Cannot INSERT data into mysql database

Question

shayan_doust 0 Newbie Poster

10 Years Ago

Hello everyone.

I have just bought a new computer and I am currently working on creating an online cloud storage service. Every thing is going fine until I am struck with a problem. At first, registering a new user works and the data is inserted into database but after testing it multiple times, it no loger works for some reason? The thing that struck me is that during the testing period, I did not tamper with any of the codes. After 5 tests, when a new user tries to register, the website says there are no errors but after going into my phpmyadmin, there is nothing there. I have tried everything and I don't know what to do! Please Help

CODE:

signup.php

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        // When the window has finished loading create our google map below
        google.maps.event.addDomListener(window, 'load', init);

        function init() {
            // Basic options for a simple Google Map
            // For more options see: https://developers.google.com/maps/documentation/javascript/reference#MapOptions
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                <div class="col-lg-8 col-lg-offset-2 mt">
                    <h1 class="animation slideDown">Please complete the following form to signup for our service.</h1>
                    <form name="register" action="registerprocess.php" method="post">
                    <tr>
            <td>USERNAME:</td>
            <td><input type="text" name="username" maxlength="20" /></td>
        </tr>
        <tr>
            <td>PASSWORD:</td>
            <td><input type="password" name="password1" /></td>
        </tr>

        <tr>
            <td>EMAIL:</td>
            <td><input type="text" name="email" id="email" /></td>
        </tr>
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            <td>&nbsp;</td>
            <td><input type="submit" value="Register" /></td>
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  </body>
</html>

registerprocess.php

<?php
$username = $_POST['username'];
$password1 = $_POST['password1'];
$email = $_POST['email'];


$con=mysqli_connect("db4free.net","-CENSORED USERNAME-","-CENSORED PASSWORD-","projectbook");
// Check connection
if (mysqli_connect_error()) {
  echo "Failed to connect to MySQL: " . mysqli_connect_error();
}

mysqli_query($con,"INSERT INTO member (name, password, email)
VALUES ('$name', '$password', '$email')");

mysqli_close($con);
?>

I WOULD APPRECIATE ANY HELP!

mysql php

6 Contributors
5 Replies
2K Views
1 Week Discussion Span
Latest Post 10 Years Ago Latest Post by matrixdevuk

gabrielcastillo 40 Web Developer

10 Years Ago

User firefox with firebug->console to see if any errors are shown. In your mysqli connection, try changing to localhost or 127.0.0.1

Isaac_4 39 Junior Poster in Training

10 Years Ago

Try adding these lines to the top of your PHP script to make sure errors show up:

define('DEBUG', true); // change true to false to disable errors showing up
if(DEBUG) {
  ini_set('display_errors', 1);
} else {
  ini_set('display_errors', 0);
}
error_reporting(E_ALL & ~E_NOTICE);

This might help you track down the error if the they are being hidden from you for some reason.

Also try to var_dump your variables to see if they have the values you expect. Combined and expanded with broj1's good recommendations this might look like this:

// sanitize the data - unless you use mysqli_prepare (mentioned again
// a few lines down) - NEVER access $_POST or $_GET without passing
// the value through mysqli_real_escape_string() or crypt() first.
// even if you do use mysqli_prepare, always use crypt() for passwords!
$username = mysqli_real_escape_string($_POST['username']);
// TODO: change second string ('*D') passed to crypt() to something unique to your
// site. if you have PHP 5.5 use password_hash instead of crypt.
$password1 = crypt($_POST['password1'], '*D'); 
//$password1 = password_hash($_POST['password1'], PASSWORD_DEFAULT);  // PHP 5.5 only
$email = mysqli_real_escape_string($_POST['email'])

echo "<br/>Username =";
var_dump($username);
echo "<br/>Password ="; // remember this will be a long string hash, that's good!
var_dump($password);
echo "<br/>Email =";
var_dump($email);

// prepare the query from sanitized data
// TODO/SECURITY RISK: instead of concatenating strings from user
// supplied values, you should consider using mysqli_prepare
// (see http://php.net/manual/en/mysqli.prepare.php) in which case
// mysqli_real_escape_string would no longer be required above
$sql = "INSERT INTO member (name, password, email) VALUES ('$name', '$password1', '$email')"

echo "<br/>SQL = ";
var_dump($sql);

// call the prepared query
mysqli_query($con,$sql);

Edited 10 Years Ago by Isaac_4

broj1 commented: Nicely explained +11

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broj1 356 Humble servant Featured Poster · Answer 1 · 2014-10-20T07:23:52+00:00

Just a few thoughts:

you should check if all the required values exist before using them in the query (otherwise the query might break)
you should sanitize the values before using them in the query (otherwise you could pass some nasty user supplied data to your DB server than can compromise security)
you should hash the password and store / use the hashed value (so users are ensured they have full control over their access)

This is example code (it can be improved, but is simplified to show what I meant):

// if any of the data is missing redirect to some error page or back to the form
if(
    !isset($_POST['username']) || empty($_POST['username']) ||
    !isset($_POST['password1']) || empty($_POST['password1']) ||
    !isset($_POST['email']) || empty($_POST['email'])
  ) {
      header("location: error.php");
      exit();
}
// sanitize the data
$username = mysqli_real_escape_string($_POST['username']);
// use crypt or similar to hash the passwords
$password1 = some_hashing_method_here($_POST['password1']);
$email = $password1 = mysqli_real_escape_string($_POST['email']);
...

msz900 0 Newbie Poster · Answer 2 · 2014-10-25T02:12:14+00:00

Dear
try these steps..
1) First check that your database column name is same as you enter in the query.

2) fix your query to this...

$smt=mysqli_query($con,"INSERT INTO member (name, password, email)VALUES('".$name."', '".$password."', '".$email."')");
if($smt)
do something here
}
else
{
show error here
}

matrixdevuk 71 6t9.me Founder · Answer 3 · 2014-10-26T15:20:09+00:00

This probably isn't your issue, but I recommend reading up on prepared statements.