[英]Firestore PHP - get one field from document
所以我有一个看起来像这样的firestore
我希望能够从文档中只获取一个值,例如从突出显示的文档中获取“uid”,我的 php 代码如下所示:
<?php
require 'vendor/autoload.php';
//putenv('upheld-pursuit-274606-237ac40b2662.json');
use Google\Cloud\Firestore\FirestoreClient;
class Firestore
{
protected $db;
protected $name;
public function __construct(string $collection)
{
$this->db = new FirestoreClient([
'keyFilePath' => 'upheld-pursuit-274606-237ac40b2662.json',
'projectId' => 'upheld-pursuit-274606'
]);
$this->name = $collection;
}
public function getWhere(string $field, string $operator, $value)
{
$arr = [];
$query = $this->db->collection($this->name)->where($field, $operator, $value)->documents()->rows();
if(!empty($query)){
foreach ($query as $q){
$arr[] = $q->data();
}
}
return $arr;
}
public function get(string $field)
{
$arr = [];
$query = $this->db->collection($this->name)->document($field)();
if(!empty($query)){
/*foreach ($query as $q){
$arr[] = $q->data();
}*/
throw new Exception('Document does not exist!');
}
return $query;
}
}
?>
调用它的代码如下所示:
<?php
session_start();
require("connect.php");
require_once 'vendor/autoload.php';
require_once 'Firestore.php';
$fs = new Firestore('users');
$tfs = new Firestore('tutors');
$username = mysqli_real_escape_string($conn, $_GET['name']);
$password = mysqli_real_escape_string($conn, $_GET['pass']);
$password = md5($password);
$data = array();
print_r($fs->get('1WINXTQdshhn4jLfhMWWaZNNdL32'));
我希望能够从文档中仅检索一个字段,以防我需要某些 if 语句或其他条件语句并用于检查目的
看来您只能使用适用于 Cloud Firestore 的 NodeJS package 来执行此操作。 在这个答案中有一个指向firestore select 的文档的链接。
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