how to ignore duplicate documents when using insertMany in mongodb php library?

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Try to replace 'continueOnError' option by 'ordered' set to false, based on the documentation, when ordered option is set to false the insertMany will continue writing, even if a single write fails.

here is the docs link: insertMany

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Behzadsh
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Behzadsh

A php/laravel developer

Updated on July 09, 2022

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  • Behzadsh
    Behzadsh almost 2 years

    I am using mongo php library, and trying to insert some old data into mongodb. I used insertMany() method and pass a huge array of document, that may have duplicate documents on unique indexes.

    Lets say I have a users collection and have these indexes:

    [
        {
            "v" : 1,
            "key" : {
                "_id" : 1
            },
            "name" : "_id_",
            "ns" : "test.users"
        },
        {
            "v" : 1,
            "unique" : true,
            "key" : {
                "email" : 1
            },
            "name" : "shop_id_1_title_1",
            "ns" : "test.users"
        }
    ]
    

    If there's a duplicate document, MongoDB\Driver\Exception\BulkWriteException would raise and stop the process. I want to find a way to ignore inserting duplicate documents (and also preventing the exception from raise) and continue inserting other documents.

    I found in php.net documentation a flag called continueOnError that do the trick but it seems that it's not working with this library.

    The example from php.net:

    <?php
    
    $con = new Mongo;
    $db = $con->demo;
    
    $doc1 = array(
            '_id' => new MongoId('4cb4ab6d7addf98506010001'),
            'id' => 1,
            'desc' => "ONE",
    );
    $doc2 = array(
            '_id' => new MongoId('4cb4ab6d7addf98506010002'),
            'id' => 2,
            'desc' => "TWO",
    );
    $doc3 = array(
            '_id' => new MongoId('4cb4ab6d7addf98506010002'), // same _id as above
            'id' => 3,
            'desc' => "THREE",
    );
    $doc4 = array(
            '_id' => new MongoId('4cb4ab6d7addf98506010004'),
            'id' => 4,
            'desc' => "FOUR",
    );
    
    $c = $db->selectCollection('c');
    $c->batchInsert(
        array($doc1, $doc2, $doc3, $doc4),
        array('continueOnError' => true)
    );
    

    And the way I tried to use the flag with mongo php library:

    <?php
    
    $users = (new MongoDB\Client)->test->users
    
    $collection->insertMany([
        [
            'username' => 'admin',
            'email' => '[email protected]',
            'name' => 'Admin User',
        ],
        [
            'username' => 'test',
            'email' => '[email protected]',
            'name' => 'Test User',
        ],
        [
            'username' => 'test 2',
            'email' => '[email protected]',
            'name' => 'Test User 2',
        ],
    ],
    [
        'continueOnError' => true    // This option is not working
    ]);
    

    The code above still raise the exception, and seems not to work. Is there other option flag or is there any way to do this?

    • s7vr
      s7vr over 7 years
      Can you be more specific as it what seems to be not working ? You can add an example to help us understand.
    • Behzadsh
      Behzadsh over 7 years
      @Veeram more details and example added as you requested