I'm making a simple registration form for a test website and for some reason it isn't sending the data to the database, and I don't get an visual error. I've searched around for a fix but haven't found any that work.
This is basically my form (I only copied the form part of the page):
<form action="includes/insert.php" method="post">
<h3>Username</h3>
<input type="text" name="username">
<br>
<br>
<h3>Email Address</h3>
<input type="email" name="email">
<br>
<br>
<h3>Password</h3>
<input type="password" name="password">
<br>
<br>
<br>
<input id="submit-btn" type="submit" name="submit" value="Submit">
</form>
As you can see everything is as its suppose to be.
and this is my insert.php
<?
define('DB_NAME', 'logindb');
define('DB_USER', 'root');
define('DB_PASSWORD', 'password');
define('DB_HOST', 'localhost');
$link = mysql_connect(DB_HOST, DB_USER, DB_PASSWORD);
if (!$link) {
die('Could not connect: ' . mysql_error());
}
$db_selected = mysql_select_db(DB_NAME, $link);
if (!$db_selected) {
die('Can\'t use ' . DB_NAME . ': ' . mysql_error());
}
$username = $_POST['username'];
$email = $_POST['email'];
$password = $_POST['password'];
$sql = "INSERT INTO `users` (`id`, `username`, `email`, `password`, `timestamp`) VALUES (NULL, '$username', '$email', '$password', CURRENT_TIMESTAMP)";
if (!mysql_query($sql)) {
die('Error: ' . mysql_error());
}
mysql_close();
?>
Opening PHP tag is
<?php
Recent versions of PHP do not enable the short code syntax by default.
Use NOW() instead of CURRENT_TIMESTAMP.
Related
I am using PHPStorm and I am trying to send my user's input from an html form to my database using PHP. My database seems to be getting the inputs since it creates a new id (which is the primary key and it is set to auto-increment) however, the fields are empty!
Here is my HTML
<!DOCTYPE html>
<html>
<head>
</head>
<body>
<form name="i_createaccount" id="createaccount" action="info.php" method="post">
<h2 class="formtitle">Create an account</h2>
<input type="text" class="firstnlast" name="i_name" placeholder="First Name" required/>
<input type="text" class="firstnlast" name="i_last" placeholder="Last Name" required/>
<input id="txtEmail" type="email" name="i_email" placeholder="Email" required/>
<input type="submit" name="submit" class="next, button" value="Join Now"/>
</form>
</body>
</html>
Here is my info.php
<?php
if(empty($_SERVER['CONTENT_TYPE'])) {
$type = "application/x-www-form-urlencoded";
$_SERVER['CONTENT_TYPE'] = $type;
}
//Connecting to sql db.
define('DB_NAME', 'nabi_data');
define('DB_USER', 'root');
define('DB_PASSWORD', '');
define('DB_HOST', 'localhost');
$link = mysql_connect(DB_HOST, DB_USER, DB_PASSWORD);
if (!$link) {
die('Could not connect: ' . mysql_error());
}
$db_selected = mysql_select_db(DB_NAME, $link);
if (!$db_selected) {
die('Cant use ' . DB_NAME . ':' . mysql_error());
}
$message = '';
$i_name = (string)filter_input(INPUT_POST, 'i_name');
$i_last = (string)filter_input(INPUT_POST, 'i_last');
$i_email = (string)filter_input(INPUT_POST, 'i_email');
$sql = "INSERT INTO nabi_instructors (i_name, i_last, i_email) VALUES ('$i_name', '$i_last', '$i_email')";
if (!mysql_query($sql)) {
die('Error: ' . mysql_error());
mysql_close();
}
I am new working on a project that has me connecting a database and a webpage and before doing the official database, I wanted to test a basic database.
I test to make sure everything is working by using phpMyAdmin. I can tell that I formed SOME kind of connection between my html and php, but that's where the issue is. whenever I press the submit button on my site, instead of getting a message telling me that everything works, I instead get a white page with all of my php code. I have tried making sure all my syntax is correct and all the variables are labeled correctly, but nothing is going through.
<!DOCTYPE php>
<?php
define('DB_NAME', 'training');
define('DB_USER', 'root');
define('DB_PASSWORD', '********');
define('DB_HOST', 'localhost');
$link = mysql_connect(DB_HOST, DB_USER, DB_PASSWORD);
if(!link){
die('Could not connect: ' .mysql_error);
}
$db_selected = mysql_select_db(DB_NAME, $link);
if (!db_selected){
die('Error: ' .mysql_error());
}
$Name = $_POST=['name'];
$Job = $_POST['job'];
$sql = "INSERT INTO employee (name,job) VALUES ('$Name', '$Job')";
if (!mysql_query($sql)){
die('Error: ' .mysql_error());
}
echo 'Success!';
mysql_close();
?>
<!DOCTYPE html>
<html>
<h1 style="text-align:center">Undergrad Sign-up</h1>
<style>
#body {background-color:#001b38;}
</style>
<body id="body" style="color:white;text-align:center" >
<form action="DBconnect.php" method="post" />
Name<br>
<input type="text" name="name" maxlength=20 required/>
<br>
Job<br>
<input type="text" name="job" maxlength=20 required/>
<br>
<button type="submit" value="submit">Submit</button>
</form>
</body>
</html>
I have written a code for form which on submit opens the demo.php tag in the browser. I want the form to produce a (Thank you) alert on same page without redirecting to demo.php. Is it possible to embed a jquery notificaiton plugin? If yes then how?
<?php
error_reporting(0);
define('DB_NAME', 'form');
define('DB_USER', '******');
define('DB_PASSWORD', '*******');
define('DB_HOST', 'localhost');
$link = mysql_connect(DB_HOST, DB_USER, DB_PASSWORD);
if (!$link) {
die('Could not connect: ' . mysql_error());
}
$db_selected = mysql_select_db(DB_NAME, $link);
if (!$db_selected) {
die('Can\'t use ' . DB_NAME . ': ' . mysql_error());
}
echo "<script type='text/javascript'>alert('It worked!')</script>";
exit();
$value = $_POST['name'];
$value2 = $_POST['email'];
$value3 = $_POST['bio'];
$sql = "INSERT INTO form (name, email, bio) VALUES ('$value', '$value2','$value3')";
if (!mysql_query($sql))
{
die('Error: ' . mysql_error());
}
mysql_close();
?>
<form method="post" action="demo.php" id="form" >
<center><legend><h1> Enter Your Details </h1></legend>
<div class="formelement">
<label for="name">Name: </label>
<input type="text" name="name" id="name" required/>
</div>
<div class="formelement">
<label for="email">Email-ID: </label>
<input type="email" name="email" id="email" required/>
</div>
<div class="formelement">
<label for="bio">Your Message: </label>
<textarea rows="1" cols="15" name="bio" id="bio"></textarea>
</div>
<div class="formelement">
<input type="submit" value="SUBMIT" class="submit"/>
</div>
</form>
try changing
<form method="post" action="demo.php" id="form" >
to
<form method="post" action="" id="form" >
it will post form to same file and so do not redirect to other page
and print alert inside-
$db_selected = mysql_select_db(DB_NAME, $link);
if (!$db_selected) {
die('Can\'t use ' . DB_NAME . ': ' . mysql_error());
}
echo "<script type='text/javascript'>alert('It worked!')</script>";
$value = $_POST['name'];
$value2 = $_POST['email'];
$value3 = $_POST['bio'];
$sql = "INSERT INTO form (name, email, bio) VALUES ('$value', '$value2','$value3')";
$result=mysql_query($sql) or die('Error: ' . mysql_error());
if ($result)
{
echo "<script type='text/javascript'>alert('success!')</script>";
}
mysql_close();
You can use jquery ajax form submission .so that you can give alert using jquery
see here
I've been trying to learn PHP and have been given a simple task to help me.
I'm trying to get a user to complete a form which has their email address in it, then save it to a database.
