Filter JSON using more than one option
I have a JSON:
location_filter = [{ title: "My Library", type: "library"},
{ title: "My Parking", type: "parking" },
{ title: "My Info", type: "info" }]
And I have a Multiple drop-down with three options: "info", "parking", "library".
I am already sending the value that I want to filter that can be one or more than one.
I know how to use the filter with just one item.
My question is: How can I use with more than one item?
splitFilter(array){
var arr = array.split(',');
return arr;
}
filter(selection) {
let stringSplit = this.splitFilter(selection.toString());
this.location_filter = JSON.parse(JSON.stringify(this.locations));
this.location_filter = this.location_filter.filter((item) => {
return item.type === selection;
});
this.loadMap(this.location_filter);
}
I am already splitting the string with the function splitFilter();
Maybe a loop?
javascript ionic3
asked Nov 14 '18 at 16:44
Low RiderLow Rider
75110
add a comment |
I have a JSON:
location_filter = [{ title: "My Library", type: "library"},
{ title: "My Parking", type: "parking" },
{ title: "My Info", type: "info" }]
And I have a Multiple drop-down with three options: "info", "parking", "library".
I am already sending the value that I want to filter that can be one or more than one.
I know how to use the filter with just one item.
My question is: How can I use with more than one item?
splitFilter(array){
var arr = array.split(',');
return arr;
}
filter(selection) {
let stringSplit = this.splitFilter(selection.toString());
this.location_filter = JSON.parse(JSON.stringify(this.locations));
this.location_filter = this.location_filter.filter((item) => {
return item.type === selection;
});
this.loadMap(this.location_filter);
}
I am already splitting the string with the function splitFilter();
Maybe a loop?
javascript ionic3
asked Nov 14 '18 at 16:44
Low RiderLow Rider
75110
As a note, this is not JSON - it's just a JavaScript Object.
– zfrisch
Nov 14 '18 at 16:56
add a comment |
I have a JSON:
location_filter = [{ title: "My Library", type: "library"},
{ title: "My Parking", type: "parking" },
{ title: "My Info", type: "info" }]
And I have a Multiple drop-down with three options: "info", "parking", "library".
I am already sending the value that I want to filter that can be one or more than one.
I know how to use the filter with just one item.
My question is: How can I use with more than one item?
splitFilter(array){
var arr = array.split(',');
return arr;
}
filter(selection) {
let stringSplit = this.splitFilter(selection.toString());
this.location_filter = JSON.parse(JSON.stringify(this.locations));
this.location_filter = this.location_filter.filter((item) => {
return item.type === selection;
});
this.loadMap(this.location_filter);
}
I am already splitting the string with the function splitFilter();
Maybe a loop?
javascript ionic3
asked Nov 14 '18 at 16:44
Low RiderLow Rider
75110
I have a JSON:
location_filter = [{ title: "My Library", type: "library"},
{ title: "My Parking", type: "parking" },
{ title: "My Info", type: "info" }]
And I have a Multiple drop-down with three options: "info", "parking", "library".
I am already sending the value that I want to filter that can be one or more than one.
I know how to use the filter with just one item.
My question is: How can I use with more than one item?
splitFilter(array){
var arr = array.split(',');
return arr;
}
filter(selection) {
let stringSplit = this.splitFilter(selection.toString());
this.location_filter = JSON.parse(JSON.stringify(this.locations));
this.location_filter = this.location_filter.filter((item) => {
return item.type === selection;
});
this.loadMap(this.location_filter);
}
I am already splitting the string with the function splitFilter();
Maybe a loop?
javascript ionic3
javascript ionic3
asked Nov 14 '18 at 16:44
Low RiderLow Rider
75110
asked Nov 14 '18 at 16:44
Low RiderLow Rider
75110
asked Nov 14 '18 at 16:44
Low RiderLow Rider
75110
asked Nov 14 '18 at 16:44
Low RiderLow Rider
75110
asked Nov 14 '18 at 16:44
Low RiderLow Rider
75110
75110
As a note, this is not JSON - it's just a JavaScript Object.
– zfrisch
Nov 14 '18 at 16:56
add a comment |
As a note, this is not JSON - it's just a JavaScript Object.
