Ajax將按鈕提交到php，然后提交到串行設備

Question

我試圖通過ajax和php來獲取在串行設備arduino上啟動功能的按鈕，但似乎無法弄清楚。

這是我的html：

 <script src="https://ajax.googleapis.com/ajax/libs/jquery/1.11.0/jquery.min.js"></script> <script type="text/javascript" src="http://code.jquery.com/jquery-1.11.0.min.js"></script> <script type="text/javascript"> $(function() { $('#contact_form').on('submit', function (e) { e.preventDefault(); $.ajax({ type: "POST", url: '/test/SubmitFormWORefresh.php', data: $('#contact_form').serialize(), success: function() { alert('form was submitted'); } }); return false; }); }); </script> <meta charset="utf-8"> <title>Enroll</title> </head> <div id="contact_form"> <form name="contact" action=""> <fieldset> <input type="submit" name="rcmd" class="button" id="submit_btn" value="Enroll" /><br /> </fieldset> </form> </div> 

這是我的PHP：

 <?php $verz="1.0"; $comPort = "/dev/ttyACM0"; /*change to correct com port */ $PHP_SELF="index.php"; //This php file locate it from root if (isset($_POST["rcmd"])) { $rcmd = $_POST["rcmd"]; switch ($rcmd) { case Stop: $fp =fopen($comPort, "w"); sleep(2); fwrite($fp, 1); /* this is the number that it will write */ fclose($fp); break; case Enroll: $fp =fopen($comPort, "w"); sleep(2); fwrite($fp, 3); /* this is the number that it will write */ fclose($fp); break; default: die('Crap, something went wrong. The page just puked.'); }/*end switch case*/ }/*end if statement*/ ?> 

當我運行它時，我得到一個對話框，提示該表單已提交，但是串行設備沒有響應。 任何幫助將不勝感激。

Answer 1

我更改監聽器以檢測單擊按鈕，然后使用$(this).val()獲取值

Javascript：

<script type="text/javascript">
$(function() {  
    $('#contact_form').on('click', '.button',function (e) { 
        e.preventDefault();
        $.ajax({    
            type: "POST",
            url: '/test/SubmitFormWORefresh.php',
            data: 'rcmd='+$(this).val(),
            success: function() {
                alert('form was submitted');

            }
        });
        return false;
    });
});
</script>

HTML：

<input type="button" class="button" id="submit_btn" value="Enroll" />