PHP insert data to Mysql

Question

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...?

You are using an obsolete database API and should use a modern replacement. You are also vulnerable to SQL injection attacks that a modern API would make it easier to defend yourself from. — Quentin
– Quentin, Commented Feb 27, 2013 at 13:05
Are you sure there are no errors? Your INSERT statement specifies more columns than it has values. — David
– David, Commented Feb 27, 2013 at 13:06
are you entering same value for new entry? Try different value because user_id is PK — عثمان غني
– عثمان غني, Commented Feb 27, 2013 at 13:09
@Tomazi — So. Step 1. Build it. Step 2. Throw it away. Step 3 . Build it again with the right tools? I strongly suggest skipping steps 1 and 2. — Quentin
– Quentin, Commented Feb 27, 2013 at 13:10

Prasanth Bendra · Accepted Answer · 2013-02-27 13:16:00Z

2

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.

edited Feb 27, 2013 at 13:16

answered Feb 27, 2013 at 13:06

Prasanth Bendra

33.1k9 gold badges56 silver badges74 bronze badges

Sign up to request clarification or add additional context in comments.

4 Comments

Prasanth Bendra Over a year ago

try echoing something inside if condition

Tomazi Over a year ago

OK u just sorted me out many thx. after amending the changes you suggested i also had an error: "Ok you got me going, but now i get an error which is "Notice: A session had already been started - ignoring session_start() in C:\xampp\htdocs\eshop\connectDB.php on line 4"" so i also hadd to remove the session_start() fromy my register.php file

Prasanth Bendra Over a year ago

There are two session_start() remove one, Also dont forget to accept the answer :P

Tomazi Over a year ago

yes i did that once again thx :) of course I will you earned it. you can like my question in return if u wish :P

HarryFink · Accepted Answer · 2013-02-27 13:09:57Z

0

Is error reporting turned on? Put this on the top of your screen:

error_reporting(E_ALL);
ini_set('display_errors', '1');

answered Feb 27, 2013 at 13:09

HarryFink

1,0001 gold badge6 silver badges6 bronze badges

6 Comments

عثمان غني Over a year ago

this is not answer. this can be told in comments also

HarryFink Over a year ago

I have no add comment button under the question.

عثمان غني Over a year ago

there is "add/show more comments" click it you will get comment box

HarryFink Over a year ago

I only have 'add comment' button under my answer.

عثمان غني Over a year ago

It is written under users question!

|

megachill · Accepted Answer · 2013-02-27 13:55:53Z

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($username == NULL) {
  echo 'No username entered, try <a href="register.php">again</a>';
  mysqli_close($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.

charanjit singh · Accepted Answer · 2017-10-13 15:08:46Z

0

<?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($username == NULL) {
  echo 'No username entered, try <a href="register.php">again</a>';
  mysqli_close($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);
  }
}
?>

answered Oct 13, 2017 at 15:08

charanjit singh

11 bronze badge

1 Comment

Ethan Field Over a year ago

Welcome to SO! It might be a good idea to explain the ideas you have used here in some sort of summary or conclusion rather than comments, so you can expand on them in a more user friendly fashion.

Collectives™ on Stack Overflow

PHP insert data to Mysql

4 Answers 4

4 Comments

6 Comments

Comments

1 Comment

Your Answer

Linked

Hot Network Questions

Collectives™ on Stack Overflow

4 Answers 4

4 Comments

6 Comments

Comments

1 Comment

Your Answer

Sign up or log in

Post as a guest

Linked

Related