Mongo function to find if superstring exists in array. Nodejs, - node.js

A document in my mongo 'companies' collection looks like this:
{
"companyName": "",
"companyIcon": "",
"domains": [
"companyDomainA.com",
"companyDomainB.dev"
],
"allowSubDomains": true
}
In my application the user enters his/her email address.
Using the Nodejs native mongo driver (https://mongodb.github.io/node-mongodb-native), I want to query (find) which company the user belongs to.
The problem is when the user enters the email as name#dept.companyDomainA.com.
I want to be able to query and find the company document of the user based on his email (subdomained 0 or more levels), ie. if the superstring of a string exists in an array in mongo.
(Caveat, I cannot store all the subdomains of the company as they are dynamic and can change at will)
Is there a regular expression way/db schema change way, to achieve this?
Thanks in advance!!

I would do it like this. First find the root domain from the email address. To do that I would split the email and fetch the domain first.
const email = "name#dept.companyDomainA.com";
const domain = email.split('#')[1]; // dept.companyDomainA.com
Now fetch the host (companyDomainA.com) from it. Follow this link.
So, I have found the root domain which is companyDomainA.com. Now run the find query.
db.collection('documents').find({"domains": "companyDomainA.com"});
I didn't test this code.

Related

How can I send an object as a query and use the $all operator in MERN stack?

I'm trying to make a filtering system for my web app using MERN stack. Currently I'm able to take in the user's requests for what to filter, take in that data, see which ones the user entered (because the user can also specify "Any" in a specific category, which means it would send an empty string or array to the backend), parse that data and turn it into one final object called request which would look something like this if a user decided to select SDG 8: Decent Work & Economic Growth as the SDG, Any as the assignment type, and Economy, People as the themes:
{
sdg: 'SDG 8: Decent Work & Economic Growth',
theme: [ 'Economy', 'People' ]
}
As you can see, the assignment_type value is not there because my backend code sees that they chose Any as the value and it returned an empty string which means I don't add it to my final request object.
So now that I have this object, I want to be able to send it to the database and return the objects that contain all of this information. I want to use the $all query statement for this because I need to make sure that if a user enters 'Economy', 'People' as the themes, an object that contains the themes 'Economy', 'People', 'Technology' also gets returned because the themes the user entered exists within that.
This is what I've tried so far in my test code but it doesn't seem to be working:
var request = {
sdg: 'SDG 8: Decent Work & Economic Growth',
theme: [ 'Economy', 'People' ]
}
const test = await Project.find( {$all: {request}}).sort({ createdAt: -1 })
console.log(test)
The code above just returns all values in my database and not the one's I've tried specifying. The main reason why I'm doing it this way is because I have to filter through the requests the user made in the backend in terms of which ones the user wanted specific values for, the user could specify "Any" for all 3 categories (sdg, theme, assignment_type) which would mean that the request object would just be empty so essentially I'd just be finding every single object in the database which is exactly what I want.
I'm not sure how I could make it so that I send these requests to the backend so it only finds those objects. I would appreciate some help

How to update multiple contacts using xero-node library

I'm using xero-node npm package & looks like it will save me a ton of time.
I want to create OR update multiple contacts but not sure how. I'm hoping someone from Xero monitors this tag.
var contacts = [];
var contact = { "Name": "ABC", ContactNumber:"code123"};
contacts.push(xeroClient.core.contacts.newContact(contact));
const retVal = await xeroClient.core.contacts.saveContacts(contacts);
If I run it once , it creates the contact ( or multiple if I add to the array ) . I want to update the contact using my code ( not the xero generated id - because then I would need to store that it my other system).
If I run it a second time , it fails. I assume that is because it is doing a PUT instead of a POST..?
Here are the docs.
https://github.com/XeroAPI/xero-node/blob/2a1ec34888e998cabd72aa79fa58a5b14f2c9cd5/docs/Contacts.md
You are correct.
Here are the docs on Contacts:
https://developer.xero.com/documentation/api/contacts
See this section:
PUT Contacts
Use this method to create one or more contact records. This method works very similar to POST Contacts but if an existing contact matches your ContactName or ContactNumber then you will receive an error.
ContactNumber is unique. So you're trying to create two contacts with the same ContactNumber.
I think saving the contact like in the example here would help: https://github.com/XeroAPI/xero-node/blob/2a1ec34888e998cabd72aa79fa58a5b14f2c9cd5/docs/Contacts.md

Error is null in Mongo? [duplicate]

Following is my user schema in user.js model -
var userSchema = new mongoose.Schema({
local: {
name: { type: String },
email : { type: String, require: true, unique: true },
password: { type: String, require:true },
},
facebook: {
id : { type: String },
token : { type: String },
email : { type: String },
name : { type: String }
}
});
var User = mongoose.model('User',userSchema);
module.exports = User;
This is how I am using it in my controller -
var user = require('./../models/user.js');
This is how I am saving it in the db -
user({'local.email' : req.body.email, 'local.password' : req.body.password}).save(function(err, result){
if(err)
res.send(err);
else {
console.log(result);
req.session.user = result;
res.send({"code":200,"message":"Record inserted successfully"});
}
});
Error -
{"name":"MongoError","code":11000,"err":"insertDocument :: caused by :: 11000 E11000 duplicate key error index: mydb.users.$email_1 dup key: { : null }"}
I checked the db collection and no such duplicate entry exists, let me know what I am doing wrong ?
FYI - req.body.email and req.body.password are fetching values.
I also checked this post but no help STACK LINK
If I removed completely then it inserts the document, otherwise it throws error "Duplicate" error even I have an entry in the local.email
The error message is saying that there's already a record with null as the email. In other words, you already have a user without an email address.
The relevant documentation for this:
If a document does not have a value for the indexed field in a unique index, the index will store a null value for this document. Because of the unique constraint, MongoDB will only permit one document that lacks the indexed field. If there is more than one document without a value for the indexed field or is missing the indexed field, the index build will fail with a duplicate key error.
You can combine the unique constraint with the sparse index to filter these null values from the unique index and avoid the error.
unique indexes
Sparse indexes only contain entries for documents that have the indexed field, even if the index field contains a null value.
In other words, a sparse index is ok with multiple documents all having null values.
sparse indexes
From comments:
Your error says that the key is named mydb.users.$email_1 which makes me suspect that you have an index on both users.email and users.local.email (The former being old and unused at the moment). Removing a field from a Mongoose model doesn't affect the database. Check with mydb.users.getIndexes() if this is the case and manually remove the unwanted index with mydb.users.dropIndex(<name>).
If you are still in your development environment, I would drop the entire db and start over with your new schema.
From the command line
➜ mongo
use dbName;
db.dropDatabase();
exit
I want to explain the answer/solution to this like I am explaining to a 5-year-old , so everyone can understand .
I have an app.I want people to register with their email,password and phone number .
In my MongoDB database , I want to identify people uniquely based on both their phone numbers and email - so this means that both the phone number and the email must be unique for every person.
However , there is a problem : I have realized that everyone has a phonenumber but not everyone has an email address .
Those that don`t have an email address have promised me that they will have an email address by next week. But I want them registered anyway - so I tell them to proceed registering their phonenumbers as they leave the email-input-field empty .
They do so .
My database NEEDS an unique email address field - but I have a lot of people with 'null' as their email address . So I go to my code and tell my database schema to allow empty/null email address fields which I will later fill in with email unique addresses when the people who promised to add their emails to their profiles next week .
So its now a win-win for everyone (but you ;-] ): the people register, I am happy to have their data ...and my database is happy because it is being used nicely ...but what about you ? I am yet to give you the code that made the schema .
Here is the code :
NOTE : The sparse property in email , is what tells my database to allow null values which will later be filled with unique values .
var userSchema = new mongoose.Schema({
local: {
name: { type: String },
email : { type: String, require: true, index:true, unique:true,sparse:true},
password: { type: String, require:true },
},
facebook: {
id : { type: String },
token : { type: String },
email : { type: String },
name : { type: String }
}
});
var User = mongoose.model('User',userSchema);
module.exports = User;
I hope I have explained it nicely .
Happy NodeJS coding / hacking!
In this situation, log in to Mongo find the index that you are not using anymore (in OP's case 'email'). Then select Drop Index
Check collection indexes.
I had that issue due to outdated indexes in collection for fields, which should be stored by different new path.
Mongoose adds index, when you specify field as unique.
Well basically this error is saying, that you had a unique index on a particular field for example: "email_address", so mongodb expects unique email address value for each document in the collection.
So let's say, earlier in your schema the unique index was not defined, and then you signed up 2 users with the same email address or with no email address (null value).
Later, you saw that there was a mistake. so you try to correct it by adding a unique index to the schema. But your collection already has duplicates, so the error message says that you can't insert a duplicate value again.
You essentially have three options:
Drop the collection
db.users.drop();
Find the document which has that value and delete it. Let's say the value was null, you can delete it using:
db.users.remove({ email_address: null });
Drop the Unique index:
db.users.dropIndex(indexName)
I Hope this helped :)
Edit: This solution still works in 2023 and you don't need to drop your collection or lose any data.
Here's how I solved same issue in September 2020. There is a super-fast and easy way from the mongodb atlas (cloud and desktop). Probably it was not that easy before? That is why I feel like I should write this answer in 2020.
First of all, I read above some suggestions of changing the field "unique" on the mongoose schema. If you came up with this error I assume you already changed your schema, but despite of that you got a 500 as your response, and notice this: specifying duplicated KEY!. If the problem was caused by schema configuration and assuming you have configurated a decent middleware to log mongo errors the response would be a 400.
Why this happens (at least the main reason)
Why is that? In my case was simple, that field on the schema it used to accept only unique values but I just changed it to accept repeated values. Mongodb creates indexes for fields with unique values in order to retrieve the data faster, so on the past mongo created that index for that field, and so even after setting "unique" property as "false" on schema, mongodb was still using that index, and treating it as it had to be unique.
How to solve it
Dropping that index. You can do it in 2 seconds from Mongo Atlas or executing it as a command on mongo shell. For the sack of simplicity I will show the first one for users that are not using mongo shell.
Go to your collection. By default you are on "Find" tab. Just select the next one on the right: "Indexes". You will see how there is still an index given to the same field is causing you trouble. Just click the button "Drop Index". Done.
So don't drop your database everytime this happens
I believe this is a better option than just dropping your entire database or even collection. Basically because this is why it works after dropping the entire collection. Because mongo is not going to set an index for that field if your first entry is using your new schema with "unique: false".
I faced similar issues ,
I Just clear the Indexes of particular fields then its works for me .
https://docs.mongodb.com/v3.2/reference/method/db.collection.dropIndexes/
This is my relavant experience:
In 'User' schema, I set 'name' as unique key and then ran some execution, which I think had set up the database structure.
Then I changed the unique key as 'username', and no longer passed 'name' value when I saved data to database. So the mongodb may automatically set the 'name' value of new record as null which is duplicate key. I tried the set 'name' key as not unique key {name: {unique: false, type: String}} in 'User' schema in order to override original setting. However, it did not work.
At last, I made my own solution:
Just set a random key value that will not likely be duplicate to 'name' key when you save your data record. Simply Math method '' + Math.random() + Math.random() makes a random string.
I had the same issue. Tried debugging different ways couldn't figure out. I tried dropping the collection and it worked fine after that. Although this is not a good solution if your collection has many documents. But if you are in the early state of development try dropping the collection.
db.users.drop();
I have solved my problem by this way.
Just go in your mongoDB account -> Atlast collection then drop your database column. Or go mongoDB compass then drop your database,
It happed sometimes when you have save something null inside database.
This is because there is already a collection with the same name with configuration..Just remove the collection from your mongodb through mongo shell and try again.
db.collectionName.remove()
now run your application it should work
I had a similar problem and I realized that by default mongo only supports one schema per collection. Either store your new schema in a different collection or delete the existing documents with the incompatible schema within the your current collection. Or find a way to have more than one schema per collection.
I got this same issue when I had the following configuration in my config/models.js
module.exports.models = {
connection: 'mongodb',
migrate: 'alter'
}
Changing migrate from 'alter' to 'safe' fixed it for me.
module.exports.models = {
connection: 'mongodb',
migrate: 'safe'
}
same issue after removing properties from a schema after first building some indexes on saving. removing property from schema leads to an null value for a non existing property, that still had an index. dropping index or starting with a new collection from scratch helps here.
note: the error message will lead you in that case. it has a path, that does not exist anymore. im my case the old path was ...$uuid_1 (this is an index!), but the new one is ....*priv.uuid_1
I have also faced this issue and I solved it.
This error shows that email is already present here. So you just need to remove this line from your Model for email attribute.
unique: true
This might be possible that even if it won't work. So just need to delete the collection from your MongoDB and restart your server.
It's not a big issue but beginner level developers as like me, we things what kind of error is this and finally we weast huge time for solve it.
Actually if you delete the db and create the db once again and after try to create the collection then it's will be work properly.
➜ mongo
use dbName;
db.dropDatabase();
exit
Drop you database, then it will work.
You can perform the following steps to drop your database
step 1 : Go to mongodb installation directory, default dir is "C:\Program Files\MongoDB\Server\4.2\bin"
step 2 : Start mongod.exe directly or using command prompt and minimize it.
step 3 : Start mongo.exe directly or using command prompt and run the following command
i) use yourDatabaseName (use show databases if you don't remember database name)
ii) db.dropDatabase()
This will remove your database.
Now you can insert your data, it won't show error, it will automatically add database and collection.
I had the same issue when i tried to modify the schema defined using mangoose. I think the issue is due to the reason that there are some underlying process done when creating a collection like describing the indices which are hidden from the user(at least in my case).So the best solution i found was to drop the entire collection and start again.
If you are in the early stages of development: Eliminate the collection. Otherwise: add this to each attribute that gives you error (Note: my English is not good, but I try to explain it)
index:true,
unique:true,
sparse:true
in my case, i just forgot to return res.status(400) after finding that user with req.email already exists
Go to your database and click on that particular collection and delete all the indexes except id.

