评估 POST 表格

Question

<!DOCTYPE html>
<meta charset="utf-8">
<?php
     echo "<h1>LED Steuerung</h1>";
     exec('gpio -1 mode 22 out');
     if(isset($_POST['newstate']))
     {
            $neu  = $_POST['newstate'];
            exec('gpio -1 write 22 $neu');
     }

     $currently = exec('gpio -1 read 22');
     if($currently == '0')
     {
            echo " <p> LED is currently off</p>";
            echo " <form action = 'led.php' method='post'>
                    <intput type ='hidden'  name = 'newstate' value= '1'>
                    <input type = 'submit' value = 'LED einschalten'>
                    </form> ";
     }
     else
     {
            echo " <p> LED is currently on</p>";
            echo " <form action = 'led.php' method='post'>
                    <intput type ='hidden'  name = 'newstate' value= '0'>
                    <input type = 'submit' value = 'LED ausschalten'>
                    </form> ";

     }

?>

使用此代码我想控制 Raspberry Pi 上的 LED。但是无论如何它都行不通。已调用表单，但未对其求值。

if 语句：

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

它returns每次都是假的。

Answer 1

将 <intput type="hidden"> 重命名为 INPUT。

这解决了你的问题。

评估 POST 表格

Evaluate POST form

php