php - 从表格发帖不起作用

Question

当我点击提交按钮时，估算的数据应该进入下面的查询，但它没有。

好像按钮的值被代替了。

有人能弄清楚为什么它不起作用。它与我的登录页面使用的代码相同，并且有效。

它转到 if 语句的 else 部分

错误:

Error: You have an error in your SQL syntax; check the manual that corresponds to your MariaDB server version for the right syntax to use near 'FROM test_set WHERE Room_Code='room'' at line 1

查询:

if (isset($_POST['room']))
    {
    require "connect.php";

    if (count($_POST) > 0)
        {
        $result = mysqli_query($conn, "SELECT id, FROM test_set WHERE Room_Code='" . $_POST["room"] . "'");
        if (!$result) {
    printf("Error: %s\n", mysqli_error($conn));
    exit();
}
        $row = mysqli_fetch_array($result);
        if (is_array($row))
            {
            $_SESSION["Room_ID"] = $row['id'];
            header("Location: ../views/student/question.php?id='" . $_SESSION["Room_ID"] . "'");
            }
          else
            {
            echo "No";
            }
        }
    }

表格:

 <form method="POST" name="room" action="../../config/functions.php">
            <label for="room" class="sr-only">Room Code</label>
            <input type="text" id="room" name="room" class="form-control" placeholder="Please Enter Room Code">
            <br>
            <input type="submit" class="waves-effect waves-light btn blue darken-3" name="room" value="room">
        </form>

编辑:

仍然不工作也以这种方式改变:

$result = mysqli_query($conn, "SELECT id FROM test_set WHERE Room_Code='" . $_POST["room"] . "'");

Answer 1

从您的查询中删除 ,:

SELECT id,

$result = mysqli_query($conn, "SELECT id FROM test_set 
WHERE Room_Code='" . $_POST["room"] . "'");