Here's my code so far:
<html>
<body>
<form action="postemail.php" method="post"> Email Address: <input type="text" name="emailaddress" /> <input type="submit" />
</form>
</body>
</html>
<?php
$connection = mysql_connect("localhost","edwardHost","password");
if (!$connection) {
die('Could not connect: ' . mysql_error());
}
mysql_select_db("my_database", $connection);
$sql="INSERT INTO Subscribers (EmailAddress) VALUES ('$_POST[emailaddress]')";
if (!mysql_query($sql,$connection)) {
die('Error: ' . mysql_error());
}
mysql_close($connection);
?>
Thanks in advance!
Change your query to this
One more thing i forget last time you are missing single quete around $_POST[emailaddress]. In your query
$sql="INSERT INTO Subscribers (EmailAddress) VALUES ('".$_POST['emailaddress']."')";
Dont use mysl function as the are deprciated
Learn mysqli_ function or PDO Or both
Check this link for mysql identifier http://dev.mysql.com/doc/refman/5.0/en/identifier-qualifiers.html
Try this example using PDO in your postemail.php
define('DB_TYPE', 'mysql');
define('DB_HOST', '127.0.0.1');
define('DB_NAME', 'dbname');
define('DB_USER', 'root');
define('DB_PASS', 'password');
try {
// create a new instance of a PDO connection
$db = new PDO(DB_TYPE.':host='.DB_HOST.';dbname='.DB_NAME, DB_USER, DB_PASS);
$db->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
}
catch(PDOException $e) {
// if the connection fails, display an error message
echo 'ERROR: ' . $e->getMessage();
}
if(isset($_POST['emailaddress']) && !empty($_POST['emailaddress'])) {
$emailaddress = $_POST['emailaddress'];
$sql = 'INSERT INTO Subscribers (EmailAddress) VALUES (:emailaddress )';
$stmt = $db->prepare($sql);
$stmt->bindValue('emailaddress ', $emailaddress);
$stmt->execute();
}
After you have totaly filled in the form, it first needs to check if the submit button is clicked, then it has to send it to a database.
You also need to give you submit button a name=""
HTML code:
<html>
<body>
<form action="postemail.php" method="post">
Email Address: <input type="text" name="emailaddress" />
<input type="submit" name="submit" value="add to database" />
</form>
</body>
</html>
PHP code:
<?php
if(isset($_POST['submit'])){
$connection = mysqli_connect("localhost","edwardHost","password","my_database");
if (!$connection) {
die('Could not connect: ' . mysql_error());
}
$email = $_POST['emailaddress'];
$sql = "INSERT INTO Subscribers (EmailAddress) VALUES ('$email')";
if (!mysqli_query($connection,$sql)) {
die('Error: ' . mysql_error());
}
mysql_close($connection);
}
?>
<html> <body>
<form action="postemail.php" method="post">
Email Address: <input type="text" name="emailaddress" />
<input type="submit" />
</form>
</body> </html>
<?php $connection = mysql_connect("localhost","username","password");
if (!$connection) { die('Could not connect: ' . mysql_error());
}
mysql_select_db("my_database", $connection);
$sql="INSERT INTO Subscribers (EmailAddress) VALUES ('$_POST[emailaddress]')";
if (!mysql_query($sql,$connection)) { die('Error: ' . mysql_error()); }
mysql_close($connection);
?>
I know there a lot of questions on MySQL and PHP but I can't seem to find an answer simple enough for me to understand what to do and why.
Here is my form script
<form name="tickets" action="tickets98829849.php" method="get">
First Name: <input type="text" name="firstname"><br>
Last Name: <input type="text" name="lastname"><br><br>
Number of Tickets: <input type="text" name="quant"><br><br>
First and Last Name of Date: <input type="text" name="date"><br>
Date a guest? <input type="checkbox" name="guest" value="Yes">Yes<br><br>
Amount paid per ticket: <br><br>
$<input type="text" name="amount" size="2"><br>
<br><input type="submit" value="Submit"></form>
and here is my PHP script
<?php
define('DB_NAME', 'ticketpurch');
define('DB_USER', 'dbuser');
define('DB_PASSWORD', 'dbpsswd');
define('DB_HOST', 'localhost');
$link = mysql_connect(DB_HOST, DB_USER, DB_PASSWORD)
if (!$link) {
die('Could not connect: ' . mysqlerror());
}
$db_selectd = mysql_select_db(DB_NAME, $link);
if (!$db_selected) {
die('Can\'t use ' . DB_NAME . ': ' . mysqlerror());
}
$value1 = $_POST['firstname']
$value2 = $_POST['lastname']
$value3 = $_POST['quant']
$value4 = $_POST['datename']
$value5 = $_POST['guest']
$value6 = $_POST['amount']
$sql = "INSERT INTO $table ticketpurch (firstname, lastname, quant, datename, guest, amount) VALUES ('$value1', '$value2', '$value3', '$value4', '$value5', '$value6')";
$result = mysql_query($sql)
if (!mysql_query($sql)) {
die('Error: ' . mysql_error());
}
mysql_close();
>
When I use the form, the data does not get entered into the table. I do not get an error message. What is wrong with it for not letting me access the table and insert data? I am very new to php, so
Your form's submit method is get but you receive post method. Change it.
In form you mention method="get" change it to method="post"
It will work :)
Remove the $table from the statement. I'm assuming the correct table is ticketpurch, so all you did with $table was confuse the database. In addition, the other answers are also correct. Your form's method is "get," yet you're using $_POST to try to get the data. You should change the form to post, because get is incredibly insecure, especially if this involves purchases.
In addition, you're using deprecated functions like mysql_connect. Instead, use the PDO object. Instead of mysql_connect(), try this:
$table = "mytable";
$sql = new PDO;
$sql->__construct(DB_NAME, DB_USER, DB_PASS);
$stmnt = $sql->prepare("INSERT INTO ticketpurch (firstname, lastname, quant, datename, guest, amount) VALUES ('$value1', '$value2', '$value3', '$value4', '$value5', '$value6')";
$sql->execute($stmnt);
That will ensure that your statements are better secured against SQL injection.
Hope that helped.
Try Like this.
HTML CODE
<form name="tickets" action="tickets98829849.php" method="post">
First Name: <input type="text" name="firstname"><br>
Last Name: <input type="text" name="lastname"><br><br>
Number of Tickets: <input type="text" name="quant"><br><br>
First and Last Name of Date: <input type="text" name="date"><br>
Date a guest? <input type="checkbox" name="guest" value="Yes">Yes<br><br>
Amount paid per ticket: <br><br>
$<input type="text" name="amount" size="2"><br>
<br><input type="submit" value="Submit"></form>
tickets98829849.php code
<?php
define('DB_NAME', 'ticketpurch');
define('DB_USER', 'dbuser');
define('DB_PASSWORD', 'dbpsswd');
define('DB_HOST', 'localhost');
$link = mysql_connect(DB_HOST, DB_USER, DB_PASSWORD)
if (!$link) {
die('Could not connect: ' . mysqlerror());
}
$db_selectd = mysql_select_db(DB_NAME, $link);
if (!$db_selected) {
die('Can\'t use ' . DB_NAME . ': ' . mysqlerror());
}
$value1 = $_POST['firstname']
$value2 = $_POST['lastname']
$value3 = $_POST['quant']
$value4 = $_POST['datename']
$value5 = $_POST['guest']
$value6 = $_POST['amount']
$sql = "INSERT INTO $table ticketpurch (firstname, lastname, quant, datename, guest, amount) VALUES ('$value1', '$value2', '$value3', '$value4', '$value5', '$value6')";
$result = mysql_query($sql)
if (!mysql_query($sql)) {
die('Error: ' . mysql_error());
}
mysql_close();
>
If you use method="get" in the form,then receive html values using $_GET['html element name'];.But for security purpose ,we generally use method="post" in html code and receive value in action page using $_POST[];.