I want to create a map of members, but every membres have 3 propreties : first name, last name, and username. How can I create like a list of liste, but with a map.
So I want to have something like :
var membres= {['lastname': 'Bonneau',
'firstname': 'Pierre',
'username': 'mariobross'],
['lastname': 'Hamel',
'firstname': 'Alex',
'username': 'Queenlatifa'],
};
As you know, this code doesn't work. But it explain pretty well what I am trying to do.
I think you are confusing the two constructs here.
Read this introduction to the language: http://www.dartlang.org/docs/dart-up-and-running/ch02.html#lists
A list is a list of elements which can be denoted with the shorthand [...] syntax:
var list = [1, 2, "foo", 3, new Date.now(), 4];
Whereas a map can be denoted with the curly brace shorthand syntax:
var gifts = { // A map literal
// Keys Values
'first' : 'partridge',
'second' : 'turtledoves',
'fifth' : 'golden rings'
};
So, let's modify your code to work:
var members = [
{
'lastname': 'Bonneau',
'firstname': 'Pierre',
'username': 'mariobross'
},
{
'lastname': 'Hamel',
'firstname': 'Alex',
'username': 'Queenlatifa'
}
];
You can, for example, print the information like this:
members.forEach((e) {
print(e['firstname']);
});
If I understand your intent correctly, you want to have a list of maps. What you have is correct except you confused [ and {. The following works:
var membres = [
{'lastname': 'Bonneau',
'firstname': 'Pierre',
'username': 'mariobross'},
{'lastname': 'Hamel',
'firstname': 'Alex',
'username': 'Queenlatifa'}
];
As an example, to get a list of all usernames:
print(membres.map((v) => v['username']));
If you don't really need a Map, what about using a class to improve the structure of your code :
class Member {
String firstname;
String lastname;
String username;
Member(this.firstname, this.lastname, this.username);
}
main() {
final members = new List<Member>();
members.add(new Member('Pierre', 'Bonneau', 'mariobross'));
members.add(new Member('Alex', 'Hamel', 'Queenlatifa'));
// use members
}
You mean like this?
// FirstName => LastName => Value
var lookup = new Map<String, Map<String, String>>();
// get / set values like this
void setValue(String firstName, String lastName, String value) {
if (!lookUp.containsKey(firstName))
lookUp[firstName] = new Map<String, String>();
lookUp[firstName][lastName] = value;
}
String getValue(String firstName, String lastName) {
if (!lookUp.containsKey(firstName)) return "";
return lookUp[firstName][lastName];
}
First of all you need to create a map with value as list. Dont forget to initialize it
then if you want to fill it you first need to use built in function like putIfAbsent as in dart to add first object in list and then use update to add items in list. therefore you will need two arrays. First to put elements and then to add elements in list with same key. Also you can use try catch to identify if the key is present or not to do that in one loop
for (var item in days) {
var date_time = DateTime.parse(item["date"] + " 00:00:00");
_events[date_time] = _events.putIfAbsent(
date_time,
() => [
{
"title": item["title"],
"date": item["date"],
"time": reUse.get_time_am_pm_format(item["time"]),
"feature": item["feature"],
}
]);
}
for (var item in days) {
var date_time = DateTime.parse(item["date"] + " 00:00:00");
_events[date_time] = _events.update(date_time, (value) {
value.add({
"title": item["title"],
"date": item["date"],
"time": reUse.get_time_am_pm_format(item["time"]),
"feature": item["feature"],
});
return value;
});
}
Related
I have to admit, this is the first time I have to ask something that I dont even myself know how to ask for it or explain, so here is my code.
It worth explains that, for specific reasons I CANNOT change the output resource, this, the metadata sent to the resource has to stay as is, otherwise it will cause a recreate and I dont want that.
currently I have a terraform code that uses static/fixed variables like this
user1_name="Ed"
user1_Age ="10"
user2_name="Mat"
user2_Age ="20"
and then those hard typed variables get used in several places, but most importanly they are passed as metadata to instances, like so
resource "google_compute_instance_template" "mytemplate" {
...
metadata = {
othervalues = var.other
user1_name = var.user1_name
user1_Age = var.user1_Age
user2_name = var.user2_name
user2_Age = var.user2_Age
}
...
