Error:
The screen goes blank when submitting the form and there is no new record in the database. Any ideas?
My files:
Conn.php
<?php
$connect =mysqli_connect ("localhost", "root", "", "gym");
if(mysqli_connect_errno($connect))
{
echo "Error Connecting to Database!";
}
?>
Form.php
<html>
<head>
<title>Gym Form</title>
<meta charset="UTF-8">
<meta name="viewport" content="width=device-width, initial-scale=1.0">
</head>
<body>
<form action="Proccess.php" method="post">
<span>Gym Membership Registration</span><br><br>
<Span>Title: </Span><input type ="text" Value =" " name ="Title" /><br>
<Span>First Name: </Span><input type ="text" Value =" " name ="Fname" /><br>
<Span>Last Name: </Span><input type ="text" Value =" " name ="Lname" /><br><br>
<Span>Gender: </Span><select name ="Gender">
<option value ="Junior">Male</option>
<option value ="Adult">Female</option>
<option value ="Senior">Private</option>
</select><br>
<Span>DOB: </Span><input type ="date" name ="DOB" /><br><br>
<Span>MembershipExpiry: </Span> <input type ="date" name ="MemX" /><br>
<Span>MembershipType: </Span><select name = "MemType">
<option value ="Junior">Junior</option>
<option value ="Adult">Adult</option>
<option value ="Senior">Senior</option>
</select><br><br>
<Span>Email Address: </Span><input type ="email" name ="Email" /><br><br>
<input type="Submit" name="submit" value ="Submit Form">
An Proccess.php (where the error seems to be):
<?php
include 'Connect.php';?>
<?php
//variables
$title = $_POST['Title'];
$fname = $_POST['Fname'];
$lname = $_POST['Lname'];
$gender = $_POST['Gender'];
$dob = $_POST['DOB'];
$memx = $_POST['MemX'];
$memtype = $_POST['MemType'];
$email = $_POST['Email'];
//query
mysqli_query($connect, "INSERT INTO records(Title,Fname,Lname,Gender,DOB,MemX,MemType,Email)
values ('$title', '$fname', '$lname', '$gender', '$dob', '$memx', '$memtype', '$email')");
Error:
The screen goes blank when submitting the form and there is no new record in the database. Any ideas?
Database table is called records and has columns:
Title
First Name
Last Name
Gender
etc
Please help me connect and upload the forms data to db.
Try to add ',' after $connect like this:
mysqli_query($connect, "INSERT INTO records(Title,Fname,Lname,Gender,DOB,MemX,MemType,Email)
values ('$title', '$fname', '$lname', '$gender', '$dob', '$memx', '$memtype', '$email')");
connection.php
<?php
$connect = mysqli_connect ("localhost", "root", "", "gym") or die('Error Connecting to Database!');
?>
process.php
<?php
//variables
$title = $_POST['Title'];
$fname = $_POST['Fname'];
$lname = $_POST['Lname'];
$gender = $_POST['Gender'];
$dob = $_POST['DOB'];
$memx = $_POST['MemX'];
$memtype = $_POST['MemType'];
$email = $_POST['Email'];
//query
$sql = 'INSERT INTO records (Title, Fname, Lname, Gender, DOB, MemX, MemType, Email) VALUES ( ?,?,?,?,?,?)';
if (! $stmt = mysqli_prepare($connection, $sql)) {
trigger_error('Error: ' . mysqli_error($connection), E_USER_WARNING);
}
mysqli_stmt_bind_param($stmt, 'ssssssss', $title, $fname, $lname, $gender, $dob, $memx, $memtype, $email);
mysqli_stmt_execute($stmt);
mysqli_stmt_close($stmt);
mysqli_close($db);
redirect('Form.php');
Check this out http://php.net/manual/en/book.mysqli.php
Related
I am new to PHP and web development, and trying to create an HTML form that will submit data into MYSQL.
