Mongodb export json without _id

I am using mongoexport to export some data into .json formatted file, however the document has a large size overhead introduced by _id:IDVALUE tuples.

I found a similar post Is there a way to retrieve data from MongoDB without the _id field? on how to omit the _id field when retrieving data from mongo, but not exporting. It is suggested to use: .Exclude("_id"). I tried to reqrite the --query parameter of mongoexport to somehow include the .Exclude("_id") parameter, but all of the attempts failed so far.

Please suggest what is the proper way of doing this, or should I revert to using some post-export techniques?

Thanks

asked Oct 19, 2012 at 14:13

There appears to be no way to exclude a field (such as _id) using mongoexport.

Here's an alternative that has worked for me on moderate sized databases:

mongo myserver/mydb --quiet --eval "db.mycoll.find({}, {_id:0}).forEach(printjson);" > out.txt

On a large database (many millions of records) it can take a while and running this will affect other operations people try to do on the system:

answered May 7, 2013 at 18:17

quux00quux00

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This works:

mongoexport --db db_name --collection collection_name | sed '/"_id":/s/"_id":[^,]*,//' > file_name.json

answered Apr 18, 2018 at 9:05

4

Pipe the output of mongoexport into jq and remove the _id field there.

mongoexport --uri=mongodb://localhost/mydb --collection=my_collection \
  | jq 'del(._id)'

Update: adding link to jq.

answered Oct 9, 2020 at 11:05

sebastiansebastian

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I know you specified you wanted to export in JSON but if you could substitute CSV data the native mongo export will work, and will be a lot faster than the above solutions

mongoexport --db  --collection  --csv --fields ",," > mongoex.csv

Peter Perháč

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answered Aug 31, 2015 at 15:35

A_funsA_funs

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mongoexport doesn't seem to have such option.

With ramda-cli stripping the _id would look like:

mongoexport --db mydb --collection mycoll -f name,age | ramda 'omit ["_id"]'

answered Jun 1, 2015 at 17:53

raineraine

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2

I applied quux00's solution but forEach(printjson) prints MongoDB Extended JSON notation in the output (for instance "last_update" : NumberLong("1384715001000").

It will be better to use the following line instead:

db.mycoll.find({}, {_id:0}).forEach(function (doc) {

    print( JSON.stringify(doc) );
});

answered Aug 27, 2014 at 15:42

noletonoleto

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mongo / --quiet --eval "db..find({}, {_id:0,:1}).forEach(printjson);" > out.txt

If you have some query to execute change "" to '' and write your condition in find with "" like find("age":13).

answered Jan 20, 2016 at 3:06

xringxring

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The simplest way to exclude the sub-document information such as the "_id" is to export it as a csv, then use a tool to convert the csv into json.

answered Dec 28, 2013 at 6:38

MattMatt

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mongoexport can not omit "_id"

sed is a powerful command to do it:

mongoexport --db mydb --collection mycoll -f name,age | sed '/"_id":/s/"_id":[^,]*,//'

The original answer is from Exclude _id field using MongoExport command

answered Apr 13, 2018 at 2:28

KiwenLauKiwenLau

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Just use --type=csv option in mongoexport command.

mongoexport --db= --collection= --type=csv --field= --out=.csv

For MongoDb version 3.4, you can use --noHeaderLine option in mongoexport command to exclude the field header in csv export too.

For Detail: https://docs.mongodb.com/manual/reference/program/mongoexport/

answered Oct 30, 2019 at 4:09

SamanSaman

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export into a file and just use replace empty value using Regular expression, in my case

"_id": "f5dc48e1-ed04-4ef9-943b-b1194a088b95"

I used "_id": "(\w|-)*",

Afsanefda

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answered Apr 25, 2020 at 21:28

Have you tried specifying your fields with the --fields flag? All fields that are not mentioned are excluded from the export.

For maintainability you can also write your fields into a seperate file and use --fieldFile.

answered Oct 19, 2012 at 14:33

zemircozemirco

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