How to check which type of access user has (Nodejs and mongodb)

I am creating an alarm tracking management system for the remote smart devices. Using nodejs and mongodb. The idea is like this when the device has the alarm I want that the user who has permission could see the alarm error (or history of errors) after he logs in. Let say I have three types of user, first type can track only one group of devices, second another group and third all devices. For the demo version I am thinking to create a user in my mongodb and add the array with the 3 types of devices group and assign true of false to each device group depending on the access urer have. I already have log in system, but I want some advise how to check if the user has access to group one then he can be redirected after login to particular route and etc. What I think create a function that checks somehow 'if(user nameOfgroup1 is true) then redirect to nameOfgroup1 view' but not sure how to write code for this condition. My user schema:
var userSchema = mongoose.Schema({
local: {
username: String,
password: String,
access:[{
nameOfgroup1: String,
available: Boolean
},
{
nameOfgroup2: String,
available: Boolean
},
{
nameOfgroup3: String,
available: Boolean
}]}});
It would be simpler to store an array of group IDs that the user should have access to. That way you'll be able to easily add more groups in the future and also checking for a given group would be easy.
For example, if the user's doc in the DB would be something like:
{ "username" : "name", "groups" : [ 0, 2 ] }
then it's easy to check which groups he has access to but also to check a specific group with:
function check(user, group) {
return user.groups.indexOf(group) > -1;
}
Also it's easy to search for users with access to a given group in the database:
db.users.find({groups:2});
Instead of numbers you can use names or ObjectIds or whatever is convenient for you.

How to find a sub document in mongoose without using _id fields but using multiple properties

I have a sample schema like this -
Comment.add({
text:String,
url:{type:String,unique:true},
username:String,
timestamp:{type:Date,default:Date}
});
Feed.add({
url:{type:String, unique:true },
username:String,
message:{type:String,required:'{PATH} is required!'},
comments:[Comment],
timestamp:{type:Date,default:Date}
});
Now, I don't want to expose the _id fields to the outside world that's why I am not sending it to the clients anywhere.
Now, I have two important properties in my comment schema (username,url)
What I want to do is update the content of the sub document that satisfies
feed.url
comment.url
comment.username
if the comment.username is same as my client value req.user.username then update the comment.text property of that record whose url was supplied by client in req.body.url variable.
One long and time consuming approach I thought is to first find the feed with the given url and then iterating over all the subdocuments to find the document which satisfies the comment.url==req.body.url and then check if the comment.username==req.user.username if so, update the comment object.
But, I think there must be an easier way of doing this?
I already tried -
db.feeds.update({"username":"harshitladdha93#gmail.com","comments.username":"harshitladdha3#gmail.com","comments.url":"test"},{$set:{"comments.$.text":"updated text 2"}})
found from http://www.tagwith.com/question_305575_how-to-find-and-update-subdocument-within-array-based-on-parent-property
but this updates even when the comments.url or comments.usernamematches other sub documents
and I also tried
db.feeds.distinct("comments._id",{"comments.url":req.body.url})
to find the _id of document associated with the url but it returns all the _id in the subdocument
First off - you should not rely on _id not being seen by the outside world in terms of security. This is a very bad idea for a multitude of reasons (primarily REST and also the fact that it's returned by default with all your queries).
Now, to address your question, what you want is the $elemMatch operator. This says that you're looking for something where the specified sub-document within an array matches multiple queries.
E.g.
db.feeds.update({
"username":"harshitladdha93#gmail.com",
comments: {
$elemMatch: {
username: "harshitladdha3#gmail.com",
url: "test"
}
}
}, {$set: {"comments.$.text":"updated text 2"}})
If you don't use $elemMatch you're saying that you're ok with the document if any of the comments match your query - i.e. if there is a comment by user "harshitladdha3#gmail.com", and separate comment has a url "test", the document will match unless you use $elemMatch

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