表单的简单是或否复选框

I have inserted a checkbox in my form.

My code:

<input type="checkbox" id="checkbox" name="checkbox" value="1"/>

if($checkbox = ($_POST['checkbox']) == '1')
        {
            $checkbox = "si";
        }
        else
        {
            $checkbox = "no";
        }

I would like that if the checkbox is checked i receive "yes" otherwise "no". Thanks.

You've written wrong if condition here, You cannot use assignment in conditions.

Also there is no need to assign value to any variable in checking condition, You can directly use $_POST['checkbox']. Like this,

if($_POST['checkbox'] == '1') {
   $checkbox = "si";
} else {
   $checkbox = "no";
}

Update:

A better option is to use isset() which determine if a variable is set and is not NULL. Like this,

if(isset($_POST['checkbox'])) {
   $checkbox = "si";
} else {
   $checkbox = "no";
}

Program will go in if condition only when user has checked the checkbox. In above case value attribute for <input> is not required. So your HTML will look something like this,

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