How to use variables from HTML form in PHP?

Question

I just want to receive the input from a HTML form in PHP. I did research for almost 2 hours trying every code I got, but it doesn't work. Here's my code:

HTML:

<!DOCTYPE html>
<html>
<head>
<title>Form site</title>
</head>
<body>
    <table width="300" border="0" align="center" cellpadding="0" cellspacing="1">
        <tr>
           <td>
              <form name="form1" method="post" action="connect.php">
              <table width="100%" border="0" cellspacing="1" cellpadding="3">
                 <tr>
                    <td colspan="3">Insert Data Into mySQL Database </td>
                 </tr>
                 <tr>
                    <td width="71">Name</td>
                    <td width="6">:</td>
                    <td width="301"><input name="name" type="text" id="name"></td>
                 </tr>
                 <tr>
                    <td>Lastname</td>
                    <td>:</td>
                    <td><input name="lastname" type="text" id="lastname"></td>
                 </tr>
                 <tr>
                    <td>Email</td>
                    <td>:</td>
                    <td><input name="email" type="text" id="email"></td>
                 </tr>
                 <tr>
                    <td colspan="3" align="center"><input type="submit" name="Submit" value="Submit"></td>
                 </tr>
              </table>
              </form>
           </td>
        </tr>
     </table></body>
</html>

PHP:

<?php
$mysqli = new mysqli("localhost", "root", "", "test");

if ($mysqli->connect_errno) {
    printf("Connect failed: %s\n", $mysqli->connect_error);
    exit();
}
$tbl_name="employees"; // Table name

// Get values from form
$name = $_POST['name'];
$lastname = $_POST['lastname'];
$email = $_POST['email'];


// if successfully insert data into database, displays message "Successful".
$sql = "INSERT INTO $tbl_name (name, lastname, email) VALUES ('$name', '$lastname', '$email')";
$mysqli->query($sql);
?>

So, when I run the HTML file in Chrome everything's normal, but if I then fill the form an click the Submit button Chrome says 'This page doesn't work'. The connection to the database works definitely. I read that it may help to set the AllowOverride option in the Apache config file (using XAMPP) to 'ALL'. It didn't. The filename specified in the HTML code matches the filename of the PHP file I'm actually using for this.

PHP is a server-side language, isn't it? So I suspect that I have to configure something to run the PHP file. PHP is already installed on my computer I checked the database connection using PHP in the command-line. Obviously I'm missing something. Does anyone have an idea?

Thanks in advance!

Thanks and shame on me but unfortunately it didn't solve my issue — Eiffelsturm, Aug 07 '19 at 10:00
@NiettheDarkAbsol Ok that could be the reason. But what is the workaround to archieve this goal? — Eiffelsturm, Aug 07 '19 at 11:26
As a side issue, you are vulnerable to SQL injection. You should check out [What is SQL injection?](https://stackoverflow.com/questions/601300/what-is-sql-injection). — Booboo, Aug 07 '19 at 13:31
@RonaldAaronson Thanks for the advice. I'll remember when I switch into production. — Eiffelsturm, Aug 07 '19 at 13:46

score 0 · Accepted Answer · answered Aug 07 '19 at 14:26

Solution: I used Apache as the webserver, but the HTML and PHP files were in another location, so PHP hadn't any webserver to run in.

Here's the working code: PHP:

?php
$mysqli = new mysqli("localhost", "root", "", "test");

if ($mysqli->connect_errno) {
    printf("Connect failed: %s\n", $mysqli->connect_error);
    exit();
}
$tbl_name="employees"; // Table name

// Get values from form
$name = $_POST['name'];
$lastname = $_POST['lastname'];
$email = $_POST['email'];

// if successfully insert data into database, displays message "Successful".
$sql = "INSERT INTO $tbl_name (name, lastname, email) VALUES (?, ?, ?)";
$statement = $mysqli->prepare($sql);
$statement->bind_param('sss', $name, $lastname, $email);

// $statement->execute();

if(!$statement->execute()) {
    echo "Query fehlgeschlagen: ".$statement->error;
}
else echo "All done!";

HTML:

<!DOCTYPE html>
<html>
<head>
<title>Form site</title>
</head>
<body>
    <table width="300" border="0" align="center" cellpadding="0" cellspacing="1">
       <tr>
       <td>
          <form name="form1" method="post" action="connect.php">
          <table width="100%" border="0" cellspacing="1" cellpadding="3">
             <tr>
                <td colspan="3">Insert Data Into mySQL Database </td>
             </tr>
             <tr>
                <td width="71">Name</td>
                <td width="6">:</td>
                <td width="301"><input name="name" type="text" id="name"></td>
             </tr>
             <tr>
                <td>Lastname</td>
                <td>:</td>
                <td><input name="lastname" type="text" id="lastname"></td>
             </tr>
             <tr>
                <td>Email</td>
                <td>:</td>
                <td><input name="email" type="text" id="email"></td>
             </tr>
             <tr>
                <td colspan="3" align="center"><input type="submit" name="Submit" value="Submit"></td>
             </tr>
           </table>
           </form>
        </td>
        </tr>
     </table></body>
</html>

score -1 · Answer 2 · answered Aug 07 '19 at 10:01

-1

First of all make sure that your html form is sending the data to your php file by adding this line to your php file just after the php tag

    echo '<pre>';
    print_r($_POST);

And you will see that you are receiving three variables i.e

name, lastname and email

now check your php code you're requesting the name as firstname.

answered Aug 07 '19 at 10:01

BlackXero

844
3
16

Do you mean '

' as a placeholder? Anyway, I pasted it in the second line and it still doesn't work. When I run it in the command-line it outputs `Array()`

– Eiffelsturm Aug 07 '19 at 10:43

so you are getting an empty array also you have to run it in browser now in terminal, terminal is not sending any data to your php code. – BlackXero Aug 07 '19 at 10:53
Normally I use the command-line for exspecting SQL Syntax Errors in the query inside PHP only. When I run it in a browser I don't get any error messages except "This page doesn't work" – Eiffelsturm Aug 07 '19 at 11:03

How to use variables from HTML form in PHP?

2 Answers2