HTML SQL server connection using PHP

Question

I have a HTML form and I want to use the results to filter the data available in a SQL file. I'm trying to connect to a local MySQL server but It doesn't work, i don't achieve to enter in the table() function.

I use the following code:

<form id="form_id" action="food_values.php" method="post" name="myform">
...
  <input onclick="table" type="button" value="Submit">

</form>

The php file contains the following code:

<html>
<body>
<?php
function table(){
    echo ("INSIDE");
    $con = mysql_connect("localhost","root"," ");
     if (mysqli_connect_errno())
        {
            echo "Failed to connect to MySQL: " . mysqli_connect_error();
        }
    else {
        echo "CONNECTED";
    }
}
?>
</body>
</html>

It doesn't show anything... Do you know where is the error?

You are treating the PHP as if it were javascript, which it is not. See the following for an example on how to properly submit a form to a PHP script: php.net/manual/en/tutorial.forms.php. — Jim Wright
– Jim Wright, Commented Jun 5, 2018 at 12:47
you have defined the function, but didn't call it. In this case you need to put table(); after the whole function table(){... } thing — cypherabe
– cypherabe, Commented Jun 5, 2018 at 12:48
Please modify you code both html and php. A lot of things are out of place. What is onclick doing there? — The Oracle
– The Oracle, Commented Jun 5, 2018 at 12:54
@cypherabe It is unclear if these are separate files or not, and I realised after what you meant so I deleted my comment. — Jim Wright
– Jim Wright, Commented Jun 5, 2018 at 12:55
Please don't mix mysql (which is deprecated!) and mysqli — Nico Haase
– Nico Haase, Commented Jun 5, 2018 at 12:56

Sasha_1879 · Accepted Answer · 2018-06-05 12:54:40Z

0

You can not execute a PHP-Function with "onclick". Execute PHP function with onClick

With a form, you load the whole .php-file if you click on the submit.

The following code should work for you:

<html>
<body>
<?php
    echo ("INSIDE");
    $con = mysql_connect("localhost","root"," ");
     if (mysqli_connect_errno())
        {
            echo "Failed to connect to MySQL: " . mysqli_connect_error();
        }
    else {
        echo "CONNECTED";
    }
?>
</body>
</html>

And the HTML

<form id="form_id" action="food_values.php" method="post" name="myform">
...
  <input type="submit" value="Submit">

</form>

answered Jun 5, 2018 at 12:54

Sasha_1879

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Comments

Suraj Libi · Accepted Answer · 2018-06-05 13:02:15Z

0

Change

<input onclick="table" type="button" value="Submit">

into

<input type="submit" name="submit" value="Submit">

As the action in form takes care of where to direct the data in php.

and in 'food_values.php' page you dont have to define function as its not javascript.

   <?php
$servername = "localhost";
$username = "username";
$password = "password";

// Create connection
$conn = mysqli_connect($servername, $username, $password);

// Check connection
if (!$conn) {
    die("Connection failed: " . mysqli_connect_error());
}
echo "Connected successfully";
?>

This will connect the database. And for validation of the form follow the SQL commands. For futher detail visit https://www.w3schools.com/php/php_form_validation.asp

edited Jun 5, 2018 at 13:02

answered Jun 5, 2018 at 12:55

Suraj Libi

5152 silver badges9 bronze badges

1 Comment

Jim Wright Over a year ago

There was a function definition, not a class.

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