Question

我想从php cron管理我的Fusion Tables。我无法让它发挥作用：

<?php
//include_once "templates/base.php";
error_reporting(E_ALL-E_WARNING-E_NOTICE-E_DEPRECATED);
ini_set('display_errors',1);
session_start();

set_include_path("src/" . PATH_SEPARATOR . get_include_path());

require_once 'Google/Client.php';
require_once 'Google/Service/Fusiontables.php';
$client_id = 'given';
$service_account_name = 'given';
$key_file_location = 'given';
$tableid='given';

$client = new Google_Client();
$client->addScope("https://www.googleapis.com/auth/fusiontables");
$service=new Google_Service_Fusiontables($client);

unset($_SESSION['service_token']);
if (isset($_SESSION['service_token'])) {
  $client->setAccessToken($_SESSION['service_token']);
}
$key = file_get_contents($key_file_location);
$cred = new Google_Auth_AssertionCredentials(
    $service_account_name,
    array('https://www.googleapis.com/auth/fusiontables'),
    $key
);
$client->setAssertionCredentials($cred);
if($client->getAuth()->isAccessTokenExpired()) {
  $client->getAuth()->refreshTokenWithAssertion($cred);
}
$_SESSION['service_token'] = $client->getAccessToken();

$result=$service->table->listTable();

$insert_statement = "SELECT tytul FROM $tableid ";
$result = $service->query->sql($insert_statement);
?>

以下是结果：

object(Google_Service_Fusiontables_TableList)#14 (7) {
  ["itemsType":protected]=>
  string(33) "Google_Service_Fusiontables_Table"
  ["itemsDataType":protected]=>
  string(5) "array"
  ["kind"]=>
  string(22) "fusiontables#tableList"
  ["nextPageToken"]=>
  NULL
  ["collection_key":protected]=>
  string(5) "items"
  ["data":protected]=>
  array(0) {
  }
  ["processed":protected]=>
  array(0) {
  }
}

Fatal error: Uncaught exception 'Google_Service_Exception' with message 'Error calling POST https://www.googleapis.com/fusiontables/v1/query: (403) Forbidden' in /funkcje/google-api/src/Google/Http/REST.php:80 Stack trace: #0 /funkcje/google-api/src/Google/Http/REST.php(44): Google_Http_REST::decodeHttpResponse(Object(Google_Http_Request)) #1 /funkcje/google-api/src/Google/Client.php(499): Google_Http_REST::execute(Object(Google_Client), Object(Google_Http_Request)) #2 /funkcje/google-api/src/Google/Service/Resource.php(195): Google_Client->execute(Object(Google_Http_Request)) #3 /funkcje/google-api/src/Google/Service/Fusiontables.php(682): Google_Service_Resource->call('sql', Array, 'Google_Service_...') #4 /funkcje/google-api/fusion-tables.php(79): Google_Service_Fusiontables_Query_Resource->sql('SELECT tytul FR...') #5 {main} thrown in /funkcje/google-api/src/Google/Http/REST.php on line 80

正如你可以看到我已经连接了API（第一个结果），但它没有返回任何表格（在开发控制台中它返回两个表格） - 好像我连接到不同的帐户......

我正在连接作为服务帐户。如果我作为Web应用程序的客户端ID连接都可以。

Answer 1

找到anwser：https://groups.google.com/d/msg/google-api-php-client/9d2lQAppTvg/o5QGhsXiAQIJ

您必须将新用户添加到文档 - 来自Google API服务的用户

在Google文档中选择文档（FT）
点击“分享”
选择“添加人员”
输入您的服务帐户电子邮件地址，例如为MyService ... @ developer.gserviceaccount.com
确保选中“可编辑”。
取消选中“通过电子邮件通知人员”
点击“分享并保存”

无法让google-api工作

1 个答案: