Question

我有一个实例，我的PHP网站上的用户可以登录，我试图在管理员发布公告的第二个实例。

我的登录工作非常精彩，但出于某种原因，我无法将我的公告提交给工作。尽管我的输入字段被正确命名，但该字段被视为＆＃34;未设置＆＃34;当我在表格上输入标题和公告时。这是我的代码：

FORM：

        <form action="announce.php">
    <input  type="text" name="title" />
    <textarea  name="announce" cols="20" rows="2" ></textarea>
    <input type="submit" name="submit" value="Announce" />
    </form>

这是我的PHP：

include_once 'creds.php';
$con=mysqli_connect("$db_hostname","$db_username","$db_password","$db_database");

if (isset($_POST['title']))
{
    $title = $_POST['title'];
    $announce = $_POST['announce'];

$query = "INSERT INTO announcements (labname, name, author, announce)
        VALUES ($lab, $title, $username, $announce)";

$insert = mysqli_query($con, $query);

mysqli_close($con);

echo "added successfully";}
else
{
echo "something went wrong";
}

我的其他形式似乎有效，但是这个没有...我的语法在某处错了吗？

Answer 1

您的表单是作为GET请求提交的，而不是POST。

更改此行：

<form action="announce.php">

到此：

<form action="announce.php" method="post">

编辑：我应该在发布之前刷新页面。所有归功于noobie-php在我发布此消息之前的comment 30分钟内获得此权限！

Answer 2

Php：

include_once 'creds.php';
$con=mysqli_connect("$db_hostname","$db_username","$db_password","$db_database");

if (isset($_POST['submit']))
{
    $title = $_POST['title'];
    $announce = $_POST['announce'];

$query = "INSERT INTO announcements (labname, name, author, announce)
        VALUES ($lab, $title, $username, $announce)";

$insert = mysqli_query($con, $query);

mysqli_close($con);

echo "added successfully";}
else
{
echo "something went wrong";
}

我认为if (isset($_POST['submit']))这就是问题所在。

if isset（$ _ POST问题

2 个答案: