Get checkbox value with php

Question

I try to determine if a checkbox is checked or not, but i get an error.

test.php

<html>
    <body>
        <form method="post" action="<?php echo $_SERVER['PHP_SELF'];?>">

            Use Proxy : <input type="checkbox" name="use_proxy"><br><br>

            <input type="submit">
        </form>

        <?php 
            $use_proxy  = $_POST['use_proxy'];

            if ($use_proxy != "on")
            {
                $use_proxy = "off";
            }

            echo "<p> use_proxy = " . $use_proxy . "</p><br>"; 
        ?>
    </body>
</html>

I get this error: Notice: Undefined index: use_proxy in C:\xampp\htdocs\mbcl\checkbox_test.php on line 11 How can i solve it?

your code seems correct.

Happy Coding
– Happy Coding

2015-09-21 11:11:38 +00:00
Commented Sep 21, 2015 at 11:11 — Happy Coding
– Happy Coding, Commented Sep 21, 2015 at 11:11

Rohit Kumar · Accepted Answer · 2015-09-21 11:11:05Z

3

This is the behaviour of checkbox until they are checked , they can not be fetched at backend(PHP) . You can try as below-

<html>
    <body>
        <form method="post" action="<?php echo $_SERVER['PHP_SELF'];?>">

            Use Proxy : <input type="checkbox" name="use_proxy" value="off"><br><br>

            <input type="submit">
        </form>

        <?php 
            $use_proxy  = isset($_POST['use_proxy'])?"on":"off";
            echo "<p> use_proxy = " . $use_proxy . "</p><br>"; 
        ?>
    </body>
</html>

answered Sep 21, 2015 at 11:11

Rohit Kumar

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6 Comments

CodeGodie Over a year ago

Obviously because you weren't using ISSET

Black Over a year ago

@CodeGodie And why is it necessary to use isset? The logik looks totaly ok.

CodeGodie Over a year ago

Because ISSET will check if POST actually exists. Initially before you submit your form, There are no POST requests, thus giving you those errors

Rohit Kumar Over a year ago

@EdwardBlack ,one more thing , what you are getting that is not "error" , they are "warning" in real . You would be overcome this by using "error_reporting(0) " too.

CodeGodie Over a year ago

You should not turn off error reporting. These warning/errors will help you make better code

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