[英]mongodb gathering all records from different collections
我試圖使用php從mongodb獲取所有記錄。 我有兩個收藏。 問題是,在實踐中,我能否為數據庫中的每個記錄寫一個簡單的句子? :
即:約翰[來自名字收集]居住在城市[來自城市收集],誰開車[來自汽車收集]。
這是上述正確的編碼嗎? 我還是新手,想一步一步學習
<?php foreach ($details as $doc) {
echo $doc['name'] . ' lives in'; }
foreach ($place as $city) {
echo $city['name'] . ' who drives a '; }
foreach ($car as $ride) {
echo $ride['name'];
echo '<br>'} ?>
歡迎您的想法
將for循環相互嵌套。
<?php
foreach ($details as $doc) {
foreach ($place as $city) {
foreach ($car as $ride) {
echo $doc['name'] . ' lives in '.$city['name'].' who drives a '.$ride['name'].'<br/>';
}
}
}
?>
那就是我試圖訪問多個集合中的數據以形成輸出時的處理方法
我希望這可以與user_id
之類的一起使用。 對此進行全表掃描將是一個非常糟糕的主意。
如果您擁有user_id
可以執行以下操作:
$users = $mongo->users->find(['_id' => ['$in' => [1,2,3,4,5,6,7,8,9]]]);
// Set some variables which will help us perform the detail queries
$cityIds = [];
$rideIds = [];
$userResults = [];
$cityResults = [];
$rideResults = [];
// Iterate through our users.
foreach($users as $_id => $user){
// We store the MongoId for use in queries
$cityIds[] = $user['city_id'];
$rideIds[] = $user['ride_id'];
// We then store the user result itself so we don't
// Do this query multiple times.
$userResults[$_id] = $user;
}
// Now, let's get our first details.
$cityResults = iterator_to_array(
$mongo->cities->find(['_id' => ['$in' => $cityIds]])
);
// And our ride details
$rideResults = iterator_to_array(
$mongo->rides->find(['_id' => ['$in' => $rideIds]])
);
// Now let's loop and echo
foreach($userResults as $k => $user){
echo $user['name'] .
' lives in ' .
$cityResults[$user['city_id']]['name'] .
' who drives a ' .
$rideResults[$user['ride_id']]['name'];
}
諸如此類的事情就可以解決問題。
在這里,我假設您的用戶架構中分別包含city
和ride
ID,並且兩個ID字段存儲city
和ride
行的ObjectId
( _id
); 這似乎是最合乎邏輯的模式。
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