表单的简单是或否复选框

Question

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.

Answer 1

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"/>