查询解决方案

Question

我在启动项目时遇到查询问题。它说

解析错误：语法错误，C：\Users\ahmed\Documents\zadacaveb\Call Center\Dateof_seminars\forma.php第 34 行中出现意外的“$sql”（T_VARIABLE）

当我按下按钮 Dodaj （engl.添加）。我没有看到我的查询有问题，所以如果你看到我的查询中哪里有错误，请告诉我。我将展示我的forma代码.php这是一个问题，以及来自文件索引.php和删除.php的代码。这两个文档工作正常，但问题是形式上的.php

索引.php

<?php

require('../util/session.php');

require('../util/db.php');

require('../Frontend/header.php');

require('../Frontend/footer.php');

if(!isset($_SESSION['user'])) {

    header('Location: /Call Center/Login/index.php');

}

$stmt = $db->query('SELECT * FROM dateof_seminars');

$rows = $stmt->fetchAll();

echo $header;

?>

<br/>

<br/>

<div class="container">

    <a href="/Call Center/Dateof_seminars/forma.php" class="btn btn-primary mt-3 mb-3 float-left">Dodaj</a>

    <br/>

    <br/>

    <table class="table">

      <thead>

        <tr>

          <th scope="col">#</th>

          <th scope="col">Počinje datuma</th>

          <th scope="col">Završava datumom</th>

          <th scope="col">Instruktor ID</th>

          <th scope="col">Seminar ID</th>

          <th scope="col">Klijent ID</th>

        </tr>

      </thead>

      <tbody>

        <?php foreach($rows as $row) { ?>

      <tr>

          <th scope="row"><?php echo $row['id']; ?></th>

          <td><?php echo $row['date_from']; ?></td>

          <td><?php echo $row['date_to']; ?></td>

          <td><?php echo $row['instructor_id']; ?></td>

          <td><?php echo $row['seminars_id']; ?></td>

          <td><?php echo $row['clients_id']; ?></td>

          <td>

          <a class="btn btn-primary" href="/Call Center/Dateof_seminars/forma.php?id=<?php echo $row['id']; ?>">Edit</a>

          <a class="btn btn-danger" href="/Call Center/Dateof_seminars/delete.php?id=<?php echo $row['id']; ?>">Delete</a>

        </tr>

        <?php } ?>

      </tbody>

    </table>

    </center>

</div>

<?php

echo $footer;

?> 

Answer 1

您在一行上缺少一个分号，即：$dateFrom = $_POST['date_from'];$dateTo = $_POST['date_to'];$instructorId = $_POST['instructor_id'];$seminarsId = $_POST['seminars_id'];$clientsId = $_POST['clients_id']&nbsp; <----$sql = 'INSERT INTO dateof_seminars (date_from, date_to, instructor_id, seminars_id, clients_id) VALUES (:date_from, :date_to, :instructor_id, :seminars_id, :clients_id)';$stmt = $db->prepare($sql);因此，下一行是出乎意料的，编译器不知道该如何处理它。$sql

Answer 2

在提到的第34行（这个：$sql = 'INSERT INTO dateof_seminars (date_from, date_to, instructor_id, seminars_id, clients_id) VALUES (:date_from, :date_to, :instructor_id, :seminars_id, :clients_id)';&nbsp;)你必须用双引号替换单引号，就像这样$sql = "INSERT INTO dateof_seminars (date_from, date_to, instructor_id, seminars_id, clients_id) VALUES (:date_from, :date_to, :instructor_id, :seminars_id, :clients_id)";据我所知，你不能在单引号中使用变量。