Question

我试图从我的HTML文件中调用PHP脚本。

以下是我的HTML代码：

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
    <title>Welcome to B.O.A.T</title>
    <link rel="stylesheet" type="text/css" href="CSS/css.css" />
    <meta name="viewport" content="width=device-width, initial-scale=1.0" />
</head>

<body>
    <!-- Navigation Bar -->



    <div class="tab_content">
        <div id="home" class="tab_active">
                <div id="banner-container">
                    <div id="banner">
                        <p id="boat-banner"> Welcome to Bike Operation Automation Technology </p>
                    </div>
                </div>



                <div id="buttons">
                    <form method="post" action="lock.php">
                    <input type="button" name="Lock" class="button" value="LOCK" />
                    </form>

                    <form method="post" action="unlock.php">
                    <input type="button" name="Lock" class="button" value="UNLOCK" />
                    </form>

                    <form method="post" action="ledOn.php">
                    <input type="button" name="Lock" class="button" value="IGNITION ON" />
                    </form>

                    <form method="post" action="ledOff.php">
                    <input type="button" name="Lock" class="button" value="IGNITION OFF" />
                    </form>


                   <form method="post" action="GPSTracker.php">
                    <input type="submit" name="GPSTracker" class="button" value="GPS TRACKER"/>
                    </form>
                </div>
                </div>

         </div>

</body>
</html>

以下是我要调用的ledOn.php文件（它是那些未被调用的PHP脚本之一）

<html>
<head>
<title>LEDON</title>
</head>
<body>
<?php
    define('HOST','mysql.hostinger.in');
    define('USER','u414932932_usr');
    define('PASS','xyz123');
    define('DB','u414932932_bdb');

    $con = mysqli_connect(HOST,USER,PASS,DB);

    //$ign = $_POST['ignition'];


    $sql = "UPDATE BOATOP SET ignition='1' WHERE user_name='boatusr'";
    if(mysqli_query($con,$sql))
    {
        echo 'SUCCESS'; ?>
        <H3> SUCESS </H3>
    <?php 
    }
    else
    {
        echo "FAILURE";
        ?>
        <H3> FAILURE </H3>
      <?php
    }
    mysqli_close($con);
?>

</body>
</html>

当我点击 Ignition On 按钮时，没有任何反应。我不明白..其他任何按钮都没有打开各自的PHP文件。我哪里错了？

Answer 1

它是一个输入type="button"，它不会提交表单。这些应该是type="submit"

<div id="buttons">

    <form method="post" action="lock.php">
      <input type="submit" name="Lock" class="button" value="LOCK" />
    </form>

    <form method="post" action="unlock.php">
      <input type="submit" name="Lock" class="button" value="UNLOCK" />
    </form>

    <form method="post" action="ledOn.php">
      <input type="submit" name="Lock" class="button" value="IGNITION ON" />
    </form>

    <form method="post" action="ledOff.php">
      <input type="submit" name="Lock" class="button" value="IGNITION OFF" />
    </form>


    <form method="post" action="GPSTracker.php">
      <input type="submit" name="GPSTracker" class="button" value="GPS TRACKER"/>
    </form>

</div>

Answer 2

编辑：有人指出，我们可以使用输入类型提交或按钮类型提交。

功能方面，它们是相同的，但使用按钮元素创建的按钮提供了更丰富的渲染可能性。

https://www.w3.org/TR/html4/interact/forms.html#h-17.5

<div id="buttons">
        <form method="post" action="lock.php">
            <button type="submit" class="button" value="lock"> LOCK </button>
        </form>
        <form method="post" action="unlock.php">
            <button type="submit" class="button" value="unlock"> UNLOCK </button>
        </form>
        <form method="post" action="ledOn.php">
            <button type="submit" class="button" value="on"> IGNITION ON </button>
        </form>
        <form method="post" action="ledOff.php">
            <button type="submit" class="button" value="off"> IGNITION OFF</button>
        </form>
        <form method="post" action="GPSTracker.php">
            <button type="submit" class="button" value="gps'> GPSTracker </button>
        </form>
    </div>

Answer 3

在我看来，使用链接代替按钮可以提供良好的服务，因为您正在使用的表单中存在绝对 no 数据。

In [296]: A=sparse.csr_matrix([0,1,2,0,0,1])

In [297]: B=sparse.csr_matrix([0,0,0,1,0,1])

In [298]: C=sparse.csr_matrix([1,0,0,0,1,0])

In [299]: sparse.vstack([A,B,C])
Out[299]: 
<3x6 sparse matrix of type '<class 'numpy.int32'>'
    with 7 stored elements in Compressed Sparse Row format>

In [300]: sparse.vstack([A,B,C]).A
Out[300]: 
array([[0, 1, 2, 0, 0, 1],
       [0, 0, 0, 1, 0, 1],
       [1, 0, 0, 0, 1, 0]], dtype=int32)

In [301]: sparse.hstack([A,B,C]).A
Out[301]: array([[0, 1, 2, 0, 0, 1, 0, 0, 0, 1, 0, 1, 1, 0, 0, 0, 1, 0]], dtype=int32)

In [302]: np.vstack([A.A,B.A,C.A])
Out[302]: 
array([[0, 1, 2, 0, 0, 1],
       [0, 0, 0, 1, 0, 1],
       [1, 0, 0, 0, 1, 0]], dtype=int32)

可以成为

<form method="post" action="ledOn.php">
    <button type="submit" class="button" value="on"> IGNITION ON </button>
</form>

Answer 4

你可以用两种方式做到这一点

1）需要像上次表格一样将按钮更改为输入类型提交： -

<input type="submit" name="GPSTracker" class="button" value="GPS TRACKER"/>

2）你需要使用jQuery，你需要在表单中添加id并点击按钮上的事件，如： -

 <form method="post" action="lock.php" id="form_id">
  <input type="button" name="Lock" class="button" value="LOCK" onclick=" $('#form_id').submit();" />

我的HTML按钮不会打开PHP链接

4 个答案: