PHP Movie::find_by_sqlの例

コード例 #1

ファイル: alert.php プロジェクト: vincentbello/api-readytowatch

 public static function find_movies($user_id = 0)
 {
     global $database;
     $sql = "SELECT m.* FROM movies m INNER JOIN alerts a ON a.id = m.id WHERE a.user_id = {$user_id}";
     $movies = Movie::find_by_sql($sql);
     foreach ($movies as $movie) {
         if ($alert = static::find($user_id, $movie->id)) {
             $movie->alert = $alert;
         }
     }
     return !empty($movies) ? $movies : false;
 }

コード例 #2

ファイル: search.php プロジェクト: vincentbello/api-readytowatch

<?php

require_once "../includes/initialize.php";
// require key and id parameter.
if (empty($_GET['q']) || empty($_GET['key'])) {
    sendResponse(400, "Invalid request. Please provide \"q\" and \"key\" parameters.");
} else {
    if ($_GET['key'] != API_KEY) {
        sendResponse(403, "Invalid API key.");
    } else {
        $searchTerm = $database->escape_value($_GET['q']);
        $sql = "SELECT * FROM movies WHERE title LIKE '% {$searchTerm}%' OR title LIKE '{$searchTerm}%' ORDER BY popularity DESC LIMIT 6";
        $movies = Movie::find_by_sql($sql);
        $results = array();
        foreach ($movies as $movie) {
            $results[] = $movie->export();
        }
        $popular = array("movies" => $results);
        sendResponse(200, json_encode($popular));
    }
}