
</form>
<?
echo $newrecord;
?>
</div>
<form action="userhome.php" method="post">
<div id="login">
<ul id="login">
<li>
<input type="submit" value="back" onclick="index.php" class="button">
</li>
</ul>
</div>
</article>
<aside>
</aside>
<div id="footer">This is my site i Made coppyrights 2013 Tomazi</div>
</div>
</body>
</html>
Furthermore:
I tried to fiddle with the code looked on web but no luck the code i have written for this script in my eyes is the best solution and the one that makes sens to me.
So i had no other option but turn to this website to look for answers, can anyone perhaps see where am going wrong with this whole thing...?
Image of the Edit Account page:
As Asked Conect_mysql.php:
<?php
$db_hoast = "127.0.0.1";
$db_username = "root";
$db_password = "";
$db_name = "eshop";
$con = mysql_connect("$db_hoast","$db_username","$db_password");
if(!$con)
{
die("Could not connect to DATABASE");
}
$db = mysql_select_db("$db_name");
if(!$db)
{
die("No database");
}
?>
the problem with your UPDATE statements are the values are not wrapped with single quotes. They are string literal and should be wrapped.
$q1 = "UPDATE users SET username='$Newusername' WHERE username='$username'";
in order to display the error,
if($Newfirstname != $firstname)
{
$q1 = "UPDATE users SET username='$Newusername' WHERE username='$username'";
$result = mysql_query($q1);
if (!$result)
{
die('Invalid query: ' . mysql_error());
}
}
Also your logical UPDATES are wrong. This causes you to updates records that matches with the conditions.
As a sidenote, the query is vulnerable with SQL Injection if the value(s) of the variables came from the outside. Please take a look at the article below to learn how to prevent from it. By using PreparedStatements you can get rid of using single quotes around values.
How to prevent SQL injection in PHP?
I am experimenting with PHP and Mysql. I have created a database and table at mu localhost using xampp. I have also created a file that suppose to populate my table by executing a query, but the strange thing is that i get no errors but at the same time no DATA has been inserted into my DataBase:
CODE:
register.php:
<?php
session_start();
if(isset($_POST['submitted'])){
include('connectDB.php');
$UserN = $_POST['username'];
$Upass = $_POST['password'];
$Ufn = $_POST['first_name'];
$Uln = $_POST['last_name'];
$Uemail = $_POST['email'];
$NewAccountQuery = "INSERT INTO users (user_id,username, password, first_name, last_name, emial) VALUES ('$UserN','$Upass', '$Ufn', '$Uln', '$Uemail')";
if(!mysql_query($NewAccountQuery)){
die(mysql_error());
}//end of nested if statment
$newrecord = "1 record added to the database";
}//end of if statment
?>
<html>
<head>
<title>Home Page</title>
<meta http-equiv="content-type" content="text/html; charset=iso-8859-1" />
<link href="style.css" rel="stylesheet" type="text/css" />
</head>
<body>
<div id="wrapper">
<header><h1>E-Shop</h1></header>
<article>
<h1>Welcome</h1>
<h1>Create Account</h1>
<div id="login">
<ul id="login">
<form method="post" action="register.php" >
<fieldset>
<legend>Fill in the form</legend>
<label>Select Username : <input type="text" name="username" /></label>
<label>Password : <input type="password" name="password" /></label>
<label>Enter First Name : <input type="text" name="first_name" /></label>
<label>Enter Last Name : <input type="text" name="last_name" /></label>
<label>Enter E-mail Address: <input type="text" name="email" /></label>
</fieldset>
<br />
<input type="submit" submit="submit" value="Create Account" class="button">
</form>
</div>
<form action="index.php" method="post">
<div id="login">
<ul id="login">
<li>
<input type="submit" value="Cancel" onclick="index.php" class="button">
</li>
</ul>
</div>
</article>
<aside>
</aside>
<div id="footer">This is my site i Made coppyrights 2013 Tomazi</div>
</div>
</body>
</html>
I have also one include file which is connectDB:
<?php
session_start();
$con = mysql_connect("127.0.0.1", "root", "");
if(!$con)
die('Could not connect: ' . mysql_error());
mysql_select_db("eshop", $con) or die("Cannot select DB");
?>
Database structure:
database Name: eshop;
only one table in DB : users;
users table consists of:
user_id: A_I , PK
username
password
first_name
last_name
email
I spend a substantial amount of time to work this out did research and looked at some tutorials but with no luck
Can anyone spot what is the root of my problem...?
