PHP: Submit checkbox to database

Question

I am having trouble submitted checkbox values and details to my database.

This is my HTML:

<form method="get" action="check.php">
   <input type="checkbox" name="checkAccount"/>
   <input type="hidden" name="table_name" value="masterChart" />
   <input type="hidden" name="column_name" value="account" />
   <p><a href='check.php'><input type="submit" class="btn btn-primary" value="Submit" /></a></p>
</form>

This is the check.php:

$table = $_GET['table_name'];
$column = $_GET['account'];
 
$dbc = mysqli_connect('localhost', 'root', 'root', 'database') or die('Connection error!');

if ($value = 1) {
 $checkbox = "INSERT INTO login_table_display(`user`, `table`, `column`, `value`) VALUES(`:user`, '$table', '$column', `$value`)";
 mysqli_query($dbc, $checkbox) or die('Database error, check!');
 }
 
header('location:index.php');

As you can see above, I used variables to get other details for that checkbox to insert into the table as well.

After I press submit if the checkbox is checked, this is what's seen in the url:

http://localhost/admin/check.php?checkAccount=on&table_name=masterChart&column_name=account

Any suggestions or help will be appreciated!

Reporter · Accepted Answer · 2013-06-11 10:03:42Z

2

The classic way to submit data is to add the value attribute to your checkboxes element in your form. On server side you have to ckeck the value for "null".

<input type="checkbox" name="checkAccount" value="putyourvaluehere"/>

answered Jun 11, 2013 at 10:03

Reporter

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3 Comments

designtocode Over a year ago

Ok I get that, so if I put value="1" it will do the insert? and if the checkbox is unchecked, it will automatically do the else statement?

Reporter Over a year ago

$ckeckAccount = $_GET['checkAccount']; if ($ckeckAccount != null) {//do something}

designtocode Over a year ago

It's working thanks, but I think there's something with my Insert statement, can you please take a look? I'll add it to the main post

Goutam Pal · Accepted Answer · 2013-06-11 10:12:30Z

2

Your Html is not ok

It should be

<form method="get" action="check.php">
            <input type="checkbox" name="checkAccount"/>
            <input type="hidden" name="table_name" value="masterChart" />
            <input type="hidden" name="column_name" value="account" />
            <p><input type="submit" class="btn btn-primary" value="Submit" /></p>
    </form>

Also
if(isset($_POST['checkAccount']) {

Should Be
if( isset($_POST['checkAccount']) ) {

answered Jun 11, 2013 at 10:12

Goutam Pal

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Comments

matino · Accepted Answer · 2013-06-11 09:54:49Z

1

Checkbox value will be submitted only when it's checked. Use isset($_GET['checkAccount']) for this:

$var= isset($_GET['checkAccount']) ? 1 : 0; // Or whatever values you use in DB

answered Jun 11, 2013 at 9:54

matino

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1 Comment

designtocode Over a year ago

I'm not too sure how to add it.. So I take out the if statement and put the $var = isset($_GET['checkAccount']) ? 1: 0; and then add the mySql Insert?

Nono · Accepted Answer · 2013-06-11 10:05:02Z

1

Try this:

First you have to edit your html code as below;

<form method="get" action="check.php">
    <input type="checkbox" name="checkAccount" value='cool'/>
    <input type="hidden" name="table_name" value="masterChart" />
    <input type="hidden" name="column_name" value="account" />
    <p><input type="submit" class="btn btn-primary" value="Submit" /></p>
</form>

you are not giving value to check box and using submit button inside a tag, it's not good practice.

answered Jun 11, 2013 at 10:05

Nono

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Comments

Radim Köhler · Accepted Answer · 2013-06-11 10:19:24Z

0

Replace:

 if(isset($_POST['checkAccount'])

to:

 if(isset($_GET['checkAccount'])

edited Jun 11, 2013 at 10:19

Radim Köhler

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answered Jun 11, 2013 at 9:58

Rajendra Yadav

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Comments

Kaviranga · Accepted Answer · 2023-01-30 06:43:16Z

// your html code shoul be like this

<form method="get" action="check.php">

        <input type="checkbox" name="checkAccount"/>

        <input type="hidden" name="table_name" value="masterChart" />

        <input type="hidden" name="column_name" value="account" />

        <p><input type="submit" class="btn btn-primary" value="Submit" /></p>
</form>

<?php

    $table = $_GET['table_name'];

    $column = $_GET['account'];

    $value = isset($_GET['checkAccount']) ? 1 : 0;

    $dbc = mysqli_connect('localhost', 'root', 'root', 'database') or die('Connection error!');

    if ($value == 1) {

        $checkbox = "INSERT INTO login_table_display('user', 'table', 'column', 'value') VALUES(':user', '$table', '$column', '$value')";

        mysqli_query($dbc, $checkbox) or die('Database error, check!');

    }

     header('location:index.php');

?>

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