根据 firebase 文档https://www.firebase.com/docs/rest-api.html,它指出:
PATCH - Updating Data
You can update specific children at a location without overwriting existing data
with a PATCH request. Named children in the data being written with PATCH will be
written, but omitted children will not be deleted. This is equivalent to the
update( ) function.
curl -X PATCH -d '{"last":"Jones"}' \
https://SampleChat.firebaseIO-demo.com/users/jack/name/.json
A successful request will be indicated by a 200 OK HTTP status code.
The response will contain the data written:
{"last":"Jones"}
现在我对此的理解是,如果我只想更新资源的一部分,那么我可以使用 PATCH 请求。
我简化的 firebase 数据库如下:
"exchange-rates" : {
"eur" : {
"fx" : 1.2,
"currency_symbol" : "€",
"updated_at" : "2014-06-13T22:49:23+0100",
},
"usd" : {
"fx" : 1.6,
"currency_symbol" : "$",
"updated_at" : "2014-06-13T22:49:23+0100",
},
"gbp" : {
"fx" : 1,
"currency_symbol" : "£",
"updated_at" : "2014-06-16T15:43:15+0100",
}
}
但是,如果我在补丁请求的有效负载中省略了currency_symbol
and updated_at
,那么 Firebase 会从数据库中删除这些属性。
$auth = 'SUPER_SECRET_CODE';
$guzzleClient = new GuzzleHttp\Client();
$url = https://DATABASE.firebaseio.com/.json;
$data['exchange-rates']['gbp']['fx'] = (float) 1;
$data['exchange-rates']['usd']['fx'] = (float) 1.66;
$data['exchange-rates']['eur']['fx'] = (float) 1.22;
$payload =
[
'query' => [ 'auth' => $auth ],
'json' => $data
];
$response = $guzzleClient->patch($url, $payload);
因此,PATCH 请求无法正常工作,或者我误解了 Firebase 应该如何处理此 PATCH 请求 - 或者我遗漏了一些东西。有什么想法吗?
另外,如果我想向 exchange-rates 对象添加一个对象,我应该可以这样做。
$data['exchange-rates']['chf']['fx'] = 2.13;
$payload =
[
'query' => [ 'auth' => $auth ],
'json' => $data
];
$response = $guzzleClient->patch($url, $payload);
然而,这一切只是覆盖了所有现有的汇率,现在我在数据库中只有 1 个汇率。