It is because if(isset($_POST['submitted'])){
you dont have input field with name submitted give the submit button name to submitted
<input name="submitted" type="submit" submit="submit" value="Create Account" class="button">
Check your insert query you have more fields than your values
Change :
$NewAccountQuery = "INSERT INTO users (user_id,username, password, first_name, last_name, email) VALUES ('$UserN','$Upass', '$Ufn', '$Uln', '$Uemail')";
to :
$NewAccountQuery = "INSERT INTO users (user_id,username, password, first_name, last_name, email) VALUES ('','$UserN','$Upass', '$Ufn', '$Uln', '$Uemail')";
Considering user_id is auto increment field.
Your email in query is written wrongly as emial.
Is error reporting turned on?
Put this on the top of your screen:
error_reporting(E_ALL);
ini_set('display_errors', '1');
Some good answers above, but I would also suggest you make use of newer MySQLi / PDO instead of outdated 2002 MySQL API.
Some examples: (i will use mysqli since you wrote your original example in procedural code)
connectDB.php
<?php
$db = mysqli_connect('host', 'user', 'password', 'database');
if (mysqli_connect_errno())
die(mysqli_connect_error());
?>
register.php -- i'll just write out an example php part and let you do the rest
<?php
//i'll always check if session was already started, if it was then
//there is no need to start it again
if (!isset($_SESSION)) {
session_start();
}
//no need to include again if it was already included before
include_once('connectDB.php');
//get all posted values
$username = $_POST['username'];
$userpass = $_POST['password'];
$usermail = $_POST['usermail'];
//and some more
//run checks here for if fields are empty etc?
//example check if username was empty
if($username == NULL) {
echo 'No username entered, try again';
mysqli_close($db);
exit();
} else {
//if username field is filled we will insert values into $db
//build query
$sql_query_string = "INSERT INTO _tablename_(username,userpassword,useremail) VALUES('$username','$userpass','$usermail')";
if(mysqli_query($db,$sql_query_string)) {
echo 'Record was entered into DB successfully';
mysqli_close($db);
} else {
echo 'Ooops - something went wrong.';
mysqli_close($db);
}
}
?>
this should work quite nicely and all you need to add is your proper posted values and build the form to post it, that's all.
<?php
$db = mysqli_connect('host', 'user', 'password', 'database');
if (mysqli_connect_errno())
die(mysqli_connect_error());
?>
register.php -- i'll just write out an example php part and let you do the rest
<?php
//i'll always check if session was already started, if it was then
//there is no need to start it again
if (!isset($_SESSION)) {
session_start();
}
//no need to include again if it was already included before
include_once('connectDB.php');
//get all posted values
$username = $_POST['username'];
$userpass = $_POST['password'];
$usermail = $_POST['usermail'];
//and some more
//run checks here for if fields are empty etc?
//example check if username was empty
if($username == NULL) {
echo 'No username entered, try again';
mysqli_close($db);
exit();
} else {
//if username field is filled we will insert values into $db
//build query
$sql_query_string = "INSERT INTO _tablename_(username,userpassword,useremail) VALUES('$username','$userpass','$usermail')";
if(mysqli_query($db,$sql_query_string)) {
echo 'Record was entered into DB successfully';
mysqli_close($db);`enter code here`
} else {
echo 'Ooops - something went wrong.';
mysqli_close($db);
}
}
?>