【发布时间】：2016-05-09 12:26:09
【问题描述】：
使用Google API PHP client 库，我使用以下代码，该代码运行良好，并打印了有关通过 OAuth2 授权我的应用程序的用户的大量信息：
<?php
require_once('google-api-php-client-1.1.7/src/Google/autoload.php');
const TITLE = 'My amazing app';
const REDIRECT = 'https://example.com/myapp/';
session_start();
$client = new Google_Client();
$client->setApplicationName(TITLE);
$client->setClientId('REPLACE_ME.apps.googleusercontent.com');
$client->setClientSecret('REPLACE_ME');
$client->setRedirectUri(REDIRECT);
$client->setScopes(array(Google_Service_Plus::PLUS_ME));
$plus = new Google_Service_Plus($client);
if (isset($_REQUEST['logout'])) {
unset($_SESSION['access_token']);
}
if (isset($_GET['code'])) {
if (strval($_SESSION['state']) !== strval($_GET['state'])) {
error_log('The session state did not match.');
exit(1);
}
$client->authenticate($_GET['code']);
$_SESSION['access_token'] = $client->getAccessToken();
header('Location: ' . REDIRECT);
}
if (isset($_SESSION['access_token'])) {
$client->setAccessToken($_SESSION['access_token']);
}
if ($client->getAccessToken() && !$client->isAccessTokenExpired()) {
try {
$me = $plus->people->get('me'); # HOW TO SPECIFY FIELDS?
$body = '<PRE>' . print_r($me, TRUE) . '</PRE>';
} catch (Google_Exception $e) {
error_log($e);
$body = htmlspecialchars($e->getMessage());
}
# the access token may have been updated lazily
$_SESSION['access_token'] = $client->getAccessToken();
} else {
$state = mt_rand();
$client->setState($state);
$_SESSION['state'] = $state;
$body = sprintf('<P><A HREF="%s">Login</A></P>',
$client->createAuthUrl());
}
?>
<!DOCTYPE HTML>
<HTML>
<HEAD>
<TITLE><?= TITLE ?></TITLE>
</HEAD>
<BODY>
<?= $body ?>
<P><A HREF="<?= REDIRECT ?>?logout">Logout</A></P>
</BODY>
</HTML>
但是我需要的信息比上述脚本返回的要少。
在People: get“API 资源管理器”中仅输入我感兴趣的字段时：
id,gender,name,image,placesLived
这再次运行良好，并且只打印指定的字段：
我的问题：
上述$me = $plus->people->get('me');调用中如何指定字段？
用代码研究1.1.7/src/Google/Service/Plus.php后：
/**
* Get a person's profile. If your app uses scope
* https://www.googleapis.com/auth/plus.login, this method is
* guaranteed to return ageRange and language. (people.get)
*
* @param string $userId The ID of the person to get the profile for. The
* special value "me" can be used to indicate the authenticated user.
* @param array $optParams Optional parameters.
* @return Google_Service_Plus_Person
*/
public function get($userId, $optParams = array())
{
$params = array('userId' => $userId);
$params = array_merge($params, $optParams);
return $this->call('get', array($params), "Google_Service_Plus_Person");
}
我尝试了以下 PHP 代码：
const FIELDS = 'id,gender,name,image,placesLived';
$me = $plus->people->get('me', array('fields' => urlencode(FIELDS)));
但由于某种原因，它会打印很多 :protected 字符串：
Google_Service_Plus_Person Object
(
[collection_key:protected] => urls
[internal_gapi_mappings:protected] => Array
(
)
[aboutMe] =>
[ageRangeType:protected] => Google_Service_Plus_PersonAgeRange
[ageRangeDataType:protected] =>
[birthday] =>
[braggingRights] =>
[circledByCount] =>
[coverType:protected] => Google_Service_Plus_PersonCover
[coverDataType:protected] =>
[currentLocation] =>
[displayName] =>
[domain] =>
[emailsType:protected] => Google_Service_Plus_PersonEmails
[emailsDataType:protected] => array
[etag] =>
[gender] => male
...
我也尝试在me 之后添加字段：
$me = $plus->people->get('me?fields=' . urlencode(FIELDS)));
但得到 404 错误：
调用 GET 时出错 https://www.googleapis.com/plus/v1/people/me%3Ffields%3Did%252Cgender%252Cname%252Cimage%252CplacesLived: (404) 未找到
更新：我在 Github 创建了Issue #948。
【问题讨论】：
标签： google-api google-plus google-api-php-client google-api-client google-oauth