Why doesnt HTML-form insert values into Mysql?

Question

So the PHP script is actually having a connection with the server, but somehow the database at localhost doesn't fill. What did I possibly do wrong?

script.php

<?php
$link = mysqli_connect("127.0.0.1", "root", "", "testdb");

if (!$link) {
echo "Error: Unable to connect to MySQL." . PHP_EOL;
echo "Debugging errno: " . mysqli_connect_errno() . PHP_EOL;
echo "Debugging error: " . mysqli_connect_error() . PHP_EOL;
exit;
}

echo "Success: A proper connection to MySQL was made! The my_db database is 
great." . PHP_EOL;
echo "Host information: " . mysqli_get_host_info($link) . PHP_EOL;

$klas = isset($_POST['klasDown']);
$naam = isset($_POST['naamTxt']);

$stmt = $link->prepare("INSERT INTO resultaten (Klas, Naam) VALUES (?, ?");
if (!$stmt) {
echo "false";
} else {
$stmt->bind_param("ss", $klas, $naam);
$stmt->execute();
}


mysqli_close($link);
?>

HTML:

<form id="myForm" method="post" action="script.php">
Naam: <input name="naamTxt" type="text" maxlength="512" 
id="naamTxt" class="searchField"/> <br>

Klas: <select name="klasDown">
  <option value="H4A" selected="selected">H4A</option>
  <option value="H4B" >H4B</option>
  <option value="H4C">H4C</option>
  <option value="H4C">H4D</option>
  <option value="H4C">V4A</option>
  <option value="H4C">V4B</option>
  <option value="H4C">V4C</option>
  <option value="H4C">V4D</option>  
</select> 


<div class="row">
<div class="col-4">1. Kies 1/5 </div>
<div class="col text-center"><input type="radio" name="v1" id="v1_1" 
value="Helemaal mee oneens"></div>
<div class="col text-center"><input type="radio" name="v1" id="v1_2" 
value="Deels oneens"></div> 
<div class="col text-center"><input type="radio" name="v1" id="v1_3" 
value="Neutraal"></div> 
<div class="col text-center"><input type="radio" name="v1" id="v1_4" 
value="Deels mee eens"></div> 
<div class="col text-center"><input type="radio" name="v1" id="v1_5" 
value="Helemaal mee eens"></div>
</div>

<input type= "submit" style="text-align: center" 
onclick="bereken()">
</form>

^ Above is the Html Form

Bereken() function:

function bereken() {

    var e = 0;
    var c = 0;


    if(document.getElementById('v1_1').checked)  {
      c = c - 0
    }else if(document.getElementById('v1_2').checked)  {
      c += 1;
    }else if(document.getElementById('v1_3').checked)  {
      c += 2;
    }else if(document.getElementById('v1_4').checked)  {
      c += 3;
    }else if(document.getElementById('v1_5').checked){
      c += 4;
    }

    var klas = li.options[li.selectedIndex].value;
    var naam = document.getElementById('searchTxt').value;

    alert(klas + ", " + naam);
}

^ This is the Javascript bereken() function . Apparently too much code, so have to add some text.

PHPMyAdmin is not a database. It is a web interface for your MySQL database. — Jay Blanchard
– Jay Blanchard, Commented Aug 22, 2017 at 21:37
Note: The object-oriented interface to mysqli is significantly less verbose, making code easier to read and audit, and is not easily confused with the obsolete mysql_query interface. Before you get too invested in the procedural style it’s worth switching over. Example: $db = new mysqli(…) and $db->prepare("…”) The procedural interface is an artifact from the PHP 4 era when mysqli API was introduced and should not be used in new code. You're mixing and matching styles here for no apparent reason. Try and use the newer form exclusively. — tadman
– tadman, Commented Aug 22, 2017 at 22:23
Consistency is important. In programming every character can matter, and all too often a simple mistake caused by confusion or a typo can take hours or days to eliminate. Try to keep things as organized as practical, as consistent as possible, and you won't make silly mistakes like that. — tadman
– tadman, Commented Aug 23, 2017 at 3:51

Felippe Duarte · Accepted Answer · 2017-08-22 21:20:31Z

2

Replace

$klas = isset($_POST['klasDown']);
$naam = isset($_POST['naamTxt']);

with

$klas = $_POST['klasDown'];
$naam = $_POST['naamTxt'];

isset will return a boolen, not a string that you are expecting.

answered Aug 22, 2017 at 21:20

Felippe Duarte

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16 Comments

Y_Lakdime Over a year ago

Thanks for you reply, but now I get the messages: Notice: Undefined index: klasDown

Y_Lakdime Over a year ago

and Notice: Undefined index: naamTxt

Felippe Duarte Over a year ago

Where those data are supposed to come from?

Y_Lakdime Over a year ago

From a html page with a form, I will update the question

Felippe Duarte Over a year ago

Everything seems to be ok. Are you using a <button type="submit"> or submitting with Ajax?

|

Hamis Hamis · Accepted Answer · 2017-08-23 04:40:24Z

0

On script.php do the following

If(isset($_POST['submit''])){

$klas = $_POST['klasDown'];

$naam = $_POST['naamTxt'];

$stmt = $link->prepare("INSERT INTO resultaten (Klas, Naam) VALUES (?, ?");

if (!$stmt) {

echo "false";

} else {

$stmt->bind_param('ss', $klas, $naam);

$stmt->execute();

}

Replace the button With the following

input type= "submit"

name="submit"

value=" Verstuur"

style="text-align: center"

onsubmit="bereken()"

edited Aug 23, 2017 at 4:40

answered Aug 23, 2017 at 4:33

Hamis Hamis

1072 silver badges9 bronze badges

10 Comments

Y_Lakdime Over a year ago

Just did that, but still no records in my DB :(. The error messages like 'Notice: Undefined index: naamTxt' are gone though

Hamis Hamis Over a year ago

just try to rename your input name and use another name

Y_Lakdime Over a year ago

if(isset($_POST['submit'])){ <- should this get the same name then?

Hamis Hamis Over a year ago

Follow The following example and apply 100% works <form id="myForm" method="post" action="index.php"> Naam: <input name="naamTxt" type="text" maxlength="512" id="naamTxt" class="searchField"/> <br> <input type="submit" name="submit"> </form> <?php If(isset($_POST['submit'])){ echo $naam = $_POST['naamTxt']; } ?>

Hamis Hamis Over a year ago

just copy and test

|

Professor Abronsius · Accepted Answer · 2017-08-23 11:48:25Z

I hope the following might prove of use in getting the database updated. The javascript function itself now submits the form ( based upon original idea ) and by doing so allows you to use an alert statement with the values to be sent. I was initially getting an error about undefined li in the javascript function - hence using querySelector to identify the form element.

In this instance the form submits to the same page but the form action could easily be set as you had previously, I only did this to write the code now. The sql uses prepared statements - not tested but should(?!) be OK.

<?php
    if( $_SERVER['REQUEST_METHOD']=='POST' ){

        try{

            $dbuser='root';
            $dbpwd='';
            $dbhost='localhost';
            $dbname='testdb';

            $link=new mysqli( $dbhost, $dbuser, $dbpwd, $dbname );


            if( !$link ) {
                $message=sprintf( "Error: Unable to connect to MySQL.\nDebugging errno: %s\nDebugging error: %s" );
                throw new Exception( $message, $link->errno, $link->error );
            }

            $klas = !empty( $_POST['klasDown'] ) ? filter_input( INPUT_POST, 'klasDown', FILTER_SANITIZE_STRING ) : false;
            $naam = !empty( $_POST['naamTxt'] ) ? filter_input( INPUT_POST, 'naamTxt', FILTER_SANITIZE_STRING ) : false;

            if( !$klas ) throw new Exception('klas cannot be empty');
            if( !$naam ) throw new Exception('naam cannot be empty');

            $stmt = $link->prepare( "insert into `resultaten` (`klas`, `naam`) values ( ?, ? )" );
            if( $stmt ){

                $stmt->bind_param( 'ss', $klas, $naam );
                $result=$stmt->execute();
                if( !$result )throw new Exception('SQL insert failed');

            } else {
                throw new Exception('Failed to prepare SQL query');
            }


            $stmt->free_result();
            $stmt->close();
            $link->close();

        }catch( Exception $e ){
            echo $e->getMessage();
        }

    }
?>
<!doctype html>
<html>
    <head>
        <title>Bereken</title>
        <script>

            function bereken() {
                var e = 0;
                var c = 0;

                if(document.getElementById('v1_1').checked)  {
                  c = c - 0
                }else if(document.getElementById('v1_2').checked)  {
                  c += 1;
                }else if(document.getElementById('v1_3').checked)  {
                  c += 2;
                }else if(document.getElementById('v1_4').checked)  {
                  c += 3;
                }else if(document.getElementById('v1_5').checked){
                  c += 4;
                }
                var li=document.querySelector('select[name="klasDown"]')
                var klas = li.options[li.selectedIndex].value;
                var naam = document.getElementById('naamTxt').value;

                alert( klas + ", " + naam );
                document.forms['myForm'].submit();
            }

        </script>
    </head>
    <body>
        <form id="myForm" method="post">
            Naam: <input name="naamTxt" type="text" maxlength="512" id="naamTxt" class="searchField"/> <br>

            Klas: <select name="klasDown">
              <option value="H4A" selected="selected">H4A</option>
              <option value="H4B" >H4B</option>
              <option value="H4C">H4C</option>
              <option value="H4C">H4D</option>
              <option value="H4C">V4A</option>
              <option value="H4C">V4B</option>
              <option value="H4C">V4C</option>
              <option value="H4C">V4D</option> 
            </select> 


            <div class="row">
                <div class="col-4">1. Kies 1/5 </div>
                <div class="col text-center">
                    <input type="radio" name="v1" id="v1_1" value="Helemaal mee oneens">
                </div>
                <div class="col text-center">
                    <input type="radio" name="v1" id="v1_2" value="Deels oneens">
                </div> 
                <div class="col text-center">
                    <input type="radio" name="v1" id="v1_3" value="Neutraal">
                </div> 
                <div class="col text-center">
                    <input type="radio" name="v1" id="v1_4" value="Deels mee eens">
                </div> 
                <div class="col text-center">
                    <input type="radio" name="v1" id="v1_5" value="Helemaal mee eens">
                </div>
            </div>
            <button type="button" style="text-align: center" onclick="bereken()">Verstuur</button>
        </form>
    </body>
</html>

You could copy the php to a new file script.php and set the form action as you had originally. I only put it all in one document like this for convenience when writing the code.

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