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Question

Answer 1

+3 A:

Use jQuery and PHP.

This is an example in jQuery:

 $.post("test.php", { isChecked:$('#checkbox_id').attr('checked') } );

This statement should be defined at either in the <head></head>, or in a separate JavaScript file.

And in your PHP code you can check for the isChecked POST variable:

if($_POST('isChecked')==1)
{
  //Set MySQL table to 1.
}
else
{
  //Set MySQL table to 0.
}

Here is a reference on how to use post. And here's how you can use PHP to update MySQL table.

Ngu Soon Hui 2009-11-16 14:48:21

Hi, thanks, could you point me to any references?

Marvin 2009-11-16 14:50:46

Thank you for your example. Ill try implement it.

Marvin 2009-11-16 14:54:26

I have a doubt the jquery part, is that code included in each check box tag ex: <input (jquery code)> or is it a generic statement that's triggered each time a check box is ticked?

Marvin 2009-11-16 15:01:17

It's a generic statement; you should define it either in the `<head></head>`, or in a separate js file

Ngu Soon Hui 2009-11-16 15:03:23

Hi, could you please take a look at the post i just made? Thank you.

Marvin 2009-11-16 15:49:54

You have an error in your code. It should be $_POST['isChecked']

mculp 2009-11-16 15:55:04

Answer 2

+1 A:

You use the checkbox's onchange event handler to issue an XmlHttpRequest to some server-side code to update your table.

V a am Y o b 2009-11-16 14:49:00

Thank you, could you please provide some reference?

Marvin 2009-11-16 14:50:14

Answer 3

A:

I'm not sure what's wrong. This is my PHP for the test.php in the example from Ngu Soon Hui:

<?php
$con = mysql_connect("localhost","root","root");
if (!$con)
{
    die('Could not connect: ' . mysql_error());
}

mysql_select_db("FACT", $con);

if($_POST('isChecked')==1) // THIS IS LINE 21
{
    mysql_query("UPDATE Orc SET CON = 'CHECKED'
    WHERE UID = '1'");
}
else
{
    mysql_query("UPDATE Orc SET CON = 'NO'
    WHERE UID = '1'");
}

mysql_close($con);

?>

Nothing is hapening and if try to open the page (test.php) I'm faced with the following error:

Fatal error: Function name must be a string in /MAMP/htdocs/Fact/change.php on line 21

I can't seem to spot what's wrong.

Note: I'm just trying to update a single record just to test if it's working. My checkboxes names are dynamically generated but I think I know how to handle that part.

Marvin 2009-11-16 15:49:22

$_POST['isChecked']

mculp 2009-11-16 15:54:20

Thank you, I was completely blind to that one.

Marvin 2009-11-16 15:58:39

Answer 4

A:

Another update:

I'm still having trouble with this,

My PHP code for the test.php:

    $con = mysql_connect("localhost","root","root");
    if (!$con)
    {
        die('Could not connect: ' . mysql_error());
    }

    mysql_select_db("FACT", $con);

    if($_POST['isChecked']==TRUE)  // SEE THIS LINE PLEASE
    {
        mysql_query("UPDATE Orc SET CON = 'CHECKED'
        WHERE UID = '1'");
    }
    else
    {
        mysql_query("UPDATE Orc SET CON = 'NO'
        WHERE UID = '1'");
    }

    mysql_close($con);

I have commented one of the above lines, originally it was "== 1" but this worked only fine for deselecting, I could not change the value back to "CHECKED". Now I tried changing for "TRUE" and now I'm unable to set it to "NO"...

What is going wrong? And how does one go about debugging these kind of errors? Is there any way to have feedback from the JavaScript being executed?

Marvin 2009-11-17 09:14:06

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HTML checkbox change value in MySQL

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