Question

如何使用PHP和MYSQL中的输入字段将表行数据提交到另一个表？

HTML

<form method="post" action="post.php">
  <input type="number" name="code" placeholder="Code..."/>
  <input type="submit" name="submit" value="Submit"/>
</form>

post.php中

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


    $getCode = "SELECT * FROM products WHERE code=$code";
    mysqli_query($connection, $getCode);
    if ($_POST['code'] == $code) {
      $migrating = "INSERT INTO managment(price) VALUES ($price) SELECT 
      price FROM products";
      mysqli_query($connection, $migrating);
      header("location: index.php");
    }
}

我的代码出了什么问题？

Answer 1

syntax是：

INSERT INTO managment(price) SELECT price FROM products

也许您想要添加WHERE子句：

INSERT INTO managment(price) SELECT price FROM products WHERE code=$code

注意：在您的代码中，$_POST['code'] == $code没有意义，因为之前有$code = $_POST['code']。

我还建议您查看How can I prevent SQL injection in PHP?以确保查询安全。

您的代码已更新（未经测试）：

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

    $getCode = "SELECT * FROM products WHERE code=$code";
    $result = mysqli_query($connection, $getCode);
    if (!$result) die("Error: " . mysqli_error($connection));
    $row = mysqli_fetch_assoc($result);
    if ($row['code'] == $code) {
        $migrating = "INSERT INTO managment(price) 
                      SELECT price FROM products WHERE code=$code";
        $result2 = mysqli_query($connection, $migrating);
        if (!$result2) die("Error: " . mysqli_error($connection));
        header("Location: index.php");
        exit;
    }
}

使用php中的输入字段将表行数据提交到另一个表

1 个答案: