Compacting mongodb queries

Working with Data
,
1

With the mongo shell, I am doing something in a way which I think should be simpler. I make a first request to get a result needed for a second one.

Here is the first command:

MongoDB -prompt-:PRIMARY> db.Collection.find({"_p_unit":"TopCollection$npsV9rp1Gg","Order":2},{"_id":0,"AUDIO":1})
{ "AUDIO" : "audio_file_key_1a2b3c_Voice.bin" }

I use this result in the next command:

MongoDB -prompt-:PRIMARY> db.Collection.updateOne({"_p_unit":"TopCollection$BTDGm5LIcX","Order":2},{"$set":{ "AUDIO" : "audio_file_key_1a2b3c_Voice.bin" }})
{ "acknowledged" : true, "matchedCount" : 1, "modifiedCount" : 1 }
MongoDB -prompt-:PRIMARY>

Now this is my question:

Is there a way to group the two commands above into one?

Though I tried a few ideas coming up my mind, nothing worked.

2

Hi @Michel_Bouchet ,

It seems that you want to update a value AUDIO in two different documents of the same collection to be the same, however how can the system know the relationship between those two.

It sounds like 2 commands is the smart way, if you need automicity of the two use transactions to do them within one transaction.

In 4.4 you can merge using aggregation and $merge of 2 data sets by $match the value you need filter, add the merging value with $addFields and merge on it to update the second document. But as you can see it is too complex so 2 steps preferred.

Thanks
Pavel

3

Yes indeed as you write “It is too complex”. Though my problem is quite simple. In the SQL world something like this would work (combining the two commands):

db.Collection.updateOne({"_p_unit":“TopCollection$BTDGm5LIcX”,“Order”:2},{"$set":(db.Collection.find({"_p_unit":“TopCollection$npsV9rp1Gg”,“Order”:2},{"_id":0,“AUDIO”:1}))})

4

Even if I have to be satisfied with two commands. Is there a way I could use a variable to store the result of the first command and then use it in the second. I tried but it did not really work.

5

Hi @Michel_Bouchet ,

Of course you can store in a var:

var result = db.Collection.find({"_p_unit":"TopCollection$npsV9rp1Gg","Order":2},{"_id":0,"AUDIO":1})



 db.Collection.updateOne({"_p_unit":"TopCollection$BTDGm5LIcX","Order":2},{"$set":{ "AUDIO" : result.AUDIO}})

Thanks
Pavel

6

I tried, but it did not work. The result is:
{ “AUDIO” : undefined } I had also tried something slightly different before which also failed.

That was:

var x = db.Collection.find({"_p_unit":"TopCollection$npsV9rp1Gg","Order":2},{"_id":0,"AUDIO":1})

db.Collection.updateOne({"_p_unit":"TopCollection$BTDGm5LIcX","Order":2},{"$set":x})
7

Hi @Michel_Bouchet ,

Oh ok so there is a different behaviour with mongo shell, as the query return a pointer to a cursor.

In the shell to overcome this there are 2 options

1:

var result = db.Collection.find({"_p_unit":"TopCollection$npsV9rp1Gg","Order":2},{"_id":0,"AUDIO":1}).toArray()

db.Collection.updateOne({"_p_unit":"TopCollection$BTDGm5LIcX","Order":2},{"$set": result[0]})

2:

var result = db.Collection.find({"_p_unit":"TopCollection$npsV9rp1Gg","Order":2},{"_id":0,"AUDIO":1})
 var updateDoc  = result.next();
 db.Collection.updateOne({"_p_unit":"TopCollection$BTDGm5LIcX","Order":2},{"$set":updateDoc});

Thanks
Pavel