Shorting list by name - Angular 5 + Firebase
I have created a service where I get all the elements of my database:
Service
getElements() {
return (this.eleList= this.firebase.list("elements"));
}
Component
eleList: Element;
getBets() {
return this.databaseService
.getElements()
.snapshotChanges()
.subscribe(item => {
this.eleList= ;
item.forEach(element => {
let x = element.payload.toJSON();
x["$key"] = element.key;
this.eleList.push(x as Element);
});
});
}
With these two methods what I do is to store all my elements in this.eleList.
I would like to create a new method, named filterByName(name), where I would update this.eleList to an array which contains only the ones that contain namein the object, for example, this.eleList[1].name
I do not know if Firebase provides a way to short it, or I need to use Javascript/Typescript for it.
list typescript firebase firebase-realtime-database angularfire2
edited Nov 14 '18 at 14:19
Frank van Puffelen
234k29380407
asked Nov 14 '18 at 12:43
MarioMario
3891423
add a comment |
I have created a service where I get all the elements of my database:
Service
getElements() {
return (this.eleList= this.firebase.list("elements"));
}
Component
eleList: Element;
getBets() {
return this.databaseService
.getElements()
.snapshotChanges()
.subscribe(item => {
this.eleList= ;
item.forEach(element => {
let x = element.payload.toJSON();
x["$key"] = element.key;
this.eleList.push(x as Element);
});
});
}
With these two methods what I do is to store all my elements in this.eleList.
I would like to create a new method, named filterByName(name), where I would update this.eleList to an array which contains only the ones that contain namein the object, for example, this.eleList[1].name
I do not know if Firebase provides a way to short it, or I need to use Javascript/Typescript for it.
list typescript firebase firebase-realtime-database angularfire2
edited Nov 14 '18 at 14:19
Frank van Puffelen
234k29380407
asked Nov 14 '18 at 12:43
MarioMario
3891423
Firebase will not return it in the sorted order. you need to use some map filters to achieve the desired output.
– Rohit.007
Nov 14 '18 at 12:52
add a comment |
I have created a service where I get all the elements of my database:
Service
getElements() {
return (this.eleList= this.firebase.list("elements"));
}
Component
eleList: Element;
getBets() {
return this.databaseService
.getElements()
.snapshotChanges()
.subscribe(item => {
this.eleList= ;
item.forEach(element => {
let x = element.payload.toJSON();
x["$key"] = element.key;
this.eleList.push(x as Element);
});
});
}
With these two methods what I do is to store all my elements in this.eleList.
I would like to create a new method, named filterByName(name), where I would update this.eleList to an array which contains only the ones that contain namein the object, for example, this.eleList[1].name
I do not know if Firebase provides a way to short it, or I need to use Javascript/Typescript for it.
list typescript firebase firebase-realtime-database angularfire2
edited Nov 14 '18 at 14:19
Frank van Puffelen
234k29380407
asked Nov 14 '18 at 12:43
MarioMario
3891423
I have created a service where I get all the elements of my database:
Service
getElements() {
return (this.eleList= this.firebase.list("elements"));
}
Component
eleList: Element;
getBets() {
return this.databaseService
.getElements()
.snapshotChanges()
.subscribe(item => {
this.eleList= ;
item.forEach(element => {
let x = element.payload.toJSON();
x["$key"] = element.key;
this.eleList.push(x as Element);
});
});
}
With these two methods what I do is to store all my elements in this.eleList.
I would like to create a new method, named filterByName(name), where I would update this.eleList to an array which contains only the ones that contain namein the object, for example, this.eleList[1].name
I do not know if Firebase provides a way to short it, or I need to use Javascript/Typescript for it.
list typescript firebase firebase-realtime-database angularfire2
list typescript firebase firebase-realtime-database angularfire2
edited Nov 14 '18 at 14:19
Frank van Puffelen
234k29380407
asked Nov 14 '18 at 12:43
MarioMario
3891423
edited Nov 14 '18 at 14:19
Frank van Puffelen
234k29380407
asked Nov 14 '18 at 12:43
MarioMario
3891423
edited Nov 14 '18 at 14:19
Frank van Puffelen
234k29380407
edited Nov 14 '18 at 14:19
Frank van Puffelen
234k29380407
edited Nov 14 '18 at 14:19
Frank van Puffelen
234k29380407
234k29380407
asked Nov 14 '18 at 12:43
MarioMario
3891423
asked Nov 14 '18 at 12:43
MarioMario
3891423
asked Nov 14 '18 at 12:43
MarioMario
3891423
3891423
Firebase will not return it in the sorted order. you need to use some map filters to achieve the desired output.
– Rohit.007
Nov 14 '18 at 12:52
add a comment |
Firebase will not return it in the sorted order. you need to use some map filters to achieve the desired output.
– Rohit.007
Nov 14 '18 at 12:52
Firebase will not return it in the sorted order. you need to use some map filters to achieve the desired output.
– Rohit.007
Nov 14 '18 at 12:52
Firebase will not return it in the sorted order. you need to use some map filters to achieve the desired output.
– Rohit.007
Nov 14 '18 at 12:52
add a comment |
1 Answer
1
active
oldest
votes
Firebase takes full advantage of the observables and async pipes.
You should take advantage of that :
eleList$ = new Subject();
getElements() {
this.this.firebase.list("elements")
.pipe(take(1))
.subscribe(list => this.eleList$.next(list));
}
getBets() {
this.databaseService
.getElements()
.snapshotChanges()
.pipe(
map(item => items.map(element => ({
...element.payload.toJSON(),
'$key': element.key
})))
)
.subscribe(elements => this.eleList$.next(list));
}
Now for a sorted list :
sortedList$ = this.eleList$.pipe(
map(elements => elements.filter(element => !!element.name))
);
answered Nov 14 '18 at 12:52
trichetrichetrichetriche
26.7k42255
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Firebase takes full advantage of the observables and async pipes.
You should take advantage of that :
eleList$ = new Subject();
getElements() {
this.this.firebase.list("elements")
.pipe(take(1))
.subscribe(list => this.eleList$.next(list));
}
getBets() {
this.databaseService
.getElements()
.snapshotChanges()
.pipe(
map(item => items.map(element => ({
...element.payload.toJSON(),
'$key': element.key
})))
)
.subscribe(elements => this.eleList$.next(list));
}
Now for a sorted list :
sortedList$ = this.eleList$.pipe(
map(elements => elements.filter(element => !!element.name))
);
answered Nov 14 '18 at 12:52
trichetrichetrichetriche
26.7k42255
add a comment |
Firebase takes full advantage of the observables and async pipes.
You should take advantage of that :
eleList$ = new Subject();
getElements() {
this.this.firebase.list("elements")
.pipe(take(1))
.subscribe(list => this.eleList$.next(list));
}
getBets() {
this.databaseService
.getElements()
.snapshotChanges()
.pipe(
map(item => items.map(element => ({
...element.payload.toJSON(),
'$key': element.key
})))
)
.subscribe(elements => this.eleList$.next(list));
}
Now for a sorted list :
sortedList$ = this.eleList$.pipe(
map(elements => elements.filter(element => !!element.name))
);
answered Nov 14 '18 at 12:52
trichetrichetrichetriche
26.7k42255
add a comment |
Firebase takes full advantage of the observables and async pipes.
You should take advantage of that :
eleList$ = new Subject();
getElements() {
this.this.firebase.list("elements")
.pipe(take(1))
.subscribe(list => this.eleList$.next(list));
}
getBets() {
this.databaseService
.getElements()
.snapshotChanges()
.pipe(
map(item => items.map(element => ({
...element.payload.toJSON(),
'$key': element.key
})))
)
.subscribe(elements => this.eleList$.next(list));
}
Now for a sorted list :
sortedList$ = this.eleList$.pipe(
map(elements => elements.filter(element => !!element.name))
);
answered Nov 14 '18 at 12:52
trichetrichetrichetriche
26.7k42255
Firebase takes full advantage of the observables and async pipes.
You should take advantage of that :
eleList$ = new Subject();
getElements() {
this.this.firebase.list("elements")
.pipe(take(1))
.subscribe(list => this.eleList$.next(list));
}
getBets() {
this.databaseService
.getElements()
.snapshotChanges()
.pipe(
map(item => items.map(element => ({
...element.payload.toJSON(),
'$key': element.key
})))
)
.subscribe(elements => this.eleList$.next(list));
}
Now for a sorted list :
sortedList$ = this.eleList$.pipe(
map(elements => elements.filter(element => !!element.name))
);
answered Nov 14 '18 at 12:52
trichetrichetrichetriche
26.7k42255
answered Nov 14 '18 at 12:52
trichetrichetrichetriche
26.7k42255
answered Nov 14 '18 at 12:52
trichetrichetrichetriche
26.7k42255
answered Nov 14 '18 at 12:52
trichetrichetrichetriche
26.7k42255
26.7k42255
add a comment |
add a comment |
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Firebase will not return it in the sorted order. you need to use some map filters to achieve the desired output.
– Rohit.007
Nov 14 '18 at 12:52