Pusher：当mysql数据改变时触发pusher

Question

我正在尝试使用 Pusher 制作实时推送通知。我想在 mysql 数据更改时触发推送器。有人可以帮我吗？我是这个推手的新手。

Answer 1

执行此操作的最简单方法是在执行数据库交互后通过 Pusher 触发事件。

例如

// Connect and interact with database
$mysqli = new mysqli("server", "username", "password", "database_name");

$value = $_POST["activity-name"];
$stmt = $mysqli->prepare("INSERT INTO activities (column) VALUES (?)");
$stmt->bind_param("s", $value);
$stmt->execute();

// TODO: check the INSERT was successful.

$stmt->close();
$mysqli->close();

// TODO: Only trigger if the DB interaction was successful

// Create Pusher instance (assuming the Pusher library has been required)
$pusher = new Pusher(APP_KEY, APP_SECRET, APP_ID);

// Trigger the event representing the new data that's been inserted
$pusher->trigger('activities', 'activity-created' ['name' => $value]);

如果数据库交互是针对 UPDATE 的，那么您可以使用不同的事件名称。

$pusher->trigger('activities', 'activity-updated' ['name' => $value]);

或者，您可以设置一个数据库触发器，让另一个脚本在发生更改时执行并与 Pusher 交互。参见 Invoking a PHP script from a MySQL trigger

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