– zfrisch
Nov 14 '18 at 16:56
As a note, this is not JSON - it's just a JavaScript Object.
– zfrisch
Nov 14 '18 at 16:56
As a note, this is not JSON - it's just a JavaScript Object.
– zfrisch
Nov 14 '18 at 16:56
add a comment |
1 Answer
1
active
oldest
votes
You could check with Array#includes.
this.location_filter = this.location_filter.filter(({ type }) => selection.includes(type));
Or with ES5 take Array#indexOf and check if not -1 for a not found item.
this.location_filter = this.location_filter.filter((item) => {
return selection.indexOf(item.type) !== -1;
});
answered Nov 14 '18 at 16:47
Nina ScholzNina Scholz
185k1596168
I would like to say thank you for your answer. The first options is working like a charm! Thank you and have a great day.
– Low Rider
Nov 14 '18 at 16:52
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
StackExchange.using("externalEditor", function () {
StackExchange.using("snippets", function () {
StackExchange.snippets.init();
});
});
}, "code-snippets");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "1"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f53305025%2ffilter-json-using-more-than-one-option%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
You could check with Array#includes.
this.location_filter = this.location_filter.filter(({ type }) => selection.includes(type));
Or with ES5 take Array#indexOf and check if not -1 for a not found item.
this.location_filter = this.location_filter.filter((item) => {
return selection.indexOf(item.type) !== -1;
});
answered Nov 14 '18 at 16:47
Nina ScholzNina Scholz
185k1596168
I would like to say thank you for your answer. The first options is working like a charm! Thank you and have a great day.
– Low Rider
Nov 14 '18 at 16:52
add a comment |
You could check with Array#includes.
this.location_filter = this.location_filter.filter(({ type }) => selection.includes(type));
Or with ES5 take Array#indexOf and check if not -1 for a not found item.
this.location_filter = this.location_filter.filter((item) => {
return selection.indexOf(item.type) !== -1;
});
answered Nov 14 '18 at 16:47
Nina ScholzNina Scholz
185k1596168
I would like to say thank you for your answer. The first options is working like a charm! Thank you and have a great day.
– Low Rider
Nov 14 '18 at 16:52
add a comment |
You could check with Array#includes.
this.location_filter = this.location_filter.filter(({ type }) => selection.includes(type));
Or with ES5 take Array#indexOf and check if not -1 for a not found item.
this.location_filter = this.location_filter.filter((item) => {
return selection.indexOf(item.type) !== -1;
});
answered Nov 14 '18 at 16:47
Nina ScholzNina Scholz
185k1596168
You could check with Array#includes.
this.location_filter = this.location_filter.filter(({ type }) => selection.includes(type));
Or with ES5 take Array#indexOf and check if not -1 for a not found item.
this.location_filter = this.location_filter.filter((item) => {
return selection.indexOf(item.type) !== -1;
});
answered Nov 14 '18 at 16:47
Nina ScholzNina Scholz
185k1596168
answered Nov 14 '18 at 16:47
Nina ScholzNina Scholz
185k1596168
answered Nov 14 '18 at 16:47
Nina ScholzNina Scholz
185k1596168
answered Nov 14 '18 at 16:47
Nina ScholzNina Scholz
185k1596168
185k1596168
I would like to say thank you for your answer. The first options is working like a charm! Thank you and have a great day.
– Low Rider
Nov 14 '18 at 16:52
add a comment |
I would like to say thank you for your answer. The first options is working like a charm! Thank you and have a great day.
– Low Rider
Nov 14 '18 at 16:52
I would like to say thank you for your answer. The first options is working like a charm! Thank you and have a great day.
– Low Rider
Nov 14 '18 at 16:52
I would like to say thank you for your answer. The first options is working like a charm! Thank you and have a great day.
– Low Rider
Nov 14 '18 at 16:52
add a comment |
Thanks for contributing an answer to Stack Overflow!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f53305025%2ffilter-json-using-more-than-one-option%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
As a note, this is not JSON - it's just a JavaScript Object.
– zfrisch
Nov 14 '18 at 16:56