php - SQL查询，JOIN另一个表语法错误

Question

我的 table :

内容:content_id、publisher_id、标题、类别、趋势

用户:ID、姓名、等级

我正在尝试加入用户并获取名称和排名

我的代码:

$query = $conn->prepare
     ("select * from contents 
     FULL OUTER JOIN users ON contents.publisher_id = users.id 
     WHERE trend =1 order by content_id desc limit 20");
$query->execute(array());

};

$data = array();
while($result= $query->fetch(PDO::FETCH_ASSOC)) {
$data[] = array("content_id"    =>$result['content_id'], 
                "publisher_id"  =>$result['publisher_id'],
                "publisher_name"=>$result['name'],
                "publisher_rank"=>$result['rank],
                "title"         =>$result['title'],
                "category"      =>$result['category'],
                "type"          =>$result['type']
                );

};
echo json_encode($data);

Answer 1

试试这个。希望这有帮助

$query = $conn->prepare("select * from contents JOIN users ON contents.publisher_id = users.id 
     WHERE trend = 1 order by content_id desc limit 20");
$query->execute();

$data = array();
while($result= $query->fetch(PDO::FETCH_ASSOC)) {
$data[] = array("content_id"    =>$result['content_id'], 
                "publisher_id"  =>$result['publisher_id'],
                "publisher_name"=>$result['name'],
                "publisher_rank"=>$result['rank'],
                "title"         =>$result['title'],
                "category"      =>$result['category'],
                "type"          =>$result['type']
                );

};
echo json_encode($data);