php - Firebase 使用 php 过滤数据

Question

我已经测试过这个 Firebase Filtering data with PHP但我收到此错误，并且我不知道如何调试此错误

Fatal error: Uncaught GuzzleHttp\Exception\ClientException: Client error: GET
https://table.firebaseio.com/tbl_admin?orderBy=%22username%22&equalTo=%22admin%22&limitToFirst=1 resulted in a 400 Bad Request response: { "error" : "Index not defined, add \".indexOn\": \"username\", for path \"/tbl_admin\", to the rules" } in D:\Xampp\htdocs\firebase\vendor\guzzlehttp\guzzle\src\Exception\RequestException.php:113 Stack trace: #0 D:\Xampp\htdocs\firebase\vendor\guzzlehttp\guzzle\src\Middleware.php(66): GuzzleHttp\Exception\RequestException::create(Object(GuzzleHttp\Psr7\Request), Object(GuzzleHttp\Psr7\Response)) #1 D:\Xampp\htdocs\firebase\vendor\guzzlehttp\promises\src\Promise.php(203): GuzzleHttp\Middleware::GuzzleHttp{closure}(Object(GuzzleHttp\Psr7\Response))

2 D:\Xampp\htdocs\firebase\vendor\guzzlehttp\promises\src\Promise.php(156):

GuzzleHttp\Promise\Promise::callHandler(1, Object(GuzzleHttp\Psr7\Response), Array) #3 D:\Xampp\htdocs\firebase\vendor\guzzlehttp\promises\src\TaskQueue.php(47): GuzzleHttp\Prom in D:\Xampp\htdocs\firebase\vendor\kreait\firebase-php\src\Firebase\Exception\QueryException.php on line 28

我将如何在 firebase 上过滤它，因为我已经在 mysql 上查询了它，如下所示:

SELECT * FROM tbl_admin WHERE username = 'admin'

Answer 1

如果您阅读错误消息，您会发现它显示:

Index not defined, add ".indexOn": "username", for path "/tbl_admin", to the rules

您可以通过导航到项目的 Firebase Web 控制台中的“数据库规则”部分(https://console.firebase.google.com/project/_/database/_/rules 应该™ 将您带到那里)并添加索引来添加这些规则，以便您的规则如下所示:

{
    "rules": {
        ".read": false,
        ".write": false,
        "tbl_admin": {
          ".indexOn": ["username"]
        }
    }
}

重要的部分是 tbl_admin 字段，如果您的规则中已有其他设置，请不要只复制并粘贴代码段 ^^。