As far as I know, with PUT one can create a resource if it doesn't exist or it is going to replace the old one with a new one.
I want to create a resource and being able to update it, not create more resources, using Node.js/Express and MongoDB.
So, I wrote this code:
app.put('/entries/:entry_id/type', (req, res) => {
const entry = new Entry (req.body);
entry.save();
res.end();
})
in Postman there is a PUT request, having the url: localhost:5000/entries/2/type
After sending it once, it creates an entry in the dabatase. All good!
But let's try to send the same request again. Now there are 2 entries in the database. I would expect to be one, because the same request was sent.
In the database they have the same data, same schema but they do have an extra field,
"_id":{"$oid":"5e8909e60c606c002axxxxxx"},, which is has the last character different.
Why are there created more entries of the same data while I was expecting to have only one entry in the database?
Mongo automatically creates a default index named _id on every collection when it is created. If you insert a Document into a collection without specifying an _id it will generate a new ObjectId as the _id field.
To get around this you can use findOneAndUpdate with upsert:
Entry.findOneAndUpdate({ entry_id: req.params.entry_id }, { <content> }, { upsert: true })
However this will update the document if it exists already instead of creating a new one. If you further wish to not change the Document at all if it already exists, then you can surround your <content> with $setOnInsert.
Following is my user schema in user.js model -
var userSchema = new mongoose.Schema({
local: {
name: { type: String },
email : { type: String, require: true, unique: true },
password: { type: String, require:true },
},
facebook: {
id : { type: String },
token : { type: String },
email : { type: String },
name : { type: String }
}
});
var User = mongoose.model('User',userSchema);
module.exports = User;
This is how I am using it in my controller -
var user = require('./../models/user.js');
This is how I am saving it in the db -
user({'local.email' : req.body.email, 'local.password' : req.body.password}).save(function(err, result){
if(err)
res.send(err);
else {
console.log(result);
req.session.user = result;
res.send({"code":200,"message":"Record inserted successfully"});
}
});
Error -
{"name":"MongoError","code":11000,"err":"insertDocument :: caused by :: 11000 E11000 duplicate key error index: mydb.users.$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.
I developed an API using Loopback framework, in that i have to insert or update to a table.
My table looks like below:
userid bbid saved id(PK)
1 5 1000 1
So when next time if(bbid = 5) it should update the above row, if bbid =5 is not there it should insert into the above table.
I tried the following:
app.models.modelname.upsert({
bbid: bb.id,
saved: Saved,
userid: 1
}, function(err, res) {});
EDIT for COMPOSITE ID:
app.models.modelname.upsert({
bbid: vvvv.bbid,
userid: 1,
saved: amountSaved
}, function(err, res) {});
Also gone through Loopback doc it says
upsert - checks if the instance (record) exists, based on the designated
ID property, which must have a unique value; if the instance already exists,
the method updates that instance. Otherwise, it inserts a new instance.
But in my case it should check for bbid not id
But its inserting everytime. Please share your ideas. Thanks in advance
EDIT FOR UPSERT:
My process is as follows:
Fetch from table1 and loop that response and calculate the saved and upsert it into another table(table2 (this is where upsert should happen)). Also the fetching from table1 will happen frequently so suppose consider if is happens 2nd time it should update the already present bbid..
You can use the findOrCreate method as follows:
app.models.modelname({where: {bbid:bb.id} }, {bbid:bb.id, saved: Saved, userid:1}, function(err, instance) {
if (err){
cb(null, err);
}
cb(null, instance);
});
There are two methods in loopback one is simple upsert and second is upsertWithWhere.To insert or update based on where condition you must use upsertWithWhere method upsert only inserts the data.
you are using
app.models.modelname.upsert({bbid:bb.id, saved: Saved, userid:1}
,function(err, res){});
instead use
app.models.modelname.upsertWithWhere({bbid:bb.id, saved: Saved, userid:1}
,function(err, res){});
this will solve your problem.
note: this only updates single instance.If multiple instances are returned in where condition result it will throw an error.
I have this:
exports.deleteSlide = function(data,callback){
customers.findOne(data.query,{'files.$':1},function(err,data2){
if(data2){
console.log(data2.files[0]);
data2.files[0].slides.splice((data.slide-1),1);
data2.files[0].markModified('slides');
data2.save(function(err,product,numberAffected){
if(numberAffected==1){
console.log("manifest saved");
var back={success:true};
console.log(product.files[0]);
callback(back);
return;
}
});
}
});
}
I get the "manifest saved" message and a callback with success being true.
When I do the console.log when I first find the data, and compare it with the console.log after I save the data, it looks like what I expect. I don't get any errors.
However, when I look at the database after running this code, it looks like nothing was ever changed. The element that I should have deleted, still appears?
What's wrong here?
EDIT:
For my query, I do {'name':'some string','files.name':'some string'}, and if the object is found, I get an array of files with one object in it.
I guess this is a subdoc.
I've looked around and it says the rules for saving subdocs are different than saving the entire collection, or rather, the subdocs are only applied when the root object is saved.
I've been going around this by grabbing the entire root object, then I do loops to find the actual subdoc I that I want, and after I manipulate that, I save the whole object.
Can I avoid doing this?
I'd probably just switch to using native drivers for this query as it is much simpler. (For that matter, I recently dropped mongoose on my primary project and am happy with the speed improvements.)
You can find documentation on getting access to the native collection elsewhere.
Following advice here:
https://stackoverflow.com/a/4588909/68567
customersNative.update(data.query, {$unset : {"slides.1" : 1 }}, function(err){
if(err) { return callback(err); }
customersNative.findAndModify(data.query, [],
{$pull: {'slides' : null } }, {safe: true, 'new' : true}, function(err, updated) {
//'updated' has new object
} );
});
How can I add a new row with the update operation
I am using following code
statuscollection.update({
id: record.id
}, {
id: record.id,
ip: value
}, {
upsert: true
}, function (err, result) {
console.log(err);
if (!err) {
return context.sendJson([], 404);
}
});
While calling this first one i will add the row
id: record.id
Then id:value then i have to add id:ggh
How can i add every new row by calling this function for each document I need to insert
By the structure of your code you are probably missing a few concepts.
You are using update in a case where you probably do not need to.
You seem to be providing an id field when the primary key for MongoDB would be _id. If that is what you mean.
If you are intending to add a new document on every call then you probably should be using insert. Your use of update with upsert has an intended usage of matching a document with the query criteria, if the document exists update the fields as specified, if not then insert a new document with the fields specified.
Unless that actually is your goal then insert is most certainly what you need. In that case you are likely to rely on the value of _id being populated automatically or by supplying your own unique value yourself. Unless you specifically want another field as an identifier that is not unique then you will likely want to be using the _id field as described before.