Question

如何获取用户表中创建的seeker_ID并将其添加到文档表中？有可能吗？

到目前为止，这是我的代码：

$link = mysqli_connect("localhost", "user", "password", "database");

$query = "INSERT INTO users (first_name, last_name) VALUES ('$FirstName','$LastName')";
$query = "INSERT INTO documents (seeker_ID, filename) VALUES ('','$filename')";

if (mysqli_multi_query($link, $query)) {
    echo "Candidate ID: $seeker_id <br>";
    echo "Candidate First Name: $FirstName <br>";
    echo "Candidate Last Name:  $LastName<br>";
    echo "Email: $Email <br>";
    echo "<br><br>Records added successfully - Refreshing In 5 Seconds<br>";
} else {
    echo "ERROR: Could not able to execute $sql. " . mysqli_error($link);
}
// close connection
mysqli_close($link);

Answer 1

这不是很有帮助。首先，我要假设是MySQL。如果是这样，我建议使用PDO。

如果包含表单数据的请求是http GET或POST，您可以执行以下操作之一：

$seeker_ID = htmlentities($_GET["seeker_ID"]);

然后您可以使用该变量来构建查询。但请使用PDO：

$pdo = new PDO("mysql:host=$dbhost;dbname=$dbname", $dbusername, $dbpass);

$stmt = $pdo->prepare("INSERT INTO documents(seeker_ID, filename) VALUES (:seeker_ID, :filename)");
$stmt->execute([
    "filename" => $filename,
    "seeker_ID" => $seeker_ID
]);

在后续插入中使用插入的ID

1 个答案: