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isset($ _ POST ['submit'])不起作用,但是$ _SERVER [“ REQUEST_METHOD”] ==“ POST”起作用

[英]isset($_POST['submit']) doesn't work but $_SERVER[“REQUEST_METHOD”] == “POST” does

I'm trying to get my button to work. 我正在尝试让我的按钮正常工作。 It works with $_SERVER["REQUEST_METHOD"] == "POST" but not with isset($_POST['submit']) . 它适用于$_SERVER["REQUEST_METHOD"] == "POST"但不适用于isset($_POST['submit']) The problem with $_SERVER["REQUEST_METHOD"] == "POST" is that it gives me multiple entries. $_SERVER["REQUEST_METHOD"] == "POST"是它给了我多个条目。

My code: 我的代码:

    if(isset($_POST['submit']))
    {
    $server_address = $_POST['submitServerIp'];
    $server_port = $_POST['submitServerPort'];


    $Query = new SourceQuery( );

    try
    {
        $Query->Connect( $server_address, $server_port, SQ_TIMEOUT, SQ_ENGINE );

        $data = $Query->GetInfo( );
    }
    catch( Exception $e )
    {
        echo $e->getMessage( );
        $error = true;
    }
    $Query->Disconnect( );
if($error == false)
{
    $errorMessage = "false";
    // Create connection
    $conn = mysqli_connect($servername, $username, $password, $dbname);
    // Check connection
    if (!$conn) {
    die("Connection failed: " . mysqli_connect_error());
}

    $stmt = $conn->prepare("INSERT INTO Servers (ipaddress, port) VALUES (?, ?)");
    $stmt->bind_param("ss", $ipaddress, $port);

    $ipaddress = $_POST['submitServerIp'];
    $port = $_POST['submitServerPort'];
    $stmt->execute();

    $stmt->execute();

    $stmt->close();
    mysqli_close($conn);
}
}
?>

HTML: HTML:

<form method="post" action="">
<div class="container-fluid search-top">
    <div class="addContainer">
            <div class="addContainer col-sm-12">
                <p class="searchText"><?php echo htmlspecialchars($errorMessage);?></p>
                <div class="container col-md-8 col-md-offset-2 col-sm-10 col-sm-offset-1 col-xs-12 ">
                    <div class="input-group stylish-input-group">
                    <input type="text" name="submitServerIp" class="form-control input-lg" placeholder="Hostname/IP" style="border:0;">
                    <input type="text" name="submitServerPort" class="form-control input-lg" placeholder="Port" style="border:0;">
                        <span class="input-group-addon">
                        <button class="btn btn-primary" type="submit" value="submit">
                            <span class="fas fa-plus" aria-hidden="true"></span>
                        </button>
                        </span>
                    </div>
                </div>
            </div>
    </div>

Can I not use the button the way I use it or am I doing something else wrong? 我可以不按我的方式使用按钮,还是我做错了其他事情?

When you use a $_POST[KEY] variable, the key is actually the name of the form element that you submitted, not the value . 当您使用$_POST[KEY]变量时,键实际上是您提交的表单元素的name ,而不是value

To solve your issue, simply add name="submit" to your submit button. 要解决您的问题,只需将name="submit"添加到您的提交按钮。

<button class="btn btn-primary" type="submit" name="submit" value="submit">
    <span class="fas fa-plus" aria-hidden="true"></span>
</button> 

As for the 2 entries in the database, you have an extra execute() function being called. 至于数据库中的2个条目,您将调用一个额外的execute()函数。

    $stmt = $conn->prepare("INSERT INTO Servers (ipaddress, port) VALUES (?, ?)");
    $stmt->bind_param("ss", $ipaddress, $port);

    $ipaddress = $_POST['submitServerIp'];
    $port = $_POST['submitServerPort'];
    $stmt->execute();

    $stmt->execute();

    $stmt->close();
    mysqli_close($conn);

You should remove one of the above lines that say $stmt->execute(); 您应该删除上面的行之一,即$stmt->execute(); .

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