Question

我正在尝试从mysql表中返回所有选定的数据。我想要的是返回数组中的数据，以便我将其显示为json数据。到目前为止我实现了以下但我不知道如何返回那个mysql数据。

public function getHome() {
    $result = mysql_query("SELECT * FROM places") or die(mysql_error());
    // check for result
    $no_of_rows = mysql_num_rows($result);
    if ($no_of_rows > 0) {
        while($row = mysql_fetch_array($result)) {
            $data[] = $row;
        }
        return $data;
    } else {
        // user not found
        return false;
    }
}

这里我称这个方法为

if($db->getHome()) {
        $data = $db->getHome();
        $response['success'] = 1;
        $response['uid'] = $data['uid'];
        $response['name'] = $data['name'];
        $response['profile_photo'] = $data['profile_photo_path'];
        $response['places']['place_photo'] = $data['place_photo_path'];
        $response['places']['created_at'] = $data['created_at'];
        echo json_encode($response);
    } else {
        echo "bye";
    }

这就是回声

{"tag":"home","success":1,"error":0,"uid":null,"name":null,"profile_photo":null,"places":{"place_photo":null,"created_at":null}}

Answer 1

您需要先将$ data定义为数组，此外您的代码看起来还不错。

$data = array();

正如您所知，可能会返回多行，您应该执行以下操作：

$data = $db->getHome(); // There's no need to call this twice
if($data) {
  foreach($data as $place) {
    // Do what you need to do with each place here
  }
}

查看$ data print_r($data);

的内容

如何从php中的mysql数据库返回选定的数据

1 个答案: