react native how to filter JSON
up vote
1
down vote
favorite
1
This is my local JSON file (JsonData.json):
{
"Phone": [
{
"num": "phone1",
"name": "galaxy",
"temp": [
"5",
"6"
],
"id": 1
},
{
"num": "phone2",
"name": "iPhone",
"temp": [
"4",
"5",
"6"
],
"id": 2
},
…
],
"Computer": [
{
"num": "computer1",
"name": "Mac",
"temp": [
"1"
],
"id": 1
},
{
"num": "computer2",
"name": "Samsung",
"temp": [
"4"
],
"id": 2
},
…
]
}
And I imported it this way (I've been correctly using it):
var JsonData = require('./JsonData.json');
I want to get a name (galaxy, iPhone...) when temp = "4", from both phone and computer data.
How can I get them?
json react-native
edited Nov 22 at 12:16
c-chavez
2,11521630
asked Nov 22 at 11:57
ello
154
add a comment |
up vote
1
down vote
favorite
1
This is my local JSON file (JsonData.json):
{
"Phone": [
{
"num": "phone1",
"name": "galaxy",
"temp": [
"5",
"6"
],
"id": 1
},
{
"num": "phone2",
"name": "iPhone",
"temp": [
"4",
"5",
"6"
],
"id": 2
},
…
],
"Computer": [
{
"num": "computer1",
"name": "Mac",
"temp": [
"1"
],
"id": 1
},
{
"num": "computer2",
"name": "Samsung",
"temp": [
