使用PHP

时间：2017-02-17 13:31:00

标签： php mongodb

我想在Mongo database明智的文档中使用PHP脚本在year中插入数据，以便它看起来像这样（一年中的所有年份）; 

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

我想添加文档，但它会提示

  

文档不能有$前缀字段名称：$ years [0]

是否可以使用PHP在Mongo中创建这样的模式？

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";


    }

?>

我不想立刻添加整车对象。我想每次都添加Year object。任何帮助都会很明显。 或 任何相对答案，以便我可以根据年份从Mongo Database获得数据？

EDIT1

第一次创作。 - 积分转到@Veram 

<?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";    
    }
?>

对于更新 - Credits转到@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";
    }
?>

此代码中的问题是它第一次成功插入数据，但是当我更新数据时，它不会更新数据。

示例： 有一个名为cars的文档。将包含year的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; （具有年份的多个对象。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
        }
    }

如果我想更新，只需将2017的对象设为2017{car=Updated-Motorolla color =Updated-blue}并插入文档中。它应该只更新文档中的2017年对象。 

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

2 个答案:

答案 0 :(得分：4)

你可以尝试这样的事情。它不可能仅仅将key作为值来执行所有Mongo数据库操作。

第一个解决方案是为了保持接近OP的设计。

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"
      }]
  }]
}

通过其价值轻松引用年份。

例如，为data 2017年的year数组添加新值。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，您可以在以下查询中运行。

使用查询位置$运算符使用查询部分查找数组索引，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

即可完成所有操作

最好将每个车辆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" }

对于您可以使用的每个条目：

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;
 }

按年份访问数据，您可以使用

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

更新了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查询，并且可以添加索引以使您的响应更快。

<强>观察：

第一个解决方案：更难以更新/插入数据，但将所有内容保存在一起，以便更轻松地读取数据。

第二个解决方案： 更清晰，更简单，可以对文档执行CRUD操作，并使用聚合管道来执行复杂查询。

答案 1 :(得分：0)

尝试更改

$document = ['_id' => new MongoDB\BSON\ObjectID, '$years[0]' => $car, '$years[1]' => $color];

类似于：

$document = ['_id' => new \MongoDB\BSON\ObjectID, $years[0] => ['car' => $car, 'color' => $color]];

它在mongo中给出了这样的结果：

{ "_id" : ObjectId("58a936ecfc11985f525a4582"), "2017" : { "car" : "Motorolla", "color" : "blue" }

如果所有汽车的数据必须在一个文件中，您需要合并最合适的数据：

$cars = [
    '2017' => [
        'car' => 'Motorolla',
        'color' => 'blue'
    ],
    '2016' => [
        'car' => 'Toyota',
        'color' => 'green'
    ],
    '2015' => [
        'car' => 'Corolla',
        'color' => 'black'
    ]
];

而不是

$document = ['_id' => new \MongoDB\BSON\ObjectID, 'cars' => $cars];

它提供了像mongo文档：

{ "_id" : ObjectId("58aabc0cfc11980f57611832"), "cars" : { "2017" : { "car" : "Motorolla", "color" : "blue" }, "2016" : { "car" : "Toyota", "color" : "green" }, "2015" : { "car" : "Corolla", "color" : "black" } } }
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