i read all answer about undefined index error but not help full for me because i'm already using isset function to check plz how to slove this problem..
<?php
$con=mysqli_connect("localhost","root","","contact");
if (mysqli_connect_errno())
{
echo "failed".mysqli_connect_error();
}
checking for submited data
if(isset($_POST['submit']))
{
$name=$_POST['name'];
$website=$_POST['website'];
$gender=$_POST['gender'];
$comment=$_POST['comment'];
}
$sql="insert into form(name,website,gender,comment) Values('$_POST[name]','$_POST[website]','$_POST[gender]','$_POST[comment]')";
if(!mysqli_query($con,$sql))
{
die('error:'.mysqli_error($con));
}
else "added";
mysqli_close($con);
?>
<html>
<body>
<form method=post action="<?php echo $_SERVER['PHP_SELF']; ?>">
Name: <input type="text" name="name"><br>
E-mail: <input type="text" name="email"><br>
Website: <input type="text" name="website"><br>
<input type="radio" name="gender" value="female">Female
<input type="radio" name="gender" value="male">Male<br>
Comment: <textarea name="comment" rows="5" cols="40"></textarea>
<input type=submit name="submit"><br>
</form>
</body>
these errors comes
Notice: Undefined index: name in H:\Wamp\Xamp\htdocs\form.php on line 15
Notice: Undefined index: website in H:\Wamp\Xamp\htdocs\form.php on line 15
Notice: Undefined index: gender in H:\Wamp\Xamp\htdocs\form.php on line 15
Notice: Undefined index: comment in H:\Wamp\Xamp\htdocs\form.php on line 15
Please try the following corrected code :
if(isset($_POST['submit']))
{
$name=isset($_POST['name']) ? $_POST['name'] : '';
$website=isset($_POST['website']) ? $_POST['website'] : '';
$gender=isset($_POST['gender']) ? $_POST['gender'] : '';
$comment=isset($_POST['comment']) ? $_POST['comment'] : '';
$sql="insert into form(name,website,gender,comment) Values('$name','$website','$gender','$comment')";
// Open the database connection here
// aka, mysqli_connect()
if(!mysqli_query($con,$sql))
{
die('error:'.mysqli_error($con));
}
else "added";
mysqli_close($con);
}
?>
<html>
<body>
<form method=post action="<?php echo $_SERVER['PHP_SELF']; ?>">
Name: <input type="text" name="name"><br>
E-mail: <input type="text" name="email"><br>
Website: <input type="text" name="website"><br>
<input type="radio" name="gender" value="female">Female
<input type="radio" name="gender" value="male">Male<br>
Comment: <textarea name="comment" rows="5" cols="40"></textarea>
<input type=submit name="submit"><br>
</form>
</body>
What I did is added validation to check if those fields are set, and if so, then set the value, if not, then set the variable (aka $name) to ''. You should probably add some further validation in the event of required fields being = '' (equal to blank).
I also adjusted your query to not use the $_POST vars, instead it uses the variables that you are assigning the $_POST values to, so you know they exist for sure.
And lastly, I moved the mysql connection code and query itself into the if(isset(submit)) statement so it does not try to process those on regular page load where the form has not been submitted yet.
Update this insert query,
$sql="insert into form(name,website,gender,comment) values('". $name ."','". $website ."','". $gender ."','". $comment ."')";
Hope this help you!
