Question

当我点击提交按钮时，推算数据应该进入下面的查询，但事实并非如此。

似乎正在使用按钮的值。

有人能弄明白为什么它不起作用。它与我的登录页面使用相同的代码，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>

修改

仍然没有以这种方式改变：

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

Answer 1

从查询中删除,：

SELECT id ,

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

Answer 2

使用按钮的值是因为您为提交按钮指定的名称是“room”：您用于输入文本的名称。从提交按钮中删除name属性，它应该可以工作。

编辑后

更改

header("Location: ../views/student/question.php?id='" . $_SESSION["Room_ID"] . "'");

到

header("Location: ../views/student/question.php?id=" . $_SESSION["Room_ID"]);

您不必在查询字符串中引用变量

从表单中发帖无效

2 个答案: