I have a collection with single DOC which I use as "source of truth" and want to be able to remove & add values to it:
{
"_id" : ObjectId("61012ada8d2ccb252be87551"),
"language" : "english",
"gui-ipv6" : "disable",
"gui-certificates" : "enable",
"gui-custom-language" : "disable",
"gui-display-hostname" : "disable",
"admin-https-ssl-versions" : "tlsv1-1 tlsv1-2 tlsv1-3",
}
I want to be able to delete a line, say language for example. Whenever I try to do this, it removes entire DOC since my query matches entire DOC e.g:
db.my_collection.remove({'language': {$exists:true}})
Any ideas on how to solve this?
Thank you.
Use operator unset for delete field
https://docs.mongodb.com/manual/reference/operator/update/unset/
db.my_collection.update(
{},
{ $unset: { language: "" },
false, true }
)
Related
What I'm trying to do is update the location UUID across many different documents, all within the same space.
The data that I'm wanting to update looks like this where I'm wanting to change the UUID.
"Location" : {
"Id" : UUID("abcd-efgh-1234-5678"),
"Name" : "Lot A",
"Code" : "A"
I'm trying to use this which is what I found through google searches but it doesn't appear to be going the way I want.
db.getCollection('collections').updateMany({UUID: {$in: [("abcd-efgh-1234-5678")]}}, {$set: {UUID: "NEW UUID"}}, false, true) )
I feel like this is close but again, all the google searches don't seem to lead me down the path. I understand how to do it if it was just UUID but the Location:ID:UUID is throwing me off.
try:
db.getCollection('collections').updateMany(
{
"Location.Id": {
$in: ["UUID1","UUID2"]
}
},
{
$set: {"Location.Id": "NEW UUID"}
}
)
exmple
I am trying to construct a Mongodb query to return a field value. My JSON looks like this:
"question" : "Global_Deployment",
"displayOrder" : 1,
"answerOptions" : {
"fieldId" : "1001",
"fieldType" : "radiobutton",
"fieldName" : "Global Deployment?",
"fieldLabel" : "Global Deployment?",
"helpText" : "Help will go here",
"emailTagFormControl" : "Global_Deployment?",
"source" : "custom",
"status" : "active",
"required" : "true",
"multiSelect" : "false",
"purgeFlag" : "false",
"enableAuditTrack" : "false",
"fields" : [],
"fieldValue" : "Yes",
"options" : [
{
"optionName" : "Yes"
},
{
"optionName" : "No"
}
],
"comments" : {
"commentId" : "C1001",
"commentDetails" : []
}
My query to reach the field with the fieldName "Global Deployment" is this:
db.getCollection('requests').find({"sections.questions.answerOptions.fieldName":"Global Deployment?"})
What I want to know is what to add to this query to return the value of "fieldValue", which is on a different line in the JSON. I am new to Mongodb. Any help would be greatly appreciated.
1) If you've multiple documents in DB with "fieldName" : "Global Deployment?", then .find() would return all the matching documents i.e; in the output what you get is an array of documents then you need to iterate through the array to get answerOptions.fieldValue for each document, Check the below scenario, as I've explained there are chances of getting multiple documents if "sections.questions.answerOptions.fieldName" is not an unique field.
db.getCollection('requests').find({"sections.questions.answerOptions.fieldName":"Global Deployment?"}, {'sections.questions.answerOptions.fieldValue':1})
Output of find :
/* 1 */
[{
"_id" : ObjectId("5d4e19826e173840500f5674"),
"answerOptions" : {
"fieldValue" : "Yes"
}
},
/* 2 */
{
"_id" : ObjectId("5d4e19826e073840500f5674"),
"answerOptions" : {}
}]
If you only need documents which has fieldValue in it then do this :
db.getCollection('requests').find({"sections.questions.answerOptions.fieldName":"Global Deployment?", 'sections.questions.answerOptions.fieldValue':{$exists: true}}, {'answerOptions.fieldValue':1})
Ok now you've array of documents then do iterate thru each to retrieve your value, check this mongoDB cursor tutorial .
2) If you think fieldName is unique across collection, then you can use .findOne() , which would exactly return one document (In case if you've multiple matching documents it would return first found doc) :
db.getCollection('requests').findOne({"sections.questions.answerOptions.fieldName":"Global Deployment?"}, {'sections.questions.answerOptions.fieldValue':1})
Output of findOne :
{
"_id" : ObjectId("5d4e19826e173840500f5674"),
"answerOptions" : {
"fieldValue" : "Yes"
}
}
If you see .find({},{}) has two arguments, second one is called projection which literally be useful if you want to retrieve only required fields in the response, By default mongoDB will return the entire document what ever you've posted in the question will be retrieved, Data in mongoDB flows as JSON's so operating will be similar to using JSON's, Here you can retrieve the required fields out of result, but for best use of network efficiency if you don't need entire document you'll only get the required fields using projection.
You can specify the second condition separated by comma. Either you are trying to filter data with $and or with $or
With simple approach:
{"sections.questions.answerOptions.fieldName":"Global Deployment?","sections.questions.answerOptions.fieldValue":"Yes" }
By using $and method:
.find(
{
$and: [
{"sections.questions.answerOptions.fieldName":"Global Deployment?"},
{"sections.questions.answerOptions.fieldValue":"Yes"}
]
}
)
Same way you can use $or method. Just replace $and with $or.
Edit:
If you want to retrieve specific value (in your case fieldValue), query would be:
db.getCollection('requests').find({
"sections.questions.answerOptions.fieldName":"Global Deployment?"
}).map(function(item){
return item.fieldValue
})
The correct answer here is the method .distinct() (docs)
In your case try it like this:
db.getCollection('requests').find({"sections.questions.answerOptions.fieldName":"Global Deployment?"}).distinct('fieldValue');
That will return only the value you want.
If you use findOne you can use dot notation.
For example, if we start with creating a collection to test using the following to get close to your sample:
db.stackOverflow.insertOne({
sections: {
questions: {
question: "Global_Deployment",
displayOrder: 1,
answerOptions: {
fieldId: "1001",
fieldType: "radiobutton",
fieldName: "Global Deployment?",
fieldLabel: "Global Deployment?",
helpText: "Help will go here",
emailTagFormControl: "Global_Deployment?",
source: "custom",
status: "active",
required: "true",
multiSelect: "false",
purgeFlag: "false",
enableAuditTrack: "false",
fields: [],
fieldValue: "Yes",
options: [
{
optionName: "Yes",
},
{
optionName: "No",
},
],
comments: {
commentId: "C1001",
commentDetails: [],
},
},
},
},
})
then, this query will return "Yes".
db.stackOverflow.findOne({}).sections.questions.answerOptions.fieldValue
I am using mongodb, I am stucked on a issue :
Data is :
{
"_id" : ObjectId("5a956e0b78d363d37f6a2ec4"),
"fieldType" : "Enter Source",
"value" : "Delhi",
"catgeory" : "Generic",
"synonym" : [
"origin name or code",
"from",
"enter source",
"from where",
"fro wher"
]
}
When I use this query
db.getCollection("Rules_DefaultValue").find(
{
"synonym" : "from where"
});
I got correct result as expected
But when I use this query
db.getCollection("Rules_DefaultValue").find(
{
"$text" : {
"$search" : "where"
}
});
I didn't got any result , So I changed it again
db.getCollection("Rules_DefaultValue").find(
{
"$text" : {
"$search" : "wher"
}
});
and this time it worked.
So I came to a conclusion that "where" is reserve keyword and I can't use it as it is. So I tried with escape char :
"$search" : "\"where\""
but again I did'nt got the result.
same thing is happening with
and , from , *
Please help me on this , How can I make query with these words.
Words like where and from are considered as stopwords in MongoDB. It means that when you create a text index those words are wiped out from the index since they appear very frequently in English while the point of FTS is to index some words that allow you to easily find the document you're looking for. To fix that you can create your text index specifying language to none, try:
db.getCollection("Rules_DefaultValue").createIndex(
{ synonym : "text" },
{ default_language: "none" }
)
Then your query should return the document mentioned in your post.
I'm new to mongo and am trying to do a very simple query in this collection:
{
"_id" : ObjectId("gdrgrdgrdgdr"),
"administrators" : {
"-HGFsfes" : {
"name" : "Jose",
"phone" : NumberLong(124324)
},
"-HGFsfqs" : {
"name" : "Peter",
"phone" : "+43242342"
}
},
"countries" : {
"-dgfgrdg : {
"lang" : "en",
"name" : "Canada"
},
"-grdgrdg" : {
"lang" : "en",
"name" : "USA"
}
}
}
How do I make a query that returns the results of administrators with name like "%Jos%" for example.
What I did until now is this: db.getCollection('coll').find({ "administrators.name": /Jos/});
And variations of this. But every thing I tried returns zero results.
What am I doing wrong?
Thanks in advance!
Your mistake is that administrators is not an array, but an object with fields that are themselves objects with name field. Right query will be
{ "administrators.-HGFsfes.name": /Jos/}
Unfortunatelly this way you're only querying -HGFsfes name field, not other administrator name field.
To achieve what you want, the only thing to do is to replace administrators object by an array, so your document will look like this :
{
"administrators" : [
{
"id" : "-HGFsfes",
"name" : "Jose",
"phone" : 124324
},
{
"id" : "-HGFsfqs",
"name" : "Peter",
"phone" : "+43242342"
}
],
countries : ...
}
This way your query will work.
BUT it will return documents where at least one entry in administrators array has the matching name field. To return only administrator matching element, and not whole document, check this question and my answer for unwind/match/group aggregation pipeline.
You need to use query like this:
db.collection_name.find({})
So if your collection name is coll, then it would be:
db.coll.find({"administrators.-HGFsfes.name": /Jos/});
Look this for like query in mongo.
Also, try with regex pattern like this:
db.coll.find({"administrators..-HGFsfes.name": {"$regex":"Jos", "$options":"i"}}});
It will give you only one result because your data is not an array as below in screenshot:
If you want multiple results, then you need to restructure your data.
Ok, think i've found a better solution for you, with aggregation framework.
Run the following query on your current collection, will return you all administrators with name "LIKE" jos (case insensitive with i option) :
db.test1.aggregate(
[
{
$project: {
administrators:{ $objectToArray: "$administrators"}
}
},
{
$unwind: {
path : "$administrators"
}
},
{
$replaceRoot: {
newRoot:"$administrators"
}
},
{
$match: {
"v.name":/jos/i
}
},
]
);
Output
{
"k" : "-HGFsfes",
"v" : {
"name" : "Jose",
"phone" : NumberLong(124324)
}
}
"k" and "v" are coming from "$objectToArray" operator, you can add a $project stage to rename them (or discard if k value doesn't matter)
Not sure for Robomongo testing but in Studio 3T, formerly Robomongo, you can either copy/paste this query in Intellishell console, or copy/import in aggregation tab, (small icon 'paste from the clipboard').
Hope it helps.
how can I return a specific value for a specific document in MongoDB? For example, I have a schema that looks like:
{
"_id" : "XfCZSje7GjynvMZu7",
"createdAt" : ISODate("2015-03-23T14:52:44.084Z"),
"services" : {
"password" : {
"bcrypt" : "$2a$10$tcb01VbDMVhH03mbRdKYL.79FPj/fFMP62BDpcvpoTfF3LPgjHJoq"
},
"resume" : {
"loginTokens" : [ ]
}
},
"emails" : {
"address" : "abc123#gmu.edu",
"verified" : true
},
"profile" : {
"companyName" : "comp1",
"flagged" : true,
"phoneNum" : "7778883333"
}}
I want to return and store the value for profile.flagged specifically for the document with _id : XfCZSje7GjynvMZu7. So far I have tried:
db.users.find({_id:'myfi3E4YTf9z6tdgS'},{admin:1})
and
db.users.find({_id: 'myfi3E4YTf9z6tdgS'}, {profile:admin});
I want the query to return true or false depending on the assigned value.
Can someone help? Thanks!
MongoDB queries always return document objects, not single values. So one way to do this is with shell code like:
var flagged =
db.users.findOne({_id: 'myfi3E4YTf9z6tdgS'}, {'profile.flagged': 1}).profile.flagged;
Note the use of findOne instead of find so that you're working with just a single doc instead of the cursor that you get with find.
The correct answer here is the method .distinct() (link here)
Use it like this:
db.users.find({_id:'myfi3E4YTf9z6tdgS'},{admin:1}).distinct('admin')
The result will be: 1 or 0