Upon checking phpmyadmin after submission of the form, it shows that there has been a row submitted,
however the row is completely blank. I had a problem before this one, that instead of a blank row, it would be "1" submitting instead of the data inserted into the HTML form. Now, no data submits into the database.
Here is the PHP:
<?php
Include("connection.php");
// HTML Identification
$lname = isset($_POST['lastname']);
$fname = isset($_POST['firstname']);
$email = isset($_POST['email']);
$phone = isset($_POST['phonenum']);
$addr = isset($_POST['address']);
$city = isset($_POST['city']);
$state = isset($_POST['state']);
$zip = isset($_POST['zipcode']);
//Database Insertion
$sql= "INSERT INTO CustomerInfo (LastName, FirstName, Email, PhoneNum, Address, City, State, ZipCode)
VALUES ('$lname', '$fname', '$email', '$phone', '$addr', '$city', '$state', '$zip')";
// Insertion
$ds= mysqli_query($conn, $sql);
// - Insertion Confirmation
if($ds)
{
print 'Row Inserted!';
print ' Response Recorded!';
}
?>
The HTML Form:
!DOCTYPE html>
<html>
<head>
<title> GS Entry Form </title>
<link rel="stylesheet" href="https://cdn.jsdelivr.net/npm/water.css#2/out/water.css" </link>
<style>
h1 {text-align: center;}
h2 {text-align: center;}
</style>
</head>
<body>
<h1>Customer Entry Form</h1>
<h2>Please Input Contact Information</h2>
<form action="database.php" method="POST">
First Name:<br />
<input type="text" name="firstname" />
<br /><br />
Last Name:<br />
<input type="text" name="lastname" />
<br /><br />
Email:<br />
<input type="text" name="email" />
<br /><br />
Phone Number:<br />
<input type="text" name="phonenum"/>
<br /><br />
Address:<br />
<input type="text" name="address"/>
<br /><br />
City:<br />
<input type="text" name="city"/>
<br /><br />
State:<br />
<input type="text" name="state"/>
<br /><br />
Zip Code:<br />
<input type="text" name="zipcode"/>
<br /><br />
<button type="button" name= "submit" value= "submit" />
</form>
</body>
</html>
Here, also, is the connection.php referenced:
<?php
$servername = "xxx";
$username = "xxx";
$password = "xxx";
$dbname = "xxx";
// Create Connection
$conn= mysqli_connect("$servername:3306","$username","$password","$dbname");
// Check Connection
if ($conn->connect_error)
{
die("Connection failed: " .$conn->connect_error);
}
else echo "Connection successful! "
?>
I don't think it has anything to do with the connection, but I figured I would post it to cover all the bases. The attached imgur picture is what my database has been looking like after submissions have been made.
I truly am not sure what to do now, any help would be greatly appreciated.
Thank you! -G
EDIT:
This is what my PHP code looks like after the changes suggested from #EinLinuus:
<?php
Include("connection.php");
// HTML Identification POST
if(isset($_POST['firstname'])) {
$fname = $_POST['firstname'];
}else{
die("Firstname is missing");
}
if(isset($_POST['lastname'])) {
$lname = $_POST['lastname'];
}else{
die("Lastname is missing");
}
if(isset($_POST['email'])) {
$email = $_POST['email'];
}else{
die("Email is missing");
}
if(isset($_POST['phone'])) {
$phone = $_POST['phone'];
}else{
die("Phone Number is missing");
}
if(isset($_POST['addr'])) {
$addr = $_POST['addr'];
}else{
die("Address is missing");
}
if(isset($_POST['city'])) {
$city = $_POST['city'];
}else{
die("City is missing");
}
if(isset($_POST['state'])) {
$state = $_POST['state'];
}else{
die("State is missing");
}
if(isset($_POST['zip'])) {
$zip = $_POST['zip'];
}else{
die("Zip Code is missing");
}
//Database Insertion
mysqli_report(MYSQLI_REPORT_ERROR | MYSQLI_REPORT_STRICT);
$stmt= $conn->prepare("INSERT INTO CustomerInfo(FirstName, LastName, Email, PhoneNum, Address, City, State, ZipCode) VALUES (?, ?, ?, ?, ?, ?, ?, ?)");
$stmt->bind_param('ssssssss', $fname, $lname, $email, $phone, $addr, $city, $state, $zip);
$stmt->execute();
// Insertion
$sql= mysqli_query($conn, $stmt);
// - Insertion Confirmation
if($ds)
{
print 'Row Inserted!';
print ' Response Recorded!';
}
$stmt->close();
$conn->close();
?>
My HTML remains the same, besides adding ID attributes to each variable to no effect. I appreciate the help!
