I am currently getting an object from my fauna database collection with an Index like so:
Get(
Match(
Index("byURL"),
"rv49z1"
)
)
I would also like to be able to increment a number variable (clicks) in the document I fetch each time I fetch it, ideally in one command.
Any help is greatly appreciated. Thank you.
Something like this should do the trick:
Let(
{ o: Get(...) },
Update(
Select("ref", Var("o")),
{
data: {
clicks: Add(1, Select(["data", "clicks"], Var("o")))
}
}
)
)
you probably want the document back before the update:
Let(
{
doc:Get(Match('byURL','rv49z1')),
ref:Select(['ref'],Var('doc')),
upd: Update(Var('ref'),{data:{clicks:Add(1,Select(['data','clicks'],Var('doc')))}})},
{
doc:Var('doc'),
upd:Var('upd')
}
)
But such operation is a read/delete/write operation at the storage engine level.
If your documents are big and you do that many times, I'd suggest moving the counter somewhere else.
Luigi
Related
I'm just learning FaunaDB and FQL and having some trouble (mainly because I come from MySQL). I can successfully query a table (eg: users) and fetch a specific user. This user has a property users.expiry_date which is a faunadb Time() type.
What I would like to do is know if this date has expired by using the function LT(Now(), users.expiry_date), but I don't know how to create this query. Do I have to create an Index first?
So in short, just fetching one of the users documents gets me this:
{
id: 1,
username: 'test',
expiry_date: Time("2022-01-10T16:01:47.394Z")
}
But I would like to get this:
{
id: 1,
username: 'test',
expiry_date: Time("2022-01-10T16:01:47.394Z"),
has_expired: true,
}
I have this FQL query now (ignore oauthInfo):
Query(
Let(
{
oauthInfo: Select(['data'], Get(Ref(Collection('user_oauth_info'), refId))),
user: Select(['data'], Get(Select(['user_id'], Var('oauthInfo'))))
},
Merge({ oauthInfo: Var('oauthInfo') }, { user: Var('user') })
)
)
How would I do the equivalent of the mySQL query SELECT users.*, IF(users.expiry_date < NOW(), 1, 0) as is_expired FROM users in FQL?
Your use of Let and Merge show that you are thinking about FQL in a good way. These are functions that can go a long way to making your queries more organized and readable!
I will start with some notes, but they will be relevant to the final answer, so please stick with me.
The Query function
https://docs.fauna.com/fauna/current/api/fql/functions/query
First, you should not need to wrap anything in the Query function, here. Query is necessary for defining functions in FQL that will be run later, for example, in the User-Defined Function body. You will always see it as Query(Lambda(...)).
Fauna IDs
https://docs.fauna.com/fauna/current/learn/understanding/documents
Remember that Fauna assigns unique IDs for every Document for you. When I see fields named id, that is a bit of a red flag, so I want to highlight that. There are plenty of reasons that you might store some business-ID in a Document, but be sure that you need it.
Getting an ID
A Document in Fauna is shaped like:
{
ref: Ref(Collection("users"), "101"), // <-- "id" is 101
ts: 1641508095450000,
data: { /* ... */ }
}
In the JS driver you can use this id by using documentResult.ref.id (other drivers can do this in similar ways)
You can access the ID directly in FQL as well. You use the Select function.
Let(
{
user: Get(Select(['user_id'], Var('oauthInfo')))
id: Select(["ref", "id"], Var("user"))
},
Var("id")
)
More about the Select function.
https://docs.fauna.com/fauna/current/api/fql/functions/select
You are already using Select and that's the function you are looking for. It's what you use to grab any piece of an object or array.
Here's a contrived example that gets the zip code for the 3rd user in the Collection:
Let(
{
page: Paginate(Documents(Collection("user")),
},
Select(["data", 2, "data", "address", "zip"], Var("user"))
)
Bring it together
That said, your Let function is a great start. Let's break things down into smaller steps.
Let(
{
oauthInfo_ref: Ref(Collection('user_oauth_info'), refId)
oauthInfo_doc: Get(Var("oathInfoRef")),
// make sure that user_oath_info.user_id is a full Ref, not just a number
user_ref: Select(["data", "user_id"], Var("oauthInfo_doc"))
user_doc: Get(Var("user_ref")),
user_id: Select("id", Var("user_ref")),
// calculate expired
expiry_date: Select(["data", "expiry_date"], Var("user_doc")),
has_expired: LT(Now(), Var("expiry_date"))
},
// if the data does not overlap, Merge is not required.
// you can build plain objects in FQL
{
oauthInfo: Var("oauthInfo_doc"), // entire Document
user: Var("user_doc"), // entire Document
has_expired: Var("has_expired") // an extra field
}
)
Instead of returning the auth info and user as separate points if you do want to Merge them and/or add additional fields, then feel free to do that
// ...
Merge(
Select("data", Var("user_doc")), // just the data
{
user_id: Var("user_id"), // added field
has_expired: Var("has_expired") // added field
}
)
)
FaunaDB's documentation covers how to update a document, but their example assumes that I'll have the id to pass into Ref:
Ref(schema_ref, id)
client.query(
q.Update(
q.Ref(q.Collection('posts'), '192903209792046592'),
{ data: { text: "Example" },
)
)
However, I'm wondering if it's possible to update a document without knowing its id. For instance, if I have a collection of users, can I find a user by their email, and then update their record? I've tried this, but Fauna returns a 400 (Database Ref expected, String provided):
client
.query(
q.Update(
q.Match(
q.Index("users_by_email", "me#example.com")
),
{ name: "Em" }
)
)
Although Bens comments are correct, (that's the way you do it), I wanted to note that the error you are receiving is because you are missing a bracket here: "users_by_email"), "me#example.com"
The error is logical if you know that Index takes an optional database reference as second argument.
To clarify what Ben said:
If you do this you'll get another error:
Update(
Match(
Index("accounts_by_email"), "test#test.com"
),
{ data: { email: "test2#test.com"} }
)
Since Match could potentially return more then one element. It returns a set of references called a SetRef. Think of setrefs as lists that are not materialized yet. If you are certain there is only one match for that e-mail (e.g. if you set a uniqueness constraint) you can materialize it using Paginate or Get:
Get:
Update(
Select(['ref'], Get(Match(
Index("accounts_by_email"), "test#test.com"
))),
{ data: { email: 'test2#test.com'} }
)
The Get returns the complete document, we need to specify that we require the ref with Select(['ref']..
Paginate:
Update(
Select(['data', 0],
Paginate(Match(
Index("accounts_by_email"), "test#test.com"
))
),
{ data: { email: "testchanged#test.com"} }
)
You are very close! Update does require a ref. You can get one via your index though. Assuming your index has a default values setting (i.e. paging a match returns a page of refs) and you are confident that the there is a single match or the first match is the one you want then you can do Select(["ref"], Get(Match(Index("users_by_email"), "me#example.com"))) to transform your set ref to a document ref. This can then be passed into update (or to any other function that wants a document ref, like Delete).
I'm having trouble getting the correct query with sequelize.
I have an array representing ids of entries lets say its like this -
userVacationsIds = [1,2,3]
i made the first query like this
Vacation.findAll({
where: {
id: {
[Op.or]: userVacationsIds
}
}
})
.then(vacationSpec => {
Vacation.findAll({
where:{
//Here i need to get all entries that DONT have the ids from the array
}
}
})
I can't get the correct query as specified in my code "comment"
I've tried referring to sequelize documentation but i can't understand how to chain these queries specifically
Also tried an online converter but that failed too.
Specified the code i have above
So i just need some help getting this query correct please.
I eventually expect to get 2 arrays - one containing all entries with the ids from the array, the other containing everything else (as in id is NOT in the array)
I figured it out.
I feel silly.
This is the query that worked
Vacation.findAll({
where: {
id: {
[Op.or]: userVacationsIds
}
}
}).then(vacationSpec => {
Vacation.findAll({
where: {
id: {
[Op.notIn]: userVacationsIds
}
}
})
Network response:
Vue instance:
Expected network response would be that the all_members would show all (in my case 12) club_members. If I query without using first/skip, it shows only 10 club_members (which is incorrect)
I currently have found a workaround with letting data.totalClubMembers.club_members override the data.club.all_members. Related discussion here: https://github.com/Akryum/vue-apollo/issues/196
(A part of) the query with arguments: $where: 1, $first: 10, $skip: 0.
query club($where: ClubWhereUniqueInput!, $first: Int, $skip: Int){
club: club(where: $where){
name
all_members: club_members {
id
}
club_members: club_members(first: $first, skip: $skip) {
id
category
club_reference_id
valid_from
valid_to
}
}
totalClubMembers: club(where: $where){
club_members {
id
}
}
}
Please advise:
if/where I made a mistake
if there is a more "optimal" solution.
Only root aliases seem to work for now (dec. 2018): https://github.com/prisma/graphql-yoga/issues/340
Refactor into 2 graphql queries will solve it.
I have a collection in which documents are all in this format:
{"user_id": ObjectId, "book_id": ObjectId}
It represents the relationship between user and book, which is also one-to-many, that means, a user can have more than one books.
Now I got three book_id, for example:
["507f191e810c19729de860ea", "507f191e810c19729de345ez", "507f191e810c19729de860efr"]
I want to query out the users who have these three books, because the result I want is not the document in this collection, but a newly constructed array of user_id, it seems complicated and I have no idea about how to make the query, please help me.
NOTE:
The reason why I didn't use the structure like:
{"user_id": ObjectId, "book_ids": [ObjectId, ...]}
is because in my system, books increase frequently and have no limit in amount, in other words, user may read thousands of books, so I think it's better to use the traditional way to store it.
This question is not restricted by MongoDB, you can answer it in relational database thoughts.
Using a regular find you cannot get back all user_id fields who own all the book_id's because you normalized your collection (flattened it).
You can do it, if you use aggregation framework:
db.collection.aggregate([
{
$match: {
book_id: {
$in: ["507f191e810c19729de860ea",
"507f191e810c19729de345ez",
"507f191e810c19729de860efr" ]
}
}
},
{
$group: {
_id: "$user_id",
count: { $sum: 1 }
}
},
{
$match: {
count: 3
}
},
{
$group: {
_id: null,
users: { $addToSet: "$_id" }
}
}
]);
What this does is filters through the pipeline only for documents which match one of the three book_id values, then it groups by user_id and counts how many matches that user got. If they got three they pass to the next pipeline operation which groups them into an array of user_ids. This solution assumes that each 'user_id,book_id' record can only appear once in the original collection.