ionic Assign HttpClient response variables
I'm learning the ionic framework and want to know how REST calls are done correctly.
I have a RestProvider class which gets me a list of products
rest.ts
getProduct(query: string) {
return this.http.get(searchUri+query)
}
And in my home page I want to assign a value from the response to a local variable
home.ts
ionViewDidLoad() {
this.restProvider.getProduct(query).subscribe(it => {
console.log(it)
this.products = it.result;
})
}
But I get an error on assignment line this.products = it.result;
[ts] Property result does not exist on type Object
How do I create models of my response and assign them to class variables?
angular ionic-framework ionic2 ionic3
asked Nov 12 '18 at 10:54
ir2pidir2pid
97011436
add a comment |
I'm learning the ionic framework and want to know how REST calls are done correctly.
I have a RestProvider class which gets me a list of products
rest.ts
getProduct(query: string) {
return this.http.get(searchUri+query)
}
And in my home page I want to assign a value from the response to a local variable
home.ts
ionViewDidLoad() {
this.restProvider.getProduct(query).subscribe(it => {
console.log(it)
this.products = it.result;
})
}
But I get an error on assignment line this.products = it.result;
[ts] Property result does not exist on type Object
How do I create models of my response and assign them to class variables?
angular ionic-framework ionic2 ionic3
asked Nov 12 '18 at 10:54
ir2pidir2pid
97011436
check if 'it' contains a result property, if so check r u accessing that in a valid way
– Sa E Chowdary
Nov 12 '18 at 10:57
'it' is a JSON response and 'it.result' isn't explicitly defined anywhere, how can I do that?
– ir2pid
Nov 12 '18 at 11:05
1
That'll be because of Typescript. Change the service tothis.http.get<any>(searchUri+query)and it should work. Better yet, if you have an interface that represents the response, use that instead ofany
– user184994
Nov 12 '18 at 11:09
add a comment |
I'm learning the ionic framework and want to know how REST calls are done correctly.
I have a RestProvider class which gets me a list of products
rest.ts
getProduct(query: string) {
return this.http.get(searchUri+query)
}
And in my home page I want to assign a value from the response to a local variable
home.ts
ionViewDidLoad() {
this.restProvider.getProduct(query).subscribe(it => {
console.log(it)
this.products = it.result;
})
}
But I get an error on assignment line this.products = it.result;
[ts] Property result does not exist on type Object
How do I create models of my response and assign them to class variables?
angular ionic-framework ionic2 ionic3
asked Nov 12 '18 at 10:54
ir2pidir2pid
97011436
I'm learning the ionic framework and want to know how REST calls are done correctly.
I have a RestProvider class which gets me a list of products
rest.ts
getProduct(query: string) {
return this.http.get(searchUri+query)
}
And in my home page I want to assign a value from the response to a local variable
home.ts
ionViewDidLoad() {
this.restProvider.getProduct(query).subscribe(it => {
console.log(it)
this.products = it.result;
})
}
But I get an error on assignment line this.products = it.result;
[ts] Property result does not exist on type Object
How do I create models of my response and assign them to class variables?
angular ionic-framework ionic2 ionic3
angular ionic-framework ionic2 ionic3
asked Nov 12 '18 at 10:54
ir2pidir2pid
97011436
asked Nov 12 '18 at 10:54
ir2pidir2pid
97011436
asked Nov 12 '18 at 10:54
ir2pidir2pid
97011436
asked Nov 12 '18 at 10:54
ir2pidir2pid
97011436
asked Nov 12 '18 at 10:54
ir2pidir2pid
97011436
97011436
check if 'it' contains a result property, if so check r u accessing that in a valid way
– Sa E Chowdary
Nov 12 '18 at 10:57
'it' is a JSON response and 'it.result' isn't explicitly defined anywhere, how can I do that?
– ir2pid
Nov 12 '18 at 11:05
1
That'll be because of Typescript. Change the service tothis.http.get<any>(searchUri+query)and it should work. Better yet, if you have an interface that represents the response, use that instead ofany
– user184994
Nov 12 '18 at 11:09
add a comment |
check if 'it' contains a result property, if so check r u accessing that in a valid way
– Sa E Chowdary
Nov 12 '18 at 10:57
'it' is a JSON response and 'it.result' isn't explicitly defined anywhere, how can I do that?
– ir2pid
Nov 12 '18 at 11:05
1
That'll be because of Typescript. Change the service tothis.http.get<any>(searchUri+query)and it should work. Better yet, if you have an interface that represents the response, use that instead ofany
– user184994
Nov 12 '18 at 11:09
check if 'it' contains a result property, if so check r u accessing that in a valid way
– Sa E Chowdary
Nov 12 '18 at 10:57
check if 'it' contains a result property, if so check r u accessing that in a valid way
– Sa E Chowdary
Nov 12 '18 at 10:57
'it' is a JSON response and 'it.result' isn't explicitly defined anywhere, how can I do that?
– ir2pid
Nov 12 '18 at 11:05
'it' is a JSON response and 'it.result' isn't explicitly defined anywhere, how can I do that?
– ir2pid
Nov 12 '18 at 11:05
1
1
That'll be because of Typescript. Change the service to
this.http.get<any>(searchUri+query) and it should work. Better yet, if you have an interface that represents the response, use that instead of any– user184994
Nov 12 '18 at 11:09
That'll be because of Typescript. Change the service to
this.http.get<any>(searchUri+query) and it should work. Better yet, if you have an interface that represents the response, use that instead of any– user184994
Nov 12 '18 at 11:09
add a comment |
1 Answer
1
active
oldest
votes
The object you get in return is a JSON parsing of the response body. The error is that Typescript does now know if the resultproperty exists on the returned type. So you should disambiguate the situation.
You declare the type of the returned object with the template argument to get as in get<MyReturnedType>. So here, you should provide a template argument containing the result property.
Either you know the type for it, then use it, or you don't, then use any and you are responsible for the object accesses you make.
You can get more details on this Angular documentation page.
Note: don't use searchUri+query as your url because it is not escaped properly. You should instead use HttpParams described here.
answered Nov 12 '18 at 11:20
MicMic
1,95711531
add a comment |
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The object you get in return is a JSON parsing of the response body. The error is that Typescript does now know if the resultproperty exists on the returned type. So you should disambiguate the situation.
You declare the type of the returned object with the template argument to get as in get<MyReturnedType>. So here, you should provide a template argument containing the result property.
Either you know the type for it, then use it, or you don't, then use any and you are responsible for the object accesses you make.
You can get more details on this Angular documentation page.
Note: don't use searchUri+query as your url because it is not escaped properly. You should instead use HttpParams described here.
answered Nov 12 '18 at 11:20
MicMic
1,95711531
add a comment |
The object you get in return is a JSON parsing of the response body. The error is that Typescript does now know if the resultproperty exists on the returned type. So you should disambiguate the situation.
You declare the type of the returned object with the template argument to get as in get<MyReturnedType>. So here, you should provide a template argument containing the result property.
Either you know the type for it, then use it, or you don't, then use any and you are responsible for the object accesses you make.
You can get more details on this Angular documentation page.
Note: don't use searchUri+query as your url because it is not escaped properly. You should instead use HttpParams described here.
answered Nov 12 '18 at 11:20
MicMic
1,95711531
add a comment |
The object you get in return is a JSON parsing of the response body. The error is that Typescript does now know if the resultproperty exists on the returned type. So you should disambiguate the situation.
You declare the type of the returned object with the template argument to get as in get<MyReturnedType>. So here, you should provide a template argument containing the result property.
Either you know the type for it, then use it, or you don't, then use any and you are responsible for the object accesses you make.
You can get more details on this Angular documentation page.
Note: don't use searchUri+query as your url because it is not escaped properly. You should instead use HttpParams described here.
answered Nov 12 '18 at 11:20
MicMic
1,95711531
The object you get in return is a JSON parsing of the response body. The error is that Typescript does now know if the resultproperty exists on the returned type. So you should disambiguate the situation.
You declare the type of the returned object with the template argument to get as in get<MyReturnedType>. So here, you should provide a template argument containing the result property.
Either you know the type for it, then use it, or you don't, then use any and you are responsible for the object accesses you make.
You can get more details on this Angular documentation page.
Note: don't use searchUri+query as your url because it is not escaped properly. You should instead use HttpParams described here.
answered Nov 12 '18 at 11:20
MicMic
1,95711531
answered Nov 12 '18 at 11:20
MicMic
1,95711531
answered Nov 12 '18 at 11:20
MicMic
1,95711531
answered Nov 12 '18 at 11:20
MicMic
1,95711531
1,95711531
add a comment |
add a comment |
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check if 'it' contains a result property, if so check r u accessing that in a valid way
– Sa E Chowdary
Nov 12 '18 at 10:57
'it' is a JSON response and 'it.result' isn't explicitly defined anywhere, how can I do that?
– ir2pid
Nov 12 '18 at 11:05
1
That'll be because of Typescript. Change the service to
this.http.get<any>(searchUri+query)and it should work. Better yet, if you have an interface that represents the response, use that instead ofany– user184994
Nov 12 '18 at 11:09