The isset function returns if the variable is declared or not -> the return type is a boolean.
$test = [
"hello" => "world"
];
var_dump(isset($test["hello"])); // bool(true)
var_dump(isset($test["something"])); // bool(false)
You can use isset to check if the field exists in the $_POST variable, but don't save the result of the isset function to the database. If you do so, the boolean will be converted to a number (true => 1, false => 0) and this number gets stored in the database.
Example:
if(isset($_POST['lastname'])) {
die("lastnameis missing");
}
$lname = $_POST['lastname'];
Security
This code is vulnerable to SQL Injections. You should never trust user input. I'd recommend to use prepared statements here:
$stmt = $mysqli->prepare("INSERT INTO CustomerInfo (LastName, FirstName, ...) VALUES (?, ?, ...)");
$stmt->execute([$lname, $fname]);
In the SQL statement, replace the actual values with ?. Now you can execute the statement and pass the values to the execute function. In the example above, $lname will replace the first ?, $fname the second, ...
html webpage screenshotphp code shown on button clickmySql database tableI need to store user login data. i am using phpMyAdmin. When I click on submit button, data is not stored. Instead the php code is shown. Both code files are given below. What I am doing wrong. Help me. I
am unable to store user data using phpmyadmin in xampp.
my html code
<html>
<head>
<title>Yahoo Signin And Signup Form</title>
</head>
<body>
<h2 style="color: midnightblue">yahoo!</h2>
<hr color="magenta">
<form method="post" action="connect.php" >
<fieldset style="background:#6495ED;">
<legend style="padding:20px 0; font-size:20px;">Signup:</legend>
<label for ="firstName">Enter First Name</label><br>
<input type="text" placeholder="First name" id="firstName" name ="firstName">
<br>
<label for ="lastName">Enter Last Name</label><br>
<input type="text" placeholder="Last name" id="lastName" name ="lastName">
<br>
<label for ="email">Enter Email</label><br>
<input type="text" placeholder="Email" id="email" name ="email"><br>
<label for ="password">Enter Password</label><br>
<input type="password" placeholder="Password" id="password" name ="password">
<br>
<label for ="number">Enter Mobile Number</label><br>
<input placeholder="03---" id="number" name ="number"><br>
<label for ="date">Enter Date of Birth</label><br>
<input type="text" placeholder="DD/MM/YY" id="date" name ="date"><br>
<label for ="gender">Enter Gender</label><br>
<input type="text" placeholder="Male/Female/Other" id="gender" name
="gender"><br>
<br><button style="background-color:orangered;border-
color:dodgerblue;color:lightyellow">Signup</button>
</fielsdet>
</form>
</body>
</html>
my connect.php
<?php
$firstName = $_POST['firstName'];
$lastName = $_POST['lastName'];
$email = $_POST['email'];
$password = $_POST['password'];
$number = $_POST['number'];
$date = $_POST['date'];
$gender = $_POST['gender'];
// Making Connection with database
$con = new mysqli('localhost','root','','phpdata');
if ($con -> connect_error) {
die('Connection failed :'.$conn -> connect_error);
}
else{
$stmt = $con->query("INSERT INTO signup(firstName, lastName, email, password,
number, date, gender)
values(?,?,?,?,?,?,?)");
$stmt->bind_param("ssssiss",$firstName, $lastName, $email, $password,
$number, $date, $gender);
$stmt->execute();
echo "Sign up successful";
$stmt->close();
$con->close();
}?>
Use prepare instead of query. All everything is ok.:
$stmt = $con->prepare("INSERT INTO signup(firstName, lastName, email, password, number, date, gender)
values(?,?,?,?,?,?,?)");
And make button type as submit:
<br><button type="submit" style="background-color:orangered;border-color:dodgerblue;color:lightyellow">Signup</button>
here is the code, it works fine with me
<?php
$firstName = $_POST['firstName'];
$lastName = $_POST['lastName'];
$email = $_POST['email'];
$password = $_POST['password'];
$number = $_POST['number'];
$date = $_POST['date'];
$gender = $_POST['gender'];
// Making Connection with database
$con = new mysqli('localhost','root','','phpdata');
if ($con -> connect_error) {
die('Connection failed :'.$conn -> connect_error);
}
else{
$stmt = $con->query("INSERT INTO signup(firstName, lastName, email, password, number, date, gender)
values("'.$firstName.'","'.$lastName.'","'.$email.'","'.$password.'","'.$number.'","'.$date.'","'.$gender.'")");
if ($con->query($stmt) === TRUE) {
echo "New record created successfully";
} else {
echo "Error: " . $sql . "<br>" . $conn->error;
}
$con->close();
}//end of else of connection
?>
Add type in your submit button.
<button type='submit' style="background-color:orangered;border-color:dodgerblue;color:lightyellow">Signup</button>
and also your question marks and params ara not matching. it should be match. otherwise data won't store your db
correct that line also
The main problem is you are not loading code via apache server try to open http://localhost/signup.html instead of C:/xmapp/htdocs/connect.php
It seems you want to user PDO but your connection string not correct
<?php
$firstName = trim($_POST['firstName']);
$lastName = trim($_POST['lastName']);
$email = trim($_POST['email']);
$password = md5(trim($_POST['password']));
$number = trim($_POST['number']);
$date = trim($_POST['date']);
$gender = trim($_POST['gender']);
$con= new PDO("mysql:host=127.0.0.1;dbname=phpdata", 'root', 'root');
$con->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
$sqli = "INSERT INTO signup(firstName, lastName, email, password, number, date, gender)
values(?,?,?,?,?,?,?)";
try {
$stmt= $con->prepare($sqli);
$stmt->bindParam(1,$firstName);
$stmt->bindParam(2,$lastName);
$stmt->bindParam(3,$email);
$stmt->bindParam(4,$password);
$stmt->bindParam(5,$number);
$stmt->bindParam(6,$date);
$stmt->bindParam(7,$gender);
$status = $stmt->execute();
echo "Sign up successful";
$stmt->close();
$con->close();
} catch(PDOException $e) {
echo "Error ".$e->getMessage();
}
?>
another problem is with your html form button type is missing
<button type="submit".... />
Here is the complete code after analyzing it for a lot of time. in your $stmt variable there was no query, it was empty. This code works fine just copy and paste it.
<?php
$firstName = $_POST['firstName'];
$lastName = $_POST['lastName'];
$email = $_POST['email'];
$password = $_POST['password'];
$number = $_POST['number'];
$date = $_POST['date'];
$gender = $_POST['gender'];
// Making Connection with database
$con = new mysqli('localhost','root','','abc');
if ($con -> connect_error) {
die('Connection failed :'.$conn -> connect_error);
}
else{
$sql = "INSERT INTO signup(firstName, lastName, email, password, number, date, gender)
values('$firstName','$lastName','$email','$password','$number','$date','$gender')";
if ($con->query($sql) === TRUE) {
echo "New record created successfully";
} else {
echo "Error: " . $sql . "<br>" . $con->error;
}
$con->close();
}//end of else of connection
?>
I wanna update just one data from a record using php form but the thing is, when i do that, the rest of the data gets removed from the record.. What do i do :/ here are my codes for updating. What is the mistake i am making.. I am very confused. Would really appreciate some help.
<?php
include('db.php');
if(isset($_POST['update']))
{
$hostname = "localhost";
$username = "root";
$password = "";
$databaseName = "winc sports";
$connect = mysqli_connect($hostname, $username, $password, $databaseName);
$id = $_POST['id'];
$fname = $_POST['fname'];
$lname = $_POST['lname'];
$age = $_POST['age'];
$country=$_POST['country'];
$phone=$_POST['phone'];
$email=$_POST['email'];
$select = "SELECT * FROM studens WHERE id = '$id'";
$selected = mysqli_query($connect, $select);
$row = mysqli_fetch_assoc($selected);
if (empty($_POST['fname'])) {$fname = $row['fname'];} else {$fname = $_POST['fname'];}
if (empty($_POST['country']))
{
$country = $row['country'];
}
else {
$country = $_POST['country'];
}
if (empty($_POST['id'])) {
$id = $row['id'];
}
else {
$id = $_POST['id'];
}
if (empty($_POST['age'])) {$age = $row['age'];} else {$age = $_POST['age'];}
if (empty($_POST['phone'])) {$phone = $row['phone'];} else {$phone = $_POST['phone'];}
if (empty($_POST['email'])) {$email = $row['email'];} else {$email = $_POST['email'];}
$query = "UPDATE students SET Fname= '$fname', Lname = '$lname', Nationality = '$country', PhoneNumber = '$phone', Email= '$email', Age = '$age' WHERE Id = '$id'";
$result = mysqli_query($connect, $query);
var_dump($result);
if($result)
{
echo 'Data Updated';
}else
{
echo 'Data Not Updated';
}
mysqli_close($connect);
}
?>
<!DOCTYPE html>
<html>
<head>
<title>PHP INSERT DATA USING PDO</title>
<meta charset="UTF-8">
<meta name="viewport" content="width=device-width, initial-scale=1.0">
</head>
<body>
<form action="updating.php" method="post">
<input type="text" name="id" placeholder="Enter new ID"><br><br>
<input type="text" name="fname" placeholder="Enter new First Name"><br><br>
<input type="text" name="lname" placeholder="Enter new Last Name"><br><br>
<input type="number" name="age" placeholder="Enter new age" min="13" max="90"><br><br>
<input type="text" name="country" placeholder="Enter new Nationality"><br><br>
<input type="number" name="phone" placeholder="Enter new Phone Number"><br><br>
<input type="text" name="email" placeholder="Enter new Email"><br><br>
<input type="submit" name="update" value="update">
</form>
</body>
</html>
The select statement is fetching data from a table called studens. This looks like a typo of the actual table so it won't actually fetch any results for you to update. Thus, the data you wind up updating the table with is empty. Rename the initial select table to students and it should properly fetch the data.
Also, please look into prepared statements or various other methods to sanitize inputs. Using POST variables directly in a query makes you extremely vulnerable to SQL Injection.
I am fairly new to PHP and I was following a simple tutorial on youtube, I followed the youtube video, double and tripple checked to make sure everything I typed was correct and data was still not being inserted.
I searched the internet for hours and I came up with a fix, sort of but I don't think it's the correct way to do it
HTML
<html>
<head>
<title>Insert Form Data In MYSQL Database Using PHP</title>
</head>
<body>
<form action="insert.php" method="post">
Name : <input type="text" name="username">
<br/>
Email : <input type="text" name="email">
<br/>
<input type="submit" value="Insert">
</form>
</body>
</html>
PHP
<?php
$con = mysqli_connect('localhost','root','');
if (!$con) {
echo 'Not Connected To Server';
}
if (!mysqli_select_db($con,'tutorial')) {
echo 'Database Not Selected';
}
if (isset($_POST['username'])){
$Name = $_POST['username'];
}
if (isset($_POST['email'])){
$Email = $_POST['email'];
}
$sql = "INSERT INTO person (Name, Email) VALUES ('John', 'john#gmail.com')";
if (!mysqli_query($con,$sql)) {
echo 'Not Inserted';
} else {
echo 'Inserted Successfully!';
}
header("refresh:10; url=index.html");
?>
I replaced '$Name' and '$Email' with John and john#gmail.com, then I type it into the html form and the data goes into the database correctly.
I then found another HTML form online with more PHP but it does the same thing(not inserting any data to the database)
HTML
<!DOCTYPE html>
<html lang="en">
<head>
<meta charset="UTF-8">
<title>Add Record Form</title>
</head>
<body>
<form action="insert1.php" method="post">
<p>
<label for="firstName">First Name:</label>
<input type="text" name="firstname" id="firstName">
</p>
<p>
<label for="lastName">Last Name:</label>
<input type="text" name="lastname" id="lastName">
</p>
<p>
<label for="emailAddress">Email Address:</label>
<input type="text" name="email" id="emailAddress">
</p>
<input type="submit" value="Submit">
</form>
</body>
</html>
PHP
<?php
/* Attempt MySQL server connection. Assuming you are running MySQL
server with default setting (user 'root' with no password) */
$link = mysqli_connect("localhost", "root", "", "demo");
// Check connection
if($link === false){
die("ERROR: Could not connect. " . mysqli_connect_error());
}
// Escape user inputs for security
$first_name = mysqli_real_escape_string($link, $_POST['firstname']);
$last_name = mysqli_real_escape_string($link, $_POST['lastname']);
$email_address = mysqli_real_escape_string($link, $_POST['email']);
// attempt insert query execution
$sql = "INSERT INTO persons (first_name, last_name, email_address) VALUES ('$first_name', '$last_name', '$email_address')";
if(mysqli_query($link, $sql)){
echo "Records added successfully.";
} else{
echo "ERROR: Could not able to execute $sql. " . mysqli_error($link);
}
// close connection
mysqli_close($link);
?>
The fields are blank, any help will be greatly appreacited!
Btw This is how the fields display I'm using xampp server.
I had used the below code and it works fine for me.
<?php
$link = mysqli_connect("localhost", "root", "", "dummy");
// Check connection
if($link === false){
die("ERROR: Could not connect. " . mysqli_connect_error());
}
/* Collect below values from $_POST
$firstname = 'John';
$lastname = 'Doe';
$email = 'test#gmail.com';
*/
// Escape user inputs for security
$first_name = mysqli_real_escape_string($link, $firstname);
$last_name = mysqli_real_escape_string($link, $lastname);
$email_address = mysqli_real_escape_string($link, $email);
// attempt insert query execution
$sql = "INSERT INTO accounts (account_firstname, account_lastname, account_email) VALUES ('$first_name', '$last_name', '$email_address')";
if(mysqli_query($link, $sql)){
echo "Records added successfully.";
} else{
echo "ERROR: Could not able to execute $sql. " . mysqli_error($link);
}
// close connection
mysqli_close($link);
?>
I am having an issue with my php not writing the values it is getting from my ajax script into my MySQL database. I know that the php script is getting the values because they are being echoed in my browser. but when i check my database, only two out of the five values are being inputted. I am sure this isn't a nuance, but I can't seem to crack this.
