Insert into Database dynamically generated value

Question

I have a simple PHP question.

I need to add inside a DB the result of this JavaScript code

function test() {
        $('#percentage').html(Math.floor((Math.random() * 100 + 1)) + '%');
    };

that generate inside a div a random number. I have a couple of textbox with some other infos and hey work properly, i just need to add this one but i can't.

the code to add the other results and then to display them is this:

ADDING ELEMENTS TO DB:

include('dbconnection.php');

$Name = $_POST['name'];
$Surname = $_POST['surname'];
$Random = $_POST['random'];
$Sentence =$_POST['sentence'];

$sql = "INSERT INTO `results`(`Name`, `Surname`, `Random`, `Sentences`) VALUES('$Name', '$Surname','$Random', '$Sentence')";

$retval = mysql_query( $sql );
if(! $retval )
{
  die('Could not enter data: ' . mysql_error());
}

header('Location: index.php');

DISPLAY DB RESULTS:

include('dbconnection.php');

         $result = mysql_query("SELECT `Name`, `Surname`, `Random`, `Sentences` FROM `results`");

         while ($row = mysql_fetch_array($result))
         {
            $name = $row['Name'];
            $surname = $row['Surname'];
            $random = $row['Random'];

            echo"Name: $name -- Surname: $surname -- Random: $random</br>";}

thanks!

Danger! - Your code is susceptible to SQL Injection because it doesn't sanitize inputs. Please, don't use mysql_* functions in new code. They are no longer maintained and are officially deprecated. See the red box? Learn about prepared statements instead, and use PDO or MySQLi - this article will help you decide which. If you choose PDO, here is a good tutorial. — Benjamin Gruenbaum
– Benjamin Gruenbaum, Commented May 26, 2014 at 21:36
Add a hidden field to your form, set its value in test function, and access it via $_POST in php. — Steve
– Steve, Commented May 26, 2014 at 21:37

Tushar Dave · Accepted Answer · 2014-05-27 07:32:56Z

You need to have a hidden variable in your form that keeps the value of number and then using this variable you can insert values in yoyr db when your form is posted.

<input type="hidden" id="gen_value" name="gen_value" value=""/>

and in the js function you need to change

function test() {
        var calculatedValue=Math.floor((Math.random() * 100 + 1);
        $('#percentage').html(calculatedValue+'%');
        $('#gen_value').val(calculatedValue + '%');
    };

changes in dbinsert code:

  include('dbconnection.php');

    $Name = $_POST['name'];
    $Surname = $_POST['surname'];
    $Random = $_POST['random'];
    $Sentence =$_POST['sentence'];
    $perCentage=$_POST['gen_value'];

    $sql = "INSERT INTO `results`(`Name`, `Surname`, `Random`, `Sentences`,`Percentage`) VALUES('$Name', '$Surname','$Random', '$Sentence',`$perCentage`)";

    $retval = mysql_query( $sql );
    if(! $retval )
    {
      die('Could not enter data: ' . mysql_error());
    }

    header('Location: index.php');

Jinesh Gandhi · Accepted Answer · 2014-05-27 08:06:06Z

0

You simply follow bellow steps :

add hidden input type in your form

<input type="hidden" name="your_field" id="your_field" class="your_field" >

add below if you want to display in form.

function test() {
  var getValues = Math.floor((Math.random() * 100 + 1));
  $('#your_field').val(getValues + '%');

  /*if your create input type text and give it name "percentage" then use below line.otherwise don't copy it.*/

  $('#percentage').html(getValues + '%');

};

And in your php file get value by using this and then put $field_name in your insert query.

$field_name = $_POST['your_field'];

and then you can use your field value to display result if it have.

answered May 27, 2014 at 8:06

Jinesh Gandhi

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