CODEWITHSUNDEEP
jsonmongodbshell
Rose
Rose

Reputation: 1498

How to get specific field from MongoDB collection using MongoDB shell?

I have indexed a json file in Mongodb in collection "clicklog" using a shell command. Below is the result of my shell command:

db.clicklogs.find().pretty()

Output:

    {
            "_id" : ObjectId("58fe78dcfbe21fa7896552e8"),
            "preview" : false,
            "offset" : 0,
            "result" : {
                    "search_term" : "484797",
                    "request_time" : "Sat Apr 01 23:58:49 -0400 2017",
                    "request_ip" : "127.0.0.1",
                    "stats_type" : "clickstats",
                    "upi" : "66024330304",
                    "unit" : "CITCS",
                    "job_title" : "IT Engineer",
                    "vpu" : "ICR",
                    "organization" : "73",
                    "location" : "MH",
                    "city" : "San Diego",
                    "country" : "USA",
                    "title" : "TOM",
                    "tab_name" : "People-Tab",
                    "page_name" : "PEOPLE",
                    "result_number" : "1",
                    "page_num" : "0",
                    "session_id" : "14e88b44576ad4fdc035bc41529762ad1",
                    "total_results" : "1",
                    "_raw":"request_time=Sat Apr 01 23:58:49 -0400 2017,request_ip=127.0.0.1,application=Search,stats_type=clickstats,upi=660243301304,unit=CITCS,job_title=IT Assistant, Client Services,vpu=ICR,location=DHAKA, BANGLADESH (IFC),organization=73,city=Dhaka,country=BANGLADESH,city_code=,search_term=484797,title=   Tom,url=http://isearch.worldbank.org/skillfinder/ppl_profile_new/000484797,tab_name=People-Tab,page_name=PEOPLE,result_number=1,page_num=0,filter=qterm=484797,total_results=1,app_environment=production,log_version=1.0,session_id=4e88b44576ad4fdc035bc41529762ad1",
                    "_time":"2017-04-01T23:58:49.000-0400"

            }
    }
{"_id" : ObjectId("58fe78dcfbe21fa7896552e9"),
        "preview" : false,
        "offset" : 0,
         "result" : {
                "search_term" : "demo",
                "request_time" : "Sat Apr 01 23:58:49 -0400 2017",
                "request_ip" : "127.0.0.1",
                 ....
                 "time":"2017-04-01T23:58:49.000-0400"
}
}

For every json document, I would like to get only the few field(id,searchterm,upi,page_name,sessionid, url(which is under _raw)). Is it possible to do it using mongo shell commands and store the result document in a new collection? Any help is appreciated.

Upvotes: 0

Views: 218

Answers (1)

s7vr
s7vr

Reputation: 75984

You can try below aggregation in 3.4 version.

The query uses $split operator couple of times to reach to url value. Rest is standard projection fields.

$out stage to write the results into new collection.

db.getCollection('clicklogs').aggregate([{
        $project: {
            searchterm: "$result.searchterm",
            upi: "$result.upi",
            page_name: "$result.page_name",
            session_id: "$result.session_id",
            url: {
                $let: {
                    vars: {
                        obj: {
                            $arrayElemAt: [{
                                $split: ["$result._raw", ',']
                            }, 1]
                        }
                    },
                    in: {
                        $arrayElemAt: [{
                            $split: ["$$obj", '=']
                        }, 1]
                    }
                }
            }
        }
    },
    {
        $out: "clicklogs_temp"
    }
])

Upvotes: 1

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