==============EDIT=============
The values that aren't being written are first name, last name, and job. ($fname, $lname, and $job respectively)
==============EDIT=============
PHP
<?php
//db connecting variables
$hostname = "foobase";
$username = "foobase";
$dbname = "contactformbase";
$password = "password";
$con = new mysqli($hostname, $username, $password, $dbname);
$tbl_name = "client_base";
//Connecting to your database
if ($con->connect_error) {
die('Connect error (' . mysqli_connect_errno() . ')' . mysqli_connect_errno());
}
echo 'success!...' . $con->host_info . "\n";
print_r($_POST);
$fname = $_POST['fname'];
$lname = $_POST['lname'];
$email = $_POST['email'];
$address = $_POST['address'];
$job = $_POST['job'];
$message = $_POST['message'];
//adding values into the database.
$sql="INSERT INTO $tbl_name (First Name, Last Name, Email, Address, Job)VALUES('POST_['first_name']', '$lname', '$address', '$email')";
$result = mysqli_query($con, $sql);
if($result){
echo "success";
}
else {
echo "error";
}
Javascript
<script type="text/javascript">
$("#submit").click(function(e) {
e.preventDefault();
var data_string = $("form#contact").serializeArray();
alert(data_string);
$.ajax({
type: "POST",
url: "database.php",
data: data_string,
success: function(){
alert(data_string);
}
});
return false;
</script>
HTML
<form action="" method="POST" id="contact">
<table>
<tbody>
<tr>
<td><h2>First Name: </h2></td>
<td><h2>Last Name: </td>
<td><h2>Email Address: </td>
</tr>
<tr>
<td><input type="text" name="first_name" placeholder="Johnny"></td>
<td><input type="text" name="last_name" placeholder="Appleseed"></td>
<td><input type="text" name="email" placeholder="johnny#email.com"></td>
</tr>
<tr>
<td><h2>Street Address:</h2></td>
<td><h2>What's Dirty?</h2></td>
</tr>
<tr>
<td><input type="text" name="address" placeholder="123 Applegrove Rd. Appletown, VA 12345"></td>
<td>
<select name="job" form="contact">
<option value="house">House</option>
<option value="roof">Roof</option>
<option value="garage-shed">Garage/shed</option>
<option value="other">Other</option>
</select>
</td>
</tr>
<tr>
<td><h2>Message: </h2></td>
</tr>
</tbody>
</table>
<textarea name="message">
</form>
You have the wrong field names in your PHP.
Change...
$fname = $_POST['fname'];
$lname = $_POST['lname'];
to
$fname = $_POST['first_name'];
$lname = $_POST['last_name'];
Also check the insert statement. It has wrong values against the fields - Address for Email, Eamil for Address.
$sql="INSERT INTO". $tbl_name ."(First Name, Last Name, Email, Address, Job) VALUES('". $_POST['first_name'] ."', '". $_POST['last_name'] ."', '". $_POST['adress'] ."', '". $_POST['email'] ."')";
You are using strange variables POST_['first_name'].
Try this query:
(Note: you have to use the quotes (' ') for the fields because you have a space in field names or use PDO to prepare and execute query.)
$sql="INSERT INTO $tbl_name (First Name, Last Name, Email, Address, Job)
VALUES('$fname', '$lname', '$address', '$email')";
Other issue is the POST array, you send first_name instead of fname and last_name instead of lname:
<tr>
<td><input type="text" name="first_name" placeholder="Johnny"></td>
<td><input type="text" name="last_name" placeholder="Appleseed"></td>
<td><input type="text" name="email" placeholder="johnny#email.com"></td>
</tr>
This will return:
$_POST = array(
"first_name" => "Johnny",
"last_name" => "Appleseed",
"email" => "johnny#email.com"
);
As you see you don't have $_POST['fname'] and $_POST['lname']
so you have to change:
$fname = $_POST['first_name'];
$lname = $_POST['last_name'];
About job, you just didn't add it in INSERT statement:
$sql="INSERT INTO $tbl_name (First Name, Last Name, Email, Address, Job)
VALUES('$fname', '$lname', '$address', '$email', '$job')";