Update multiple elements in an array in mongodb [d

问题:

This question already has an answer here:

• How to Update Multiple Array Elements in mongodb 10 answers

I have a document like following in my stocks collection in mongodb.

{ _id: 'xRMuRBhqRLgASyQyW',
  History: 
     [ { orderId: '12032017001877',
         status: 'COMPLETED',
        },
       { orderId: '22122017002450', 
        status: 'PROCESSED',
        },
        { orderId: '22122018002450', 
        status: 'DEPOSIT',
        }  
     ] 
 }

I want to iterate through all document in stocks collection and add a field flag: true if status is not 'PROCESSED'.

回答1:

You can achieve it using script.

db.stocks.find().forEach(function(e){var modified=false;e.History.forEach(function(o){if(o.status!="PROCESSED"){o.flag=true;modified=true}});if(modified){db.stocks.save(e)}});

回答2:

You need to use the all $[] positional operator to update each element in the array

db.collection.update(
   { "History": { "$elemMatch": { "status": { "$ne": "PROCESSED" } } } }, 
   { "$set": { "History.$[].flag": false } },
   { "multi": true }
)

回答3:

You can simply do it by using '$' operator. Something like:

 db.stocks.update( {"history.status" : { $not: "PROCESSED" } , 
                { $set: {"history.$.flag": true }} , 
                false , 
                true);

回答4:

My answer is very similar @Anthony but a two extra parameter( for upsert and multi) are added. For reference You can check official document.

db.stocks.update(
    { "History": { "$elemMatch": { "status": { "$ne": "PROCESSED" } } } }, 
    { "$set": { "History.$[].flag": false }},
     false,              // show that if there is not any document with specified criteria , it will not create a new document.
     ture              // according to document of mongodb if multi is true then more than one document will be updated, other wise only one document will be updated.
)