How to manually set the Firestore document ID saving a new document calling Firestore API? - firebase

I am finding the following problem trying to save a new document into Firestore database by calling the related POST API and manually setting the document ID.
I am using Python but I suppose that the problem is not related to the language.
I try to explain what I have done and what is not working. My first attempt (that works but automatically set the document ID on Firestore) was:
First of all, I created this JSON document that will be the payload of my API:
# Convert the record to a dictionary
doc = {
'fields': {
'surname': {'stringValue':record[2]},
'firstName': {'stringValue':record[1]},
'socialSecurityCode': {'stringValue':codici_fiscali_list_as_string},
'city': {'stringValue':record[4]},
'personalPhone': {'stringValue':record[5]},
'personalPhone2': {'stringValue':record[6]},
'personalEmail': {'stringValue':emails_list_as_string},
'pazPres': {'stringValue':record[7]},
'pazNotes': {'stringValue':record[8]},
'pazMemo': {'stringValue':record[9]},
'isArchived': {'booleanValue':isArchived},
'isMigrated': {'booleanValue':True},
#'decomposition_keyword_search_list':{'arrayValue':{'values':decomposition_keyword_search_list}}
"decomposition_keyword_search_list":{
"arrayValue":{
"values":[
]
}
}
}
}
then I perform the API call by these lines:
api_endpoint = 'https://firestore.googleapis.com/v1/projects/MY-PROJECT-NAME/databases/(default)/documents/test/'
response = requests.post(api_endpoint, json=doc)
It works fine and it put the expected document into my test collection. But in this way, the ID was automatically generated by Firestore. For some reason, I have to use the content of a variable as ID (my ID must be the value of my record[0] that is a unique string)
So I tried to change the previous API endpoint in the following way:
api_endpoint = 'https://firestore.googleapis.com/v1/projects/MY-PROJECT-NAME/databases/(default)/documents/test/'+ record[0]
I expected that it creates a document using the record[0] as a document ID but it seems that I am wrong since I am obtaining the following error message:
Error saving document: {
"error": {
"code": 400,
"message": "Document parent name \"projects/MY-PROJECT-NAME/databases/(default)/documents/test\" lacks \"/\" at index 71.",
"status": "INVALID_ARGUMENT"
}
}
So, what is wrong? What am I missing? How can correctly manually set the ID of the document that I am creating calling the previous API?

Take a look at the documentation for creating documents. If you want to specify a document ID, it says you should pass that as a query parameter called documentId:
The client-assigned document ID to use for this document.
Optional. If not specified, an ID will be assigned by the service.

Related

Can I run a firestore query on a key that doesn't exist? [duplicate]

Let's say I have a data model with some optional properties. This could be for example a user object with a "firstname", a "lastname" and an optional "website" property.
In Cloud Firestore only user documents with a known website would have the "website" property set, for all other user documents this property would not exist.
My questions is now, how to query for all user documents without a "website" property?
Documents can contain properties with a null value data type (see data types documentation). This will then allow you to construct a query to limit results where the website property is null.
This is not quite the same as a missing property, but if you use custom objects to write data to Firestore, empty properties will automatically be saved as null rather than not at all. You can also manually/programmatically write a null value to the database.
In Android, I tested this using the following:
FirebaseFirestore.getInstance().collection("test").whereEqualTo("website", null).get();
Where my database structure looked like:
This returned only the document inuwlZOvZNTHuBakS6GV, because document 9Hf7uwORiiToOKz6zcsX contains a string value in the website property.
I believe you usually develop in Swift, where unfortunately custom objects aren't supported, but you can use NSNull() to write a null value to Firestore. For example (I'm not proficient in Swift, so feel free to correct any issues):
// Writing data
let docData: [String: Any] = [
"firstname": "Example",
"lastname": "User",
"website": NSNull()
]
db.collection("data").document("one").setData(docData) { err in
if let err = err {
print("Error writing document: \(err)")
} else {
print("Document successfully written!")
}
}
// Querying for null values
let query = db.collection("test").whereField("website", isEqualTo: NSNull())
The documentation doesn't mention a method to query for values that don't exist, so this seems like the next best approach. If anyone can improve or suggest alternatives, please do.

Is it possible to go through documents in cloud firestore to see if a value of a property is equal to a comparing one?

I have website written in plain javascript to keep daily to-do tasks and the app crashed lately because different tasks of the same date was created on accident. My question is...
how can i write an if statement that checks if a document from a collection has a property (in my case the date) that is equal to the one in the input field of my form. i guess it should check after i click submit? if it exists, creation should be denyed, if not, ok to proceed.
i am using cloud firestore by the way... many thanks in advance for the help!
First, make a query to get a document that has same date:
var query = db.collection("yourCollectionName").where("date", "==", dateInInputfield);
query.get().then(function(querySnapshot) {
if (querySnapshot.empty) {
//empty
} else {
// not empty
}
});
If empty{you can proceed}, if notEmpty{some other task already exist on same date}
If you are making an app like this, a cleaner approach will be to name the id of a document as it's date, for eg. if a task is created at timestamp of 1234567, create a document named 1234567 and inside it, store all the necessary information.
By following this approach, if you create a new task, simply fetch a document by the name in inputfield,
var docRef = db.collection("yourCollectionName").doc("date");
docRef.get().then(function(doc) {
if (doc.exists) {
//this means some other document already exists
} else {
//safe to create a new document by this date.
}
}).catch(function(error) {
console.log("Error:", error);
});

How can I make a firestore falsey query [duplicate]

Let's say I have a data model with some optional properties. This could be for example a user object with a "firstname", a "lastname" and an optional "website" property.
In Cloud Firestore only user documents with a known website would have the "website" property set, for all other user documents this property would not exist.
My questions is now, how to query for all user documents without a "website" property?
Documents can contain properties with a null value data type (see data types documentation). This will then allow you to construct a query to limit results where the website property is null.
This is not quite the same as a missing property, but if you use custom objects to write data to Firestore, empty properties will automatically be saved as null rather than not at all. You can also manually/programmatically write a null value to the database.
In Android, I tested this using the following:
FirebaseFirestore.getInstance().collection("test").whereEqualTo("website", null).get();
Where my database structure looked like:
This returned only the document inuwlZOvZNTHuBakS6GV, because document 9Hf7uwORiiToOKz6zcsX contains a string value in the website property.
I believe you usually develop in Swift, where unfortunately custom objects aren't supported, but you can use NSNull() to write a null value to Firestore. For example (I'm not proficient in Swift, so feel free to correct any issues):
// Writing data
let docData: [String: Any] = [
"firstname": "Example",
"lastname": "User",
"website": NSNull()
]
db.collection("data").document("one").setData(docData) { err in
if let err = err {
print("Error writing document: \(err)")
} else {
print("Document successfully written!")
}
}
// Querying for null values
let query = db.collection("test").whereField("website", isEqualTo: NSNull())
The documentation doesn't mention a method to query for values that don't exist, so this seems like the next best approach. If anyone can improve or suggest alternatives, please do.

Vue-Firebase: Updating children & writing to child

I started working with Firebase and Vue also with VueFire and i dont understand how to update child nodes at Firebase.
I opened a firebase project and connected to it and i can push data to it.
Firebase
I made a vue component
import db from '../FireBase'
let team = db.ref('Teams');//Reference to Teams at firebase
let miss = db.ref().child('Teams'); //Attempt to get to the children of Teams
export default {
name: "App",
firebase: {
Teams_loc: db.ref('Teams'),
Mission: this.Teams_loc.child('Teams'),
missionKey: db.ref().child('Teams').push("").key,
},
...
I manage to get the Teams from firebase and send data to it:
this.$firebaseRefs.Teams_loc.push({
"test": "tester"
});
Which works but when i try to update the children inside
this.miss.push({
"where": "am i"
})
I get the following error
Cannot read property 'child' of undefined
And when i try to update a child
this.$firebaseRefs.missionKey.update(arr[0]);//arr[0] is an object
I tried looking at quite a few places but nothing seems to do the trick.
Thanks,
When you do the following you are doing an error at the second line.
Teams_loc: db.ref('Teams'),
Mission: this.Teams_loc.child('Teams'),
There is no child of the Teams node that has a key with the value `Teams.
So if you want to update an item, you first have to get its key (e.g. -LEzOBT-mp.....) and do as follows, as explained in the doc:
updateItem: function (item) {
// create a copy of the item
const copy = {...item}
// remove the .key attribute
delete copy['.key']
//possibly update (or add) some values of (to) the item
this.$firebaseRefs.Teams_loc.child(item['.key']).set(copy)
}
Also (if I am not mistaking) doing db.ref() will generate an error because you have to pass a value to ref().
I suggest that you study a bit more the doc and the example: https://github.com/vuejs/vuefire and https://github.com/vuejs/vuefire/blob/master/examples/todo-app/index.html
Update following your comment. Details on how to "know the random generated key"
According to the documentation:
Each record in the bound array will contain a .key property which
specifies the key where the record is stored. So if you have data at
/Teams/-LEzOBT-mp...../, the record for that data will have a .key of
"-LEzOBT-mp.....".
Look at this part of the doc: https://github.com/vuejs/vuefire#array-bindings.
So with this you will get all the keys of the Teams object. You have now to choose the item you want to update.
You could also declare a query in your firebase object, like:
firebase: {
team21483: this.database
.ref('Teams')
.orderByChild('teamCode')
.equalTo('21483')
}
and you would get an array with only one team, the one with TeamCode = 21483.
The best approach, in this latest case, is to manually bind to this Firebase query with the $bindAsArray (or possibly the $bindAsObject) instance methods, using a variable that you pass to equalTo().

Firestore: How to query for non-existing (undefined) fields? [duplicate]

Let's say I have a data model with some optional properties. This could be for example a user object with a "firstname", a "lastname" and an optional "website" property.
In Cloud Firestore only user documents with a known website would have the "website" property set, for all other user documents this property would not exist.
My questions is now, how to query for all user documents without a "website" property?
Documents can contain properties with a null value data type (see data types documentation). This will then allow you to construct a query to limit results where the website property is null.
This is not quite the same as a missing property, but if you use custom objects to write data to Firestore, empty properties will automatically be saved as null rather than not at all. You can also manually/programmatically write a null value to the database.
In Android, I tested this using the following:
FirebaseFirestore.getInstance().collection("test").whereEqualTo("website", null).get();
Where my database structure looked like:
This returned only the document inuwlZOvZNTHuBakS6GV, because document 9Hf7uwORiiToOKz6zcsX contains a string value in the website property.
I believe you usually develop in Swift, where unfortunately custom objects aren't supported, but you can use NSNull() to write a null value to Firestore. For example (I'm not proficient in Swift, so feel free to correct any issues):
// Writing data
let docData: [String: Any] = [
"firstname": "Example",
"lastname": "User",
"website": NSNull()
]
db.collection("data").document("one").setData(docData) { err in
if let err = err {
print("Error writing document: \(err)")
} else {
print("Document successfully written!")
}
}
// Querying for null values
let query = db.collection("test").whereField("website", isEqualTo: NSNull())
The documentation doesn't mention a method to query for values that don't exist, so this seems like the next best approach. If anyone can improve or suggest alternatives, please do.

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