php - 使用PHP在Mongodb中的文档内部添加数据

我想在Mongo database明智的文档中使用PHP脚本在year中插入数据，以便看起来像这样(所有年份都在一个文档中)；

cars{
        2017{
            car=Motorolla
            color = blue
        }
        2016{
            car=Toyota
            color = green
        }
        2015{
            car=Corolla
            color = black
        }
    }

我想添加文档，但提示



是否可以使用PHP在Mongo中制作这样的架构？

代码

<?php
    try {
        $car = 'Motorolla';
        $color = 'blue';

        //$car = 'Toyota';
        //$color = 'green';

        //$car = 'Corolla';
        //$color = 'black';

        $years = array(2017, 2016, 2015);
        $manager = new MongoDB\Driver\Manager("mongodb://localhost:27017");
        $bulkWriteManager = new MongoDB\Driver\BulkWrite;
        $document = ['_id' => new MongoDB\BSON\ObjectID, '$years[0]' => $car, '$years[1]' => $color];   // Making a query type

        try {
            $bulkWriteManager->insert($document);   // Inserting Document
            echo 1;
        } catch(MongoCursorException $e) {
            /* handle the exception */
            echo 0;
        }

        $manager->executeBulkWrite('dbName.carsCol', $bulkWriteManager);  // Going to DB and Collection

    } catch (MongoDB\Driver\Exception\Exception $e) {
        $filename = basename(__FILE__);
        echo "The $filename script has experienced an error.\n";
        echo "It failed with the following exception:\n";
        echo "Exception:", $e->getMessage(), "\n";


    }

?>

我不想立即添加整个汽车对象。我想每次添加Year object。任何帮助将是可观的。
或
有什么相对的答案，以便我可以根据年份从Mongo Database中获取数据吗？

编辑1

首次创建。 -积分归@Veeram

<?php
    try {
        $car = 'Malibu';
        $color = 'red';
        $years = array(2017);

        $manager = new MongoDB\Driver\Manager("mongodb://localhost:27017");
        $bulkWriteManager = new MongoDB\Driver\BulkWrite;
        //{"car":"chevy", "color":"black", year: 2017}
        $insert = ['car' => $car, 'color' => $color, 'year' => $years[0]];
        try {
            $bulkWriteManager -> insert($insert); // Inserting Document
            echo 1;
        } catch (MongoCursorException $e) {
            echo 0;
        }
        $manager->executeBulkWrite('dbName.mycol', $bulkWriteManager);  // Going to DB and Collection
    } catch (MongoDB\Driver\Exception\Exception $e) {
        $filename = basename(__FILE__);
        echo "The $filename script has experienced an error.\n";
        echo "It failed with the following exception:\n";
        echo "Exception:", $e->getMessage(), "\n";
        echo "In file:", $e->getFile(), "\n";
        echo "On line:", $e->getLine(), "\n";
    }
?>

要进行更新-致谢@Veeram

<?php
    try {
        $car = 'ChangedCar';
        $color = 'changedColor';
        $years = array(2017);

        $manager = new MongoDB\Driver\Manager("mongodb://localhost:27017");
        $bulkWriteManager = new MongoDB\Driver\BulkWrite;

        $query = ['cars.year' => $years[0]];

        //{ $push: { "cars.$.data": { "car":"chevy", "color":"black"} }}
        $update = ['$push'=> ['cars.$.data'=>['car' => $car, 'color' => $color]]];
        try {
            $bulkWriteManager->update($query, $update);   // Inserting Document
        } catch(MongoCursorException $e) {

        }
        $manager->executeBulkWrite('dbName.mycol', $bulkWriteManager);  // Going to DB and Collection
    } catch (MongoDB\Driver\Exception\Exception $e) {
        $filename = basename(__FILE__);
        echo "The $filename script has experienced an error.\n";
        echo "It failed with the following exception:\n";
        echo "Exception:", $e->getMessage(), "\n";
    }
?>

这段代码中的问题是，它成功地首次插入了数据，但是当我更新数据时，它没有更新。

示例:
有一个名为cars的文档。将年份对象的数据插入到一个文档中。假设对象是2017，它包含颜色和汽车模型。如下图所示； (带有多年的多个对象。年份在整个文档中是唯一的。)

cars{
        2017{
            car=Motorolla
            color = blue
        }
        2016{
            car=Toyota
            color = green
        }
        2015{
            car=Corolla
            color = black
        }
    }

如果要更新，只需创建一个像2017这样的2017{car=Updated-Motorolla color =Updated-blue}对象，然后插入文档中即可。它仅应更新文档中的2017年对象。

 cars{
            2017{
                car=Updated-Motorolla
                color =Updated-blue
            }
            2016{
                car=Toyota
                color = green
            }
            2015{
                car=Corolla
                color = black
            }
        }

最佳答案

您可以尝试这样的事情。仅基于键作为值无法执行所有Mongo数据库操作。

编写第一个解决方案是为了保持与OP的设计接近。

假设您可以向year添加密钥。

{
  "cars": [{
      "year": "2017",
      "data": [{
          "car": "Motorolla",
          "color": "blue"
      }]
  }, {
      "year": "2016",
      "data": [{
          "car": "Toyota",
          "color": "green"
      }]
  }]
}

通过其值可以轻松引用年份。

例如，将新值添加到data 2017的year数组中。您可以尝试以下代码。

使用更新位置$运算符。
query部分引用存储2017记录的数组。
update部分使用push将新的car记录添加到data行的现有2017数组中。

<?php
    try {
        $car = 'Malibu';
        $color = 'blue';
        $years = [2017];

        $manager = new MongoDB\Driver\Manager("mongodb://localhost:27017");
        $bulkWriteManager = new MongoDB\Driver\BulkWrite;

        //{"cars.year":2017}
        $query = ['cars.year' => $years[0]];

        //{ $push: { "cars.$.data": { "car":"chevy", "color":"black"} }}
        $update = ['$push'=> ['cars.$.data'=>['car' => $car, 'color' => $color]]];

        try {
            $bulkWriteManager->update($query, $update);  // Update Document
            echo 1;
        } catch(MongoCursorException $e) {
            /* handle the exception */
            echo 0;
        }

        $manager->executeBulkWrite('dbName.carsCol', $bulkWriteManager);  // Going to DB and Collection

    } catch (MongoDB\Driver\Exception\Exception $e) {
        $filename = basename(__FILE__);
        echo "The $filename script has experienced an error.\n";
        echo "It failed with the following exception:\n";
        echo "Exception:", $e->getMessage(), "\n";
    }

?>

要按年份访问数据，可以在查询下面运行。

使用查询位置 $ 运算符可通过查询部分找到数组索引，并在投影部分中引用该值。

db.collection.find({"cars.year":2017}, {"cars.$.data":1});

替代解决方案:

只需插入即可完成所有工作

最好将每个汽车条目保存在自己的文档中。

{ "year" : 2017, "car" : "Motorolla", "color" : "blue" }
{ "year" : 2016, "car" : "Toyota", "color" : "green" }
{ "year" : 2015, "car" : "Corolla", "color" : "black" }

对于每个条目，您可以使用:

db.collection.insert({"year":2017,  "car":"Motorolla", "color":"blue"});

PHP代码:

 //{"car":"chevy", "color":"black", year: 2017}
 $insert = ['car' => $car, 'color' => $color, 'year' => $years[0]];

 try {
    $bulkWriteManager - > insert($insert); // Inserting Document
    echo 1;
 } catch (MongoCursorException $e) {
    /* handle the exception */
    echo 0;
 }

对于按年份访问数据，您可以使用

db.collection.find({"year":2017});

更新的PHP代码:

<?php
try {
  $cars = ['Motorolla','Toyota', 'Corolla'] ;
  $colors = ['blue', 'green', 'black'];

  $years = [2017, 2016, 2015];
  $manager = new MongoDB\Driver\Manager("mongodb://localhost:27017");
  $bulkWriteManager = new MongoDB\Driver\BulkWrite;

  $query1 =["year" => $years[0]];
  $query2 =["year" => $years[1]];
  $query3 =["year" => $years[2]];

  $update1 = ['$set' => ['car' => $cars[0], 'color' => $colors[0]]];
  $update2 = ['$set' => ['car' => $cars[1], 'color' => $colors[1]]];
  $update3 = ['$set' => ['car' => $cars[2], 'color' => $colors[2]]];

  try {
    $bulkWriteManager->update($query1, $update1, ["upsert" => true]);
    $bulkWriteManager->update($query2, $update2, ["upsert" => true]);
    $bulkWriteManager->update($query3, $update3, ["upsert" => true]);
     echo 1;
  } catch(MongoCursorException $e) {
  /* handle the exception */
  echo 0;
  }

  $manager->executeBulkWrite('dbName.carsCol', $bulkWriteManager); // Going to DB and Collection

  } catch (MongoDB\Driver\Exception\Exception $e) {
    $filename = basename(__FILE__);
    echo "The $filename script has experienced an error.\n";
    echo "It failed with the following exception:\n";
    echo "Exception:", $e->getMessage(), "\n";
  }
?>

您可以使用聚合管道执行复杂的查询，还可以添加索引以加快响应速度。

观察值:

第一个解决方案:难以更新/插入数据，但将所有内容保持在一起，因此更易于读取数据。

第二个解决方案:
对文档进行CRUD操作更干净，更简单，并使用聚合管道执行复杂的查询。