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.
Related
I don't find much information about this problem to solve.
On my mongodb I create a collection every 60 seconds with the name "test "+ date.now(). So far everything works ok. It creates me different collections with the name test XXXXXX1, test XXXXX2 etc.
I have problems with the mongoose.find() method. I can't find my last created collection.
let test = mongoose.model('test' + date.now(), Schema);
test.find({}, function (err, response) {});
How do I find the latest collection in stream? Thank you!
Mongo ,By default , does not support sequence .
for that purpose you're going to have to add specific field for sorting or sort your fields based on your current field properties.
After that you have to use .sort() cursor method :
Collection.find().sort([...]);
Read this article for more info
I had a problem using query with MongoDB.
The problem was solved but I wanted to check if there was any other approach I could have taken.
At first, my model (Ad) had a property of price: {type: String}, and I tried to find by queries $gte and $lt to get ads with a price within a given range.
After reading online I figured that query operations are not working on String type properties.
Then even after changing the type to Number - price: {type: Number} - the find function didn't work properly on the price, even though on other properties which were type Number it worked as it should.
In the end, I just deleted the whole database and reupload it, and then everything worked properly (haven't changed a thing).
Has anyone had this kind of problem and solved it differently?
I'll first start by assuming you're using mongoose as the "types" you've pasted look like mongoose schema types.
You need to separate these two concepts:
The schema that represents data at the app level
The actual data in the DB.
Let's say I have this schema for a certain collection:
{ name: String }
But in the actual database there is only one document in that collection that looks like this:
{ price: 5, product_id: 1 }
Then when I query the data what do you expect to happen? do you expect mongoose to automatically generate a name for that document and delete the actual fields?
The reason it didn't "work" as you intended was that all the values were saved as string, changing the Schema does not retroactively update the database, so when you use $lt and $gte it uses string comparison which means "10" is less than "9" because that's how string comparison work.
The schema does help with newly inserted data and can cast it to the right type if supported, for that you should check the docs with what values are available.
I have a Mongo collection that has two fields, let's say "name" and "randomString".
I want to create a random string for a name, only if it doesn't exist already. So the first request for { name: "SomeName" } will result in saving e.g. { name: "someName", randomString: "abc" }. The second request will do nothing.
Is there a mongo command for this? All I could find are things like findOneAndUpdate, replaceOne etc, who all support an optional "upsert" but their behavior on match is to update, I want the behavior on match to be do nothing.
I'm not looking for an if-then solution like in this question, as I have a race condition issue - I need to be able to get multiple requests simultaneously without updating the document or failing any of the requests.
Yes there is a command for this you can do this by using $addToSet method.
For more info please go through the given link: https://docs.mongodb.com/manual/reference/operator/update/addToSet/
PS: If you still have any confusion regarding this question please feel free to comment further.
Thanks
This is the solution I found in the end:
CustomerRandomString.findOneAndUpdate(
{ name: "someName" },
{
$setOnInsert: { randomString: generateRandomString() },
},
{ upsert: true },
);
The setOnInsert operator only applies when creating a new document, which is exactly what I needed.
EDIT: per the docs, this solution requires a unique index on the field in order to fully avoid duplicates.
You can easily do it using the $exists command to check for randomString field and then use $set in an aggregation pipeline to upsert that field.
db.collection.updateMany({"name":someName,"randomString":{$exists: false}},[{$set:{"randomString":"abcd"}}],{upsert:true})
If the condition query doesn't match with any documents, then it returns null.
Note: Aggregation pipeline works in updateMany() only from MongoDB version 4.2 and above.
I have a sample schema like this -
Comment.add({
text:String,
url:{type:String,unique:true},
username:String,
timestamp:{type:Date,default:Date}
});
Feed.add({
url:{type:String, unique:true },
username:String,
message:{type:String,required:'{PATH} is required!'},
comments:[Comment],
timestamp:{type:Date,default:Date}
});
Now, I don't want to expose the _id fields to the outside world that's why I am not sending it to the clients anywhere.
Now, I have two important properties in my comment schema (username,url)
What I want to do is update the content of the sub document that satisfies
feed.url
comment.url
comment.username
if the comment.username is same as my client value req.user.username then update the comment.text property of that record whose url was supplied by client in req.body.url variable.
One long and time consuming approach I thought is to first find the feed with the given url and then iterating over all the subdocuments to find the document which satisfies the comment.url==req.body.url and then check if the comment.username==req.user.username if so, update the comment object.
But, I think there must be an easier way of doing this?
I already tried -
db.feeds.update({"username":"harshitladdha93#gmail.com","comments.username":"harshitladdha3#gmail.com","comments.url":"test"},{$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
I'd like the unique _id field in one of my models to be relatively short: 8 letters/numbers, instead of the usual Mongo _id which is much longer. Having a short unique-index like this helps elsewhere in my code, for reasons I'll skip over here. I've successfully created a schema that does the trick (randomString is a function that generates a string of the given length):
new Schema('Activities', {
'_id': { type: String, unique: true, 'default': function(){ return randomString(8); } },
// ... other definitions
}
This works well so far, but I am concerned about duplicate IDs generated from the randomString function. There are 36^8 possible IDs, so right now it is not a problem... but as the set of possible IDs fills up, I am worried about insert commands failing due to a duplicate ID.
Obviously, I could do an extra query to check if the ID was taken before doing an insert... but that makes me cry inside.
I'm sure there's a better way to be doing this, but I'm not seeing it in the documentation.
This shortid lib https://github.com/dylang/shortid is being used by Doodle or Die, seems to be battle tested.
By creating a unique index on _id you'll get an error if you try to insert a document with a duplicate key. So wrap error handling around any inserts you do that looks for the error and then generates another ID and retries the insert in that case. You could add a method to your schema that implements this enhanced save to keep things clean and DRY.