I'm trying to do a very easy task, add an item to a dictionary using "append".
This is the dict:
var myDictionary: [String:Int] = [
"Apple" : 1,
"Banana" : 2,
"Strawberry" : 3
]
I've tried this
myDictionary+=["Raspberry":4]
Here I get the error message:
binary operator cannot be apllied to two operands
and also I tried:
myDictionary.append("Raspberry":4)
and
myDictionary.append[("Raspberry":4)]
as well, but I get the error that it has :
no member 'append.
What exactly is the problem, how could I add the 4th item ?
Thanks for your help
append is the wrong tool here. You just want to set the value:
myDictionary["Raspberry"] = 4
append applies to things that conform to RangeReplaceableCollection. Dictionary does not. When you insert new things into a dictionary, they are not being appended to the end. They're being inserted into the appropriate buckets (perhaps replacing things already in place). If you use append to add something, you should reasonably expect last to then return that thing, but that's not promised (or even very likely) in a dictionary. Set is similar, and also has no append.
Related
Firebase Database
I tried using this bit of code but it doesn't seem to work. I take the name the user selects and store it in nameList.
Lets say I store Blake Wodruff in nameList[0].
How do I remove only that name?
var nameList = [String](repeating: "", count:100)
func remove() {
print(nameList[countAddNames])
let usernameRef = Database.database().reference().child("Candidate 1").child("alton").child(nameList[countAddNames]);
usernameRef.removeValue();
}
To write to or delete a node, you must specify its entire path. So to delete node 0 from your JSON, you'd do:
let usernameRef = Database.database().reference().child("Candidate 1").child("alton").child("0");
usernameRef.removeValue();
Or a bit shorter:
let usernameRef = Database.database().reference().child("Candidate 1/alton/0");
usernameRef.removeValue();
If you only know the name of the user you want to remove, you'll need to first look up its index/full path before you can remove it. If you have the data in your application already, you can do it in that code. Otherwise you may have to use a database query (specifically .queryOrderedByValue and .queryEqualToValue) to determine where the value exists in the database.
Also see: Delete a specific child node in Firebase swift
Once you remove a value from your JSON structure, Firebase may no longer recognize it as an array. For this reason it is highly recommended to not use arrays for the structure that you have. In fact, I'd model your data as a set, which in JSON would look like:
"alton": {
"Jake Jugg": true,
"Blake Wodruff": true,
"Alissa Sanchez": true
}
This would automatically:
Prevent duplicates, as each name can by definition only appear once.
Make removing a candidate by their name as easy as Database.database().reference().child("Candidate 1/alton/Jake Jugg").removeValue()
For more on this, also see my answer to Firebase query if child of child contains a value
I currently have two maps: -
mapBuffer = Map[String, ListBuffer[(Int, String, Float)]
personalMapBuffer = Map[mapBuffer, String]
The idea of what I'm trying to do is create a list of something, and then allow a user to create a personalised list which includes a comment, so they'd have their own list of maps.
I am simply trying to print information as everything is good from the above.
To print the Key from mapBuffer, I use: -
mapBuffer.foreach(line => println(line._1))
This returns: -
Sample String 1
Sample String 2
To print the same thing from personalMapBuffer, I am using: -
personalMapBuffer.foreach(line => println(line._1.map(_._1)))
However, this returns: -
List(Sample String 1)
List(Sample String 2)
I obviously would like it to just return "Sample String" and remove the List() aspect. I'm assuming this has something to do with the .map function, although this was the only way I could find to access a tuple within a tuple. Is there a simple way to remove the data type? I was hoping for something simple like: -
line._1.map(_._1).removeDataType
But obviously no such pre-function exists. I'm very new to Scala so this might be something extremely simple (which I hope it is haha) or it could be a bit more complex. Any help would be great.
Thanks.
What you see if default List.toString behaviour. You build your own string with mkString operation :
val separator = ","
personalMapBuffer.foreach(line => println(line._1.map(_._1.mkString(separator))))
which will produce desired result of Sample String 1 or Sample String 1, Sample String 2 if there will be 2 strings.
Hope this helps!
I have found a way to get the result I was looking for, however I'm not sure if it's the best way.
The .map() method just returns a collection. You can see more info on that here:- https://www.geeksforgeeks.org/scala-map-method/
By using any sort of specific element finder at the end, I'm able to return only the element and not the data type. For example: -
line._1.map(_._1).head
As I was writing this Ivan Kurchenko replied above suggesting I use .mkString. This also works and looks a little bit better than .head in my mind.
line._1.map(_._1).mkString("")
Again, I'm not 100% if this is the most efficient way but if it is necessary for something, this way has worked for me for now.
I work with an array of string, each string var is a coded object.
I want to decode the object, when I print a string var I get something structured like that :
"firstName=\"Elliot\" lastName=\"Alderson\" gender=\"male\" age=\"33\",some description I also need to get"
Is that a standard format to store key value properties ? I can't find anything on internet. The keys are always the same so that's not a big deal to get theses values as a dictionary but I would like to know if there is like a best practice method to get theses data instead of just searching for each key and then reach value from the first quote to the second one (for each value)
Because my file is 30000 lines so I better choose the more optimized way.
Thanks !
I am trying to update a string with firebase swift but I am getting an error that I do not know how to get rid of.
I have this code part that is getting an error:
self.dbRef.child("feed-items/\(dataPathen)/likesForPost").updateChildValues("likesForPost": "7")
The error I am getting is expected "," seperator just before the :. I am using dbRef in another code part so I know i works and the dataPathen is being printed just before the above code part, so that is working too.
Can anyone help me with this bug?
Just change
self.dbRef.child("feed-items/\(dataPathen)/likesForPost").updateChildValues("likesForPost": "7")
To
self.dbRef.child("feed-items/\(dataPathen)/likesForPost").updateChildValues(["likesForPost": "7"])
And if you are only looking for incrementing a particular value at a specific node you might wanna check my answer's :- https://stackoverflow.com/a/39465788/6297658, https://stackoverflow.com/a/39471374/6297658
PS Prefer runTransactionBlock: to update properties like likeForPosts as there might be a moment when two users try to like same post at the same moment (Highly Unlikely, but still a possibility...),using updateChildValues might end up just updating like only from one user. But runTransactionBlock: keep firing until the changes of that thread have been committed to the node
updateChildValues accepts [AnyHashable:Any] dictionary:
self.dbRef.child("feed-items/\(dataPathen)/likesForPost")
.updateChildValues(["likesForPost": "7"])
Whenever updating values at any reference in Firebase Database, you need to pass a dictionary parameter for updateChildValues method as [AnyHashable: Any] for your path reference. So just update your code of line as below:
self.dbRef.child("feed-items/(dataPathen)/likesForPost").updateChildValues("likesForPost": "7")
Also if you need to update more than 1 key-value pairs then you can pass those key-value pairs inside dictionary by seperating using comma as below:
self.dbRef.child("feed-items/(dataPathen)/likesForPost").updateChildValues(["likesForPost": "7", "otherKey": "OtherKeyValue"])
thanks for your time and attention on this beginner's issue
The book I've read claims that "You can use the first and last propeties, which return one of the elements in the set"
However, I've tried it and seems it is not working. I would very appreciate if someone can explain it for me.
You cannot ask for the last item of a Set. Indeed, a Set is not ordered, so there is not last item (nor last property).
Don't hesitate to take a look at the Swift Doc on Set.
If you want to find the last element, you need an ordered collection like an Array:
var myArray = [1, 2, 3, 4, 5]
print(myArray.last) // Display 5