}
I am not an expert on terraform, thus asking, but I know for fact this is 100% ugly and wrong, and I need to use lists or array or whatever, so I am changing my declarations to this:
users = [
{ "name" : "yo", "age" : "10", "last" : "other" },
{ "name" : "El", "age" : "20", "last" : "other" }
]
but then, how do i get around to generate the same result for that resource? The resulting resource has to still have the same metadata as shown.
Assuming of course that the order of the users will be used as the "index" of the value, first one gets user1_name and so on...
I assume I need to use a for_each loop in there but cant figure out how to get around a loop inside properties of a resource
Not sure if I make myself clear on this, probably not, but didn't found a better way to explain.
From your example it seems like your intent is for these to all ultimately appear as a single map with keys built from two parts.
Your example doesn't make clear what the relationship is between user1 and Ed, though: your first example shows that "user1's" name is Ed, but in your example of the data structure you want to create there is only one "name" and it isn't clear to me whether that name would replace "user1" or "Ed" from your first example.
Instead, I'm going to take a slightly different variable structure which still maintains both the key like "user1" and the name attribute, like this:
variable "users" {
type = map(object({
name = string
age = number
})
}
locals {
# First we'll transform the map of objects into a
# flat set of key/attribute/value objects, because
# that's easier to work with when we generate the
# flattened map below.
users_flat = flatten([
for key, user in var.users : [
for attr, value in user : {
key = key
attr = attr
value = value
}
]
])
}
resource "google_compute_instance_template" "mytemplate" {
metadata = merge(
{
othervalues = var.other
},
{
for vo in local.users_flat : "${vo.key}_${vo.attr}" => vo.value
}
)
}
local.users_flat here is an intermediate data structure that flattens the two-level heirarchy of keys and object attributes from the input. It would be shaped something like this:
[
{ key = "user1", attr = "name", value = "Ed" },
{ key = "user1", attr = "age", value = 10 },
{ key = "user2", attr = "name", value = "Mat" },
{ key = "user2", attr = "age", value = 20 },
]
The merge call in the metadata argument then merges a directly-configured mapping of "other values" with a generated mapping derived from local.users_flat, shaped like this:
{
"user1_name" = "Ed"
"user1_age" = 10
"user2_name" = "Mat"
"user2_age" = 20
}
From the perspective of the caller of the module, the users variable should be defined with the following value in order to get the above results:
users = {
user1 = {
name = "Ed"
age = 10
}
user2 = {
name = "Mat"
age = 20
}
}
metadata is not a block, but a regular attribute of type map. So you can do:
# it would be better to use map, not list for users:
variable "users"
default {
user1 = { "name" : "yo", "age" : "10", "last" : "other" },
user2 = { "name" : "El", "age" : "20", "last" : "other" }
}
}
resource "google_compute_instance_template" "mytemplate" {
for_each = var.users
metadata = each.value
#...
}
I have two List<dynamic> and I am trying to figure out how I can check if there is a same value in the id field
List list1 = [
{"id": 2, "name": "test1"},
{"id": 3, "name": "test3"}
];
List list2 = [
{"id": 2, "name": "test1"}
];
I tried this but it returns me false
var isMatch = (list1.toSet().intersection(list2.toSet()).length > 0);
You can not compare like that because you can't compare dynamic as Boken said, you need to create a class for your object and implement a basic search , you can convert list2 into a set to make the search less complex (contains function)
void main() {
List list1 = [
MyObject(2,"test"),
MyObject(3,"test1")
];
List list2 = [
MyObject(4,"test")
];
for(int i=0;i<list1.length;i++){
if(list2.contains(list1[i])){
// do your logic
print(true);
break;
}
}
}
class MyObject{
int id;
String name;
MyObject(int id,String name){
this.id = id;
this.name = name;
}
// redifine == operator
bool operator ==(o) => (o as MyObject).id == this.id;
}
I am using flutter/dart and I have run into following problem.
I have a list of map like this.
var questions = [
{
'questionText': 'What\'s your favorite color?',
'answer': ['Black', 'Red', 'Green', 'White']
},
{
'questionText': 'What\'s your favorite animal?',
'answer': ['Deer', 'Tiger', 'Lion', 'Bear']
},
{
'questionText': 'What\'s your favorite movie?',
'answer': ['Die Hard', 'Due Date', 'Deep Rising', 'Dead or Alive']
},
];
Now suppose I need to get the string Tiger from this list. How do I do that? Dart is seeing this as List<Map<String, Object>> questions
Maybe a more portable way with a function:
String getAnswer(int question, int answer) {
return (questions[question]['answer'] as List<String>)[answer];
}
// Get 'Tiger'
String a = getAnswer(1, 1);
You can convert object in list in following way and then use index to get any value.
var p = questions[1]['answer'] as List<String>;
print(p[1]);
I am so sorry, but after one day researching and trying all different combinations and npm packages, I am still not sure how to deal with the following task.
Setup:
MongoDB 2.6
Node.JS with Mongoose 4
I have a schema like so:
var trackingSchema = mongoose.Schema({
tracking_number: String,
zip_code: String,
courier: String,
user_id: Number,
created: { type: Date, default: Date.now },
international_shipment: { type: Boolean, default: false },
delivery_info: {
recipient: String,
street: String,
city: String
}
});
Now user gives me a search string, a rather an array of strings, which will be substrings of what I want to search:
var search = ['15323', 'julian', 'administ'];
Now I want to find those documents, where any of the fields tracking_number, zip_code, or these fields in delivery_info contain my search elements.
How should I do that? I get that there are indexes, but I probably need a compound index, or maybe a text index? And for search, I then can use RegEx, or the $text $search syntax?
The problem is that I have several strings to look for (my search), and several fields to look in. And due to one of those aspects, every approach failed for me at some point.
Your use case is a good fit for text search.
Define a text index on your schema over the searchable fields:
trackingSchema.index({
tracking_number: 'text',
zip_code: 'text',
'delivery_info.recipient': 'text',
'delivery_info.street': 'text',
'delivery_info.city': 'text'
}, {name: 'search'});
Join your search terms into a single string and execute the search using the $text query operator:
var search = ['15232', 'julian'];
Test.find({$text: {$search: search.join(' ')}}, function(err, docs) {...});
Even though this passes all your search values as a single string, this still performs a logical OR search of the values.
Why just dont try
var trackingSchema = mongoose.Schema({
tracking_number: String,
zip_code: String,
courier: String,
user_id: Number,
created: { type: Date, default: Date.now },
international_shipment: { type: Boolean, default: false },
delivery_info: {
recipient: String,
street: String,
city: String
}
});
var Tracking = mongoose.model('Tracking', trackingSchema );
var search = [ "word1", "word2", ...]
var results = []
for(var i=0; i<search.length; i++){
Tracking.find({$or : [
{ tracking_number : search[i]},
{zip_code: search[i]},
{courier: search[i]},
{delivery_info.recipient: search[i]},
{delivery_info.street: search[i]},
{delivery_info.city: search[i]}]
}).map(function(tracking){
//it will push every unique result to variable results
if(results.indexOf(tracking)<0) results.push(tracking);
});
Okay, I came up with this.
My schema now has an extra field search with an array of all my searchable fields:
var trackingSchema = mongoose.Schema({
...
search: [String]
});
With a pre-save hook, I populate this field:
trackingSchema.pre('save', function(next) {
this.search = [ this.tracking_number ];
var searchIfAvailable = [
this.zip_code,
this.delivery_info.recipient,
this.delivery_info.street,
this.delivery_info.city
];
for (var i = 0; i < searchIfAvailable.length; i++) {
if (!validator.isNull(searchIfAvailable[i])) {
this.search.push(searchIfAvailable[i].toLowerCase());
}
}
next();
});
In the hope of improving performance, I also index that field (also the user_id as I limit search results by that):
trackingSchema.index({ search: 1 });
trackingSchema.index({ user_id: 1 });
Now, when searching I first list all substrings I want to look for in an array:
var andArray = [];
var searchTerms = searchRequest.split(" ");
searchTerms.forEach(function(searchTerm) {
andArray.push({
search: { $regex: searchTerm, $options: 'i'
}
});
});
I use this array in my find() and chain it with an $and:
Tracking.
find({ $and: andArray }).
where('user_id').equals(userId).
limit(pageSize).
skip(pageSize * page).
exec(function(err, docs) {
// hooray!
});
This works.
var thename = 'Andrew';
db.collection.find({'name':thename});
How do I query case insensitive? I want to find result even if "andrew";
Chris Fulstow's solution will work (+1), however, it may not be efficient, especially if your collection is very large. Non-rooted regular expressions (those not beginning with ^, which anchors the regular expression to the start of the string), and those using the i flag for case insensitivity will not use indexes, even if they exist.
An alternative option you might consider is to denormalize your data to store a lower-case version of the name field, for instance as name_lower. You can then query that efficiently (especially if it is indexed) for case-insensitive exact matches like:
db.collection.find({"name_lower": thename.toLowerCase()})
Or with a prefix match (a rooted regular expression) as:
db.collection.find( {"name_lower":
{ $regex: new RegExp("^" + thename.toLowerCase(), "i") } }
);
Both of these queries will use an index on name_lower.
You'd need to use a case-insensitive regular expression for this one, e.g.
db.collection.find( { "name" : { $regex : /Andrew/i } } );
To use the regex pattern from your thename variable, construct a new RegExp object:
var thename = "Andrew";
db.collection.find( { "name" : { $regex : new RegExp(thename, "i") } } );
Update: For exact match, you should use the regex "name": /^Andrew$/i. Thanks to Yannick L.
I have solved it like this.
var thename = 'Andrew';
db.collection.find({'name': {'$regex': thename,$options:'i'}});
If you want to query for case-insensitive and exact, then you can go like this.
var thename = '^Andrew$';
db.collection.find({'name': {'$regex': thename,$options:'i'}});
With Mongoose (and Node), this worked:
User.find({ email: /^name#company.com$/i })
User.find({ email: new RegExp(`^${emailVariable}$`, 'i') })
In MongoDB, this worked:
db.users.find({ email: { $regex: /^name#company.com$/i }})
Both lines are case-insensitive. The email in the DB could be NaMe#CompanY.Com and both lines will still find the object in the DB.
Likewise, we could use /^NaMe#CompanY.Com$/i and it would still find email: name#company.com in the DB.
MongoDB 3.4 now includes the ability to make a true case-insensitive index, which will dramtically increase the speed of case insensitive lookups on large datasets. It is made by specifying a collation with a strength of 2.
Probably the easiest way to do it is to set a collation on the database. Then all queries inherit that collation and will use it:
db.createCollection("cities", { collation: { locale: 'en_US', strength: 2 } } )
db.names.createIndex( { city: 1 } ) // inherits the default collation
You can also do it like this:
db.myCollection.createIndex({city: 1}, {collation: {locale: "en", strength: 2}});
And use it like this:
db.myCollection.find({city: "new york"}).collation({locale: "en", strength: 2});
This will return cities named "new york", "New York", "New york", etc.
For more info: https://jira.mongodb.org/browse/SERVER-90
... with mongoose on NodeJS that query:
const countryName = req.params.country;
{ 'country': new RegExp(`^${countryName}$`, 'i') };
or
const countryName = req.params.country;
{ 'country': { $regex: new RegExp(`^${countryName}$`), $options: 'i' } };
// ^australia$
or
const countryName = req.params.country;
{ 'country': { $regex: new RegExp(`^${countryName}$`, 'i') } };
// ^turkey$
A full code example in Javascript, NodeJS with Mongoose ORM on MongoDB
// get all customers that given country name
app.get('/customers/country/:countryName', (req, res) => {
//res.send(`Got a GET request at /customer/country/${req.params.countryName}`);
const countryName = req.params.countryName;
// using Regular Expression (case intensitive and equal): ^australia$
// const query = { 'country': new RegExp(`^${countryName}$`, 'i') };
// const query = { 'country': { $regex: new RegExp(`^${countryName}$`, 'i') } };
const query = { 'country': { $regex: new RegExp(`^${countryName}$`), $options: 'i' } };
Customer.find(query).sort({ name: 'asc' })
.then(customers => {
res.json(customers);
})
.catch(error => {
// error..
res.send(error.message);
});
});
To find case Insensitive string use this,
var thename = "Andrew";
db.collection.find({"name":/^thename$/i})
I just solved this problem a few hours ago.
var thename = 'Andrew'
db.collection.find({ $text: { $search: thename } });
Case sensitivity and diacritic sensitivity are set to false by default when doing queries this way.
You can even expand upon this by selecting on the fields you need from Andrew's user object by doing it this way:
db.collection.find({ $text: { $search: thename } }).select('age height weight');
Reference: https://docs.mongodb.org/manual/reference/operator/query/text/#text
You can use Case Insensitive Indexes:
The following example creates a collection with no default collation, then adds an index on the name field with a case insensitive collation. International Components for Unicode
/*
* strength: CollationStrength.Secondary
* Secondary level of comparison. Collation performs comparisons up to secondary * differences, such as diacritics. That is, collation performs comparisons of
* base characters (primary differences) and diacritics (secondary differences). * Differences between base characters takes precedence over secondary
* differences.
*/
db.users.createIndex( { name: 1 }, collation: { locale: 'tr', strength: 2 } } )
To use the index, queries must specify the same collation.
db.users.insert( [ { name: "Oğuz" },
{ name: "oğuz" },
{ name: "OĞUZ" } ] )
// does not use index, finds one result
db.users.find( { name: "oğuz" } )
// uses the index, finds three results
db.users.find( { name: "oğuz" } ).collation( { locale: 'tr', strength: 2 } )
// does not use the index, finds three results (different strength)
db.users.find( { name: "oğuz" } ).collation( { locale: 'tr', strength: 1 } )
or you can create a collection with default collation:
db.createCollection("users", { collation: { locale: 'tr', strength: 2 } } )
db.users.createIndex( { name : 1 } ) // inherits the default collation
This will work perfectly
db.collection.find({ song_Name: { '$regex': searchParam, $options: 'i' } })
Just have to add in your regex $options: 'i' where i is case-insensitive.
To find case-insensitive literals string:
Using regex (recommended)
db.collection.find({
name: {
$regex: new RegExp('^' + name.replace(/[-\/\\^$*+?.()|[\]{}]/g, '\\$&') + '$', 'i')
}
});
Using lower-case index (faster)
db.collection.find({
name_lower: name.toLowerCase()
});
Regular expressions are slower than literal string matching. However, an additional lowercase field will increase your code complexity. When in doubt, use regular expressions. I would suggest to only use an explicitly lower-case field if it can replace your field, that is, you don't care about the case in the first place.
Note that you will need to escape the name prior to regex. If you want user-input wildcards, prefer appending .replace(/%/g, '.*') after escaping so that you can match "a%" to find all names starting with 'a'.
Regex queries will be slower than index based queries.
You can create an index with specific collation as below
db.collection.createIndex({field:1},{collation: {locale:'en',strength:2}},{background : true});
The above query will create an index that ignores the case of the string. The collation needs to be specified with each query so it uses the case insensitive index.
Query
db.collection.find({field:'value'}).collation({locale:'en',strength:2});
Note - if you don't specify the collation with each query, query will not use the new index.
Refer to the mongodb doc here for more info - https://docs.mongodb.com/manual/core/index-case-insensitive/
The following query will find the documents with required string insensitively and with global occurrence also
db.collection.find({name:{
$regex: new RegExp(thename, "ig")
}
},function(err, doc) {
//Your code here...
});
An easy way would be to use $toLower as below.
db.users.aggregate([
{
$project: {
name: { $toLower: "$name" }
}
},
{
$match: {
name: the_name_to_search
}
}
])