Change
if(isset($_POST['submit']))
{
$name=$_POST['name'];
$website=$_POST['website'];
$gender=$_POST['gender'];
$comment=$_POST['comment'];
}
$sql="insert into form(name,website,gender,comment) Values('$_POST[name]','$_POST[website]','$_POST[gender]','$_POST[comment]')";
if(!mysqli_query($con,$sql))
{
die('error:'.mysqli_error($con));
}
else "added";
mysqli_close($con);
?>
to
if(isset($_POST['submit']))
{
$name=$_POST['name'];
$website=$_POST['website'];
$gender=$_POST['gender'];
$comment=$_POST['comment'];
$sql="insert into form(name,website,gender,comment) values ('$name','$website','$gender','$comment')";
if(!mysqli_query($con,$sql))
{
die('error:'.mysqli_error($con));
}
else "added";
mysqli_close($con);
}?>
than friends problem is slove
What wrong?
if(isset($_POST['submit']))
{
$name=$_POST['name'];
$website=$_POST['website'];
$gender=$_POST['gender'];
$comment=$_POST['comment'];
$sql="insert into form(name,website,gender,comment) Values('". $name . "','" . $website . "','" . $gender . "','" . $comment . "')";
//these also in if(isset()) Block
if(!mysqli_query($con,$sql))
{
die('error:'.mysqli_error($con));
}
else "added";
}
thanx to all
Related
I'm not that good with PHP but i need to do a user registration form and this is what I haven done so far
<!doctype html>
<html>
<head>
<title>Register</title>
</head>
<body>
<p>Register | Login</p>
<h3>Formulario de registro</h3>
<form action="" method="POST">
Usuario: <input type="text" name="login"><br />
Contrasena: <input type="password" name="password"><br />
Email: <input type="text" name="email"><br />
Sexo: <input type="text" name="sex"><br />
Edad: <input type="text" name="age"><br />
Telefono: <input type="text" name="phone"><br />
Pais: <input type="text" name="country"><br />
Ciudad: <input type="text" name="city"><br />
Direccion: <input type="text" name="address"><br />
<input type="submit" value="Register" name="submit" />
</form>
<?php
if(isset($_POST["submit"])){
if(!empty($_POST['login']) && !empty($_POST['password'])) {
$login=$_POST['login'];
$password=$_POST['password'];
$email=$_POST['email'];
$sex=$_POST['sex'];
$age=$_POST['age'];
$phone=$_POST['phone'];
$country=$_POST['country'];
$city=$_POST['city'];
$address=$_POST['address'];
$con=mysqli_connect('localhost','root','','registro_usuarios') or die(mysql_error());
$sql="SELECT * FROM usuario ";
$result = mysqli_query($con,$sql);
if(mysqli_num_rows($result)>=0)
{
echo "I'm in";
$sql="INSERT INTO usuario (login,password,email,sex,age,phone,country,city,address)
VALUES('$login','$password','$password','$email','$sex','$age','$country','$city','$address')";
$result=mysqli_query($con,$sql);
if($result){
echo "Account Successfully Created";
} else {
echo "Failure!";
}
} else {
echo "That username already exists! Please try again with another.";
}
} else {
echo "All fields are required!";
}
}
?>
</body>
</html>
I can see the message "I'm in" in my screen but I'm also get the "Failure" message because it's not inserting anything into the the database. I have been struggling trying to find the error but nothing so far. The error should be around here
$sql="INSERT INTO usuario (login,password,email,sex,age,phone,country,city,address)
VALUES('$login','$password','$password','$email','$sex','$age','$country','$city','$address')";
$result=mysqli_query($con,$sql);
I have checked all the values from my database in case of a typo just in case you guys wonder and there is no typo.
Thanks
You wrong in sql insert.
your sql :
$sql="INSERT INTO usuario (login,password,email,sex,age,phone,country,city,address) VALUES('$login','$password','$password','$email','$sex','$age','$country','$city','$address')";
$result=mysqli_query($con,$sql);
look at the value row doesn't match with the insert fields. You insert twice $password, and no $phone.
If still error, you can add this to know what wrong with your query:
if(mysqli_query($con,$sql))
{
echo 'Success'.'</br>';
}
else
{
echo 'Fail';
echo "Error Description : ".mysqli_error($con);
}
I have the tables users and register in my database. I've created a login page which starts a session using the users table, then people fill out a form to insert data into the register table. I've used the following code to insert the data. My code doesn't have errors but the thing is it is not inserted to my table. Please help me. Thanks.
<?php
include("includes/db.php");
session_start();
if(!isset($_SESSION['user_name'])){
header("location: login.php");
}
else { ?>
<html>
<body>
<h2>New users Signup!</h2>
<form action="login.php" method="post">
<input type="text" name = "firstname" placeholder="Firstname"/>
<input type="text" name = "lastname" placeholder="Lastname"/>
<input type="text" name = "address" placeholder="Address"/>
<input type="text" name = "contact" placeholder="Contact"/>
<input type="text" name = "email" placeholder="Email Address"/>
<input type="password" name = "password" placeholder="Password"/>
<div class = "bttn">
<button type="submit" name = "submit" class="btn btn-default">Signup</button>
</div>
</form>
</body>
</html>
<?php
if(isset($_POST['submit']))
{
$users_firstname = $_POST['firstname'];
$users_lastname = $_POST['lastname'];
$users_address = $_POST['address'];
$users_contact = $_POST['contact'];
$users_email = $_POST['email'];
$users_password = $_POST['password'];
$users_date = date('Y-m-d');
if($users_firstname=='' or $users_lastname=='' or $users_address=='' or $users_contact=='' or $users_email=='' or $users_password=='')
{
echo "<script>alert('Any of the fields is empty')</script>";
exit();
} else {
$insert_query = mysql_query("insert into users (users_firstname,users_lastname,users_address,users_contact,users_email,users_password,users_date) values ('$users_firstname','$users_lastname','$users_address','$users_contact','$users_email','$users_password','$users_date')");
$users_id=mysql_insert_id();
if(mysql_query($insert_query)) {
echo "<script>alert('post published successfuly')</script>";
}
}
}
} ?>
Now try this code:
<?php
include("includes/db.php");
session_start();
if(!isset($_SESSION['user_name'])){
header("location: login.php");
}
else {
?>
<html>
<body>
<?php
if(isset($_POST['submit']))
{
$users_firstname = $_POST['firstname'];
$users_lastname = $_POST['lastname'];
$users_address = $_POST['address'];
$users_contact = $_POST['contact'];
$users_email = $_POST['email'];
$users_password = $_POST['password'];
$users_date = date('Y-m-d');
if($users_firstname=='' or $users_lastname=='' or $users_address=='' or $users_contact=='' or $users_email=='' or $users_password=='')
{
echo "<script>alert('Any of the fields is empty')</script>";
exit();
}
else
{
$insert_query = mysql_query("INSERT INTO `users` (users_firstname, users_lastname, users_address, users_contact, users_email, users_password, users_date) values ('$users_firstname', '$users_lastname', '$users_address', '$users_contact', '$users_email', '$users_password', '$users_date')");
$users_id=mysql_insert_id();
echo "<script>alert('post published successfuly')</script>";
}
}
?>
<h2>New users Signup!</h2>
<form action="" method="post">
<input type="text" name="firstname" placeholder="Firstname"/>
<input type="text" name="lastname" placeholder="Lastname"/>
<input type="text" name="address" placeholder="Address"/>
<input type="text" name="contact" placeholder="Contact"/>
<input type="text" name="email" placeholder="Email Address"/>
<input type="password" name="password" placeholder="Password"/>
<div class="bttn">
<button type="submit" name="submit" class="btn btn-default">Signup</button>
</div>
</form>
</body>
</html>
<?php } ?>
I have:
Repositioned your PHP code for inserting to be at the top of the form
changed <form action="login.php" to <form action="" because we are executing from the same page
Your query has already run so removed the if(mysql_query...
