Uploading data from form to MySQL Database

Question

I am trying to upload data from a form to a MySQL Database and it is not working.

It just stays there and does nothing, and it does not insert into the database.

Please review my code below.

<?php
        if(isset($_POST['submit'])) {
        $first_name = $_POST['first_name'];
        $last_name = $_POST['last_name'];
        $username = $_POST['username'];
        $password = md5($_POST['password']);

        $query  = "INSERT INTO users (";
            $query .= "  first_name, last_name, username, password";
            $query .= ") VALUES (";
            $query .= " '{$first_name}', '{$last_name}', '{$username}', '{$password}'";
            $query .= ")";
            $result = mysqli_query($connection, $query);


        if($result) {
            echo "It worked";
        } else {
            echo "It did not work";
        }
        }   


?>

This is my html

           <div class="col-md-8 col-xs-12">
                    <div class="white-box">
                        <form class="form-horizontal form-material" method="post">
                            <div class="form-group">
                                <label class="col-md-12">First Name</label>
                                <div class="col-md-12">
                                    <input type="text" placeholder="John" name="first_name" class="form-control form-control-line"> </div>
                            </div>
                            <div class="form-group">
                                <label class="col-md-12">Last Name</label>
                                <div class="col-md-12">
                                    <input type="text" placeholder="Doe" name="last_name" class="form-control form-control-line"> </div>
                            </div>

                            <div class="form-group">
                                <label class="col-md-12">Username</label>
                                <div class="col-md-12">
                                    <input type="text" placeholder="Username" name="username" class="form-control form-control-line"> </div>
                            </div>
                            <div class="form-group">
                                <label class="col-md-12">Password</label>
                                <div class="col-md-12">
                                    <input type="password" placeholder="Password" name="password" class="form-control form-control-line"> </div>
                            </div>
                            <div class="form-group">
                                <div class="col-sm-12">
                                    <button class="btn btn-success">Create User</button>
                                </div>
                            </div>
                        </form>
                    </div>
                </div>
            </div>
        </div>

what do you mean by "it is not working" ? does it make an error? does it insert new row but no values? , please be more descriptive about the problem — Accountant م, Jul 20 '17 at 20:34
It does nothing it just stays the same and it does not insert a new row. — , Jul 20 '17 at 20:35
Possible duplicate of [Inserting data to table (mysqli insert)](https://stackoverflow.com/questions/16835753/inserting-data-to-table-mysqli-insert) — Muhammad Usman, Jul 20 '17 at 20:36
Which message is echoed ? "It worked" or "It did not work" ? — Accountant م, Jul 20 '17 at 20:36
@ShiylohBenFisher Then your PHP script is probably not executed at all, make `die('script is executed');` at the first line in your file and see what happens — Accountant م, Jul 20 '17 at 20:39
Array ( [first_name] => Shiyloh [last_name] => mk; [username] => lm,;,; [password] => ;lm;l,;l ) — , Jul 20 '17 at 20:44
you are checking $_POST['submit'] which is not set. since the condition failed, the code in if condition is not executed. Can you update the question with html form code — Ravinder Reddy, Jul 20 '17 at 20:47

score 1 · Accepted Answer · answered Jul 20 '17 at 20:57

1

Since you are checking isset($_POST["submit"]), you need to add the input tag with type submit in your html form.

Replace

<button class="btn btn-success">Create User</button>

with

<input class="btn btn-success" type="submit" name="submit" value="Create User">

answered Jul 20 '17 at 20:57

Ravinder Reddy

3,869
1
13
22

It's working. Array ( [first_name] => sdffsdf [last_name] => fdfdf [username] => dfdfdf [password] => dfdf [submit] => Create User ) It worked – Jul 20 '17 at 21:04
I can not upvote unless i get an upvote where i can upvote. – Jul 20 '17 at 21:06
If you give @ShiylohBenFisher one more upvote he will be able to upvote your answer – Accountant م Jul 20 '17 at 21:14

score 0 · Answer 2 · answered Jul 20 '17 at 20:52

0

<?php
if (isset($_POST['submit'])) {
 $first_name = $_POST['first_name'];
        $last_name = $_POST['last_name'];
        $username = $_POST['username'];
        $password = ($_POST['password']);



$query=mysqli_query($conn,"INSERT INTO users(first_name, last_name, username, password) VALUES ('$first_name','$last_name','$username','$password')");
   if(!$query){
        die('could not enter data:'. mysqli_error($conn));
    }
    else{
        echo "Entered data successfully";
    }


}
?>

answered Jul 20 '17 at 20:52

sharjeel habib

1
2

No it does not work and it still does not show a message. – Jul 20 '17 at 20:57
@ShiylohBenFisher dont worry i will just send you whole code after correcting it – sharjeel habib Jul 20 '17 at 21:41

score 0 · Answer 3 · answered Jul 20 '17 at 20:58

<?php
        if(isset($_POST['submit'])) {
            $first_name = $_POST['first_name'];
            $last_name = $_POST['last_name'];
            $username = $_POST['username'];
            $password = md5($_POST['password']);

            $query  = "INSERT INTO users (first_name, last_name, username, password) VALUES ('$first_name','$last_name','$username','$password')";
            $result = mysqli_query($connection, $query);


            if($result) {
                echo "It worked";
            } else {
                echo "It did not work";
            }
        }   


?>

Did you replace"" with and also
– Mike Jul 20 '17 at 21:09 — Mike, Jul 20 '17 at 21:09

score 0 · Answer 4 · answered Jul 20 '17 at 22:02

<!DOCTYPE html>
<html>
<body>

<form  method="post" action="">

                                First Name

                                    <input type="text"  name="first_name" > 

                                Last Name

                                    <input type="text"  name="last_name" class="form-control form-control-line"> 

                                Username

                                    <input type="text" placeholder="Username" name="username" > 

                                Password

                                    <input type="password" placeholder="Password" name="password" > 
                         <input type="submit" name="update" value="create user">

                        </form>
</body>
</html>

<?php
    $host = "localhost";
$username = "root";
$password = "";


$conn = mysqli_connect($host,$username,$password);
mysqli_select_db($conn,'db');            //database name is "db"
if (!$conn){
    die("could not create connection");
}
else
    echo"conection made <br>";
if (isset($_POST['update'])) {
        $first_name = $_POST['first_name'];
        $last_name = $_POST['last_name'];
        $username = $_POST['username'];
        $password = ($_POST['password']);



$query=mysqli_query($conn,"INSERT INTO users(first_name, last_name, username, password) VALUES ('$first_name','$last_name','$username','$password')");
   if(!$query){
        die('could not enter data:'. mysqli_error($conn));
    }
    else{
        echo "Entered data successfully";
    }


}
?>

Uploading data from form to MySQL Database

4 Answers4