Removed the spaces in the form e.g. name = " nameofform" to name="nameofform"
I don't see any reason for having this $users_id=mysql_insert_id();, YOu should use auto-increment for the userID on your database
But since we don't know how you have connected to your database, because also that can be an issue: you can also try this way
<?php
//connect to DB
$hostname_localhost = "localhost"; //hostname if it is not localhost
$database_localhost = "databasename";
$username_localhost = "root"; //the username if it is not root
$password_localhost = "password_if_any"; //if no password leave empty
$localhost = mysql_pconnect($hostname_localhost, $username_localhost, $password_localhost) or trigger_error(mysql_error(),E_USER_ERROR);
?>
<?php
// include("includes/db.php");
if (!isset($_SESSION)) {
session_start();
}
if(!isset($_SESSION['user_name'])){
header("location: login.php");
}
else {
?>
<html>
<body>
<?php
if(isset($_POST['submit']))
{
$users_firstname = $_POST['firstname'];
$users_lastname = $_POST['lastname'];
$users_address = $_POST['address'];
$users_contact = $_POST['contact'];
$users_email = $_POST['email'];
$users_password = $_POST['password'];
$users_date = date('Y-m-d');
if($users_firstname=='' or $users_lastname=='' or $users_address=='' or $users_contact=='' or $users_email=='' or $users_password=='')
{
echo "<script>alert('Any of the fields is empty')</script>";
exit();
}
else
{
$insert_query = sprintf("INSERT INTO `users` (users_firstname, users_lastname, users_address, users_contact, users_email, users_password, users_date) values ('$users_firstname', '$users_lastname', '$users_address', '$users_contact', '$users_email', '$users_password', '$users_date')");
mysql_select_db($database_localhost, $localhost);
$Result = mysql_query($insert_query, $localhost) or die(mysql_error());
echo "<script>alert('post published successfuly')</script>";
}
}
?>
<h2>New users Signup!</h2>
<form action="" method="post">
<input type="text" name="firstname" placeholder="Firstname"/>
<input type="text" name="lastname" placeholder="Lastname"/>
<input type="text" name="address" placeholder="Address"/>
<input type="text" name="contact" placeholder="Contact"/>
<input type="text" name="email" placeholder="Email Address"/>
<input type="password" name="password" placeholder="Password"/>
<div class = "bttn">
<button type="submit" name="submit" class="btn btn-default">Signup</button>
</div>
</form>
</body>
</html>
<?php } ?>
You should removed the whitespaces in your html-code:
Wrong
<input type="text" name = "firstname" placeholder="Firstname"/>
^^^^^
Correct
<input type="text" name="firstname" placeholder="Firstname"/>
Do not put the variables in single quotes:
$insert_query = mysql_query("INSERT INTO users
(users_firstname,users_lastname,users_address,users_contact,users_email,users_password,users_date)
VALUES
($users_firstname,$users_lastname,$users_address,$users_contact,$users_email,$users_password,$users_date)");
Update: This was wrong. The whole string is in double-quotes so the OP did correct and my notice was wrong. For information-purposes i will let stand the link to the documentation here.
Note: Unlike the double-quoted and heredoc syntaxes, variables and escape sequences for special characters will not be expanded when they occur in single quoted strings.
Read more about single- and double-quotes in the PHP documentation.
Do not double-run the query/perform wrong function-call
$insert_query = mysql_query(".......");
........
if(mysql_query($insert_query)){
echo "<script>alert('post published successfuly')</script>";
}
You already have run the query in the first line. If you want to check if it was successful, you have to use if($insert_query) {}. The call mysql_query($insert_query) is wrong because mysql_query() returns a ressource instead of the sql-query.
Do not use mysql_*() function calls. You mysqli_*() instead.
Warning
This extension was deprecated in PHP 5.5.0, and it was removed in PHP 7.0.0. Instead, the MySQLi or PDO_MySQL extension should be used. See also MySQL: choosing an API guide and related FAQ for more information. Alternatives to this function include:
mysqli_query()
PDO::query()
Check your use of session.
You are checking the $_SESSION for user_name and if it is not set, you are redirecting via header("location: login.php").
The problem is, that you are never inserting the user_name into the session, so it will always be not set.
You can set the value via $_SESSION['user_name'] = $_POST['user_name']. Have in mind that you have to set the session before checking the session-value. ;-)
remove action
Try this
<form action="" method="post">
Hi guys I am learning PHP and the code below sends information to my MySQL database but it does not match with the data i just gave him to insert, for example it inserts 1|blank|blank for the code below. I can´t solve it :
insert.php file:
<?php
$con=mysqli_connect("localhost","root","");
// Check connection
if (mysqli_connect_errno()) {
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
// escape variables for security
$firstname = mysqli_real_escape_string($con, $_POST['firstname']);
$age = mysqli_real_escape_string($con, $_POST['age']);
$sql="INSERT INTO Person (PID,Name, Age)
VALUES ('1','$firstname', '$age')";
if (!mysqli_query($con,$sql)) {
die('Error: ' . mysqli_error($con));
}
echo "1 record added to database";
mysqli_close($con);
?>
It keeps me saying this:
Notice: Undefined index: firstname in E:\XAMPP\htdocs\Bioinformatics\insert.php on line 11
Notice: Undefined index: age in E:\XAMPP\htdocs\Bioinformatics\insert.php on line 12
code for html form:
<!DOCTYPE html>
<html>
<body>
<form name="input" action="insert.PHP" method="get">
First name: <input type="text" name="FirstName" value="Ronaldo"><br>
Age: <input type="text" name="Age" value="28"><br>
<input type="submit" value="Submit">
</form>
</body>
</html>
Thanks for any help guys!
It looks like you have forgotten to post the needed data.
$_POST['firstname'] and $_POST['age'] are not defined in your html.
You are missing the following inputs or have misspelled them:
<input type="text" name="firstname" />
<input type="text" name="age" />
cause you have not values for $_POST['firstname'] and $_POST['age'] try to check it's have values or not.
if(!empty($_POST['Firstname'])) {
$firstname = mysqli_real_escape_string($con, $_POST['Firstname']);
}
else {
echo 'firstname empty';
}
if(!empty($_POST['Age'])) {
$age= mysqli_real_escape_string($con, $_POST['Age']);
}
else {
echo 'age empty';
}
also same for age
For check post values try print_r($_POST); on this page
change
$firstname = mysqli_real_escape_string($con, $_POST['firstname']);
$age = mysqli_real_escape_string($con, $_POST['age']);
to
$firstname = $_POST['firstname'] ? mysqli_real_escape_string($con, $_POST['firstname']) : "";
$age = $_POST['age'] ? mysqli_real_escape_string($con, $_POST['age']) : "";
echo "<pre>";print_r($_REQUEST);exit; // to check values in all variable.
Place your php code below html code to get rid of undefined index notices or use isset() function eg. if(isset($_POST['age'])) or initialize those variables with null value. like $age = "";
also see answer by #karlingen
I have tried a number of different ways of inserting data into my DB I have got a little further it used to just say error but now when you submit the form it loads a blank page, the data from the form isn't added to the table however ;/
<form name="datainsert" method="post" action="dataInsert.php">
<label>Server Name: </label>
<input type="text" name="name" placeholder="Enter Server Name" style="margin-left:90px; width:160px; padding:5px; margin-top:10px;"><br />
<label>Server Location:</label>
<input type="text" name="location" placeholder="Enter Server Location" style="margin-left:71px; width:160px; padding:5px; margin-top:10px;"><br />
<label>Server Operating System:</label>
<input type="text" name="os" placeholder="Enter Server OS" style="margin-left:16px; width:160px; padding:5px; margin-top:10px;"><br/>
<input style="margin-top:10px;" name="submit" value="submit" type="submit">
</form>
<?php
include 'dbconnect.php';
$name = $_POST['name'];
$location = $_POST['location'];
$os = $_POST['os'];
)
mysql_query("INSERT INTO fostvm (name, location, os) VALUES ('$name', '$location', '$os')");
$result=mysql_query($sql);
if($result){
echo "Data Added Successfully";
} else {
echo "Error";
}
?>
Can anyone see a syntax error or where I might be going wrong
thanks!
Try this. your code is Ok just comment this line ($result=mysql_query($sql);). use this code. why you try mysql_qury() two times in your code.
<?php
include 'dbconnect.php';
$name = $_POST['name'];
$location = $_POST['location'];
$os = $_POST['os'];
)
$result = mysql_query("INSERT INTO fostvm (name, location, os) VALUES ('$name', '$location', '$os')");
//$result=mysql_query($sql);
if($result){
echo "Data Added Successfully";
} else {
echo "Error";
}
?>
You have no $sql variable and you are using it mysql_query try this,
<?php
include 'dbconnect.php';
if(isset($_POST['submit'])){ // check for form submit
$name = $_POST['name'];
$location = $_POST['location'];
$os = $_POST['os'];
$result=mysql_query("INSERT INTO fostvm (name, location, os) VALUES ('$name', '$location', '$os')");
if($result){
echo "Data Added Successfully";
} else {
echo "Error";
}
}
?>
Also you should use mysqli as mysql is deprecated
I am trying to make Sign Up Now! area for a restaurant website and want to insert data of new members in the members_t table of database members with all running on localhost. I am using PHP and HTML for the purpose. Moreover, I am doing form validation using javaScript in a separate file which is working perfectly!
Code for PHP:
<?php
$user="root";
$password="";
$database="members";
$con = mysql_connect('localhost',$user,$password);
if (!$con)
{
die('Could not connect: ' . mysql_error());
}
mysql_select_db($database, $con) or die( "Unable to select database");
if(isset($_POST['sign_up']) && !empty($_POST['sign_up']))
{
$sql = "INSERT INTO members_t('Name', 'Email', 'Password', 'Phone', 'Address', 'Sex', 'More') VALUES('".$_POST['username']."','".$_POST['email']."','".$_POST['passid_1']."','".$_POST['zip']."','".$_POST['address']."','".$_POST['sex']."','".$_POST['desc']."');";
$resultDI = mysql_query($sql, $con) or die(mysql_error());
mysql_close($con);
echo "Successfolly run database query!";
}
else
{
echo("Failed to update database!!!");
}
?>
Code for HTML:
<html>
<body>
<h2 class="letter_spacing">Not a Member?<span><br>Sign Up Now:</br></span></h2>
<form id = "register" name="registration" method = "post" onSubmit="return formValidation();">
<ul>
<li><label for="username">* Full Name:</label></li>
<li><input type="text" name="username" size="50" /></li>
<li><label for="email">* Email:</label></li>
<li><input type="text" name="email" size="50" /></li>
<li><label for="passid_1">* Desired Password:</label></li>
<li><input type="password" name="passid_1" size="12" /></li>
<li><label for="passid_2">* Re-Enter Password:</label></li>
<li><input type="password" name="passid_2" size="12" /></li>
<li><label for="zip">* Contact Number:</label></li>
<li><input type="text" name="zip" /></li>
<li><label for="address">* Address:</label></li>
<li><input type="text" name="address" size="50" /></li>
<li><label id="gender">* Sex:</label></li>
<li><input type="radio" name="msex" value="Male" /><span>Male</span></li>
<li><input type="radio" name="fsex" value="Female" /><span>Female</span></li>
<li><label for="desc">Anything More:</label></li>
<li><textarea name="desc" id="desc" cols="40" rows="4"></textarea></li>
<li><label for="note" ><h6>Note: All feilds marked with * are necessary</h6></label></li>
<li><input class="button1" type="submit" name="sign_up" value="Sign Up!" /></li>
</ul>
</form>
</body>
</html>
I have tried to keep the code in a separate file called insert.php and added the action field to the HTML form tag yet of no use.
I am never able to insert data into the database. It seems the PHP code never goes into the
if(isset($_POST['sign_up']) && !empty($_POST['sign_up']))
block.
Try this:
<form id="register" action="" method="post" name="registration" onSubmit="return formValidation();">
<input type="text" name="username" size="50" />
<input type="text" name="email" size="50" />
<input type="password" name="passid_1" size="12" />
<input type="password" name="passid_2" size="12" />
<input type="text" name="zip" />
<input type="text" name="address" size="50" />
<input type="radio" name="sex" value="Male" /><span>Male</span>
<input type="radio" name="sex" value="Female" /><span>Female</span>
<textarea name="desc" id="desc" cols="40" rows="4"></textarea>
<input class="button1" type="submit" name="sign_up" value="Sign Up!" />
</form>
<?php
if (isset($_POST['sign_up']) && !empty($_POST['sign_up'])) {
// escape all submitted data before inserting into database
foreach ($_POST as $key => $value) {
$_POST[$key] = mysql_real_escape_string(strip_tags($value));
}
$result = mysql_query("
INSERT INTO members_t (Name, Email, Password, Phone, Address, Sex, More)
VALUES ('{$_POST['username']}', '{$_POST['email']}', '{$_POST['passid_1']}', '{$_POST['zip']}', '{$_POST['address']}', '{$_POST['sex']}', '{$_POST['desc']}')
") or die(mysql_error());
if (mysql_affected_rows() == 1) {
echo "Successfully run database query!";
} else {
echo("Failed to update database!!!");
}
}
?>
Note that name of the radio buttons should be the same "sex" not "msex" and "fsex" as in your code. And I have added the action attribute in the form tag plus some other modifications you can easily notice.
First of all, i have cleaned up the code a little so it looks nice and smooth. Then i have removed the !empty part you made, cant see the reason why you want to verify that it actually is empty when you already used isset.
HTML:
<?php
$hostname = "";
$user = "root";
$password = "";
$database = "members";
$desc = $_POST['desc'];
$con = mysql_connect($hostname, $user, $password);
if (!$con)
{
die('Could not connect: ' . mysql_error());
}
mysql_select_db($database, $con) or die( "Unable to select database");
if(isset($_POST['sign_up']))
{
if(isset($_POST['username'])){
$username = $_POST['username'];
}
else {
echo "The username is not set"; die;
}
if(isset($_POST['email'])){
$email = $_POST['email'];
}
else {
echo "The email is not set"; die;
}
if(isset($_POST['zip'])){
$zip = $_POST['zip'];
}
else {
echo "The zip code is not set"; die;
}
if(isset($_POST['address'])){
$address = $_POST['address'];
}
else {
echo "The gender is not set"; die;
}
if(isset($_POST['sex'])){
$sex = $_POST['sex'];
}
else {
echo "The gender is not set"; die;
}
if(isset($_POST['passid_1'])){
$passid = $_POST['passid_1'];
}
else {
echo "The password is not set"; die;
}
if(isset($_POST['passid_2'])){
$passid2 = $_POST['passid_2'];
}
else {
echo "The re-entered password is not set"; die;
}
if($passwid == $passid2){
$correctpid = $passwid;
}
else {
echo "The passwords do not match"; die;
}
$sql = "INSERT INTO members_t('Name', 'Email', 'Password', 'Phone', 'Address', 'Sex', 'More') VALUES('$username', '$email','$correctpid', '$zip', '$address', '$sex', '$desc');";
mysql_query($sql) or die(mysql_error());
mysql_close($con);
echo "Successfolly run database query!";
}
else
{
echo("Failed to update database!!!");
}
?>
I have made the php code to check if all the fields are filled with data. If not the site die and gives them a error message. It kills the website before it can set anything into the database.
-- I made some more changes to the code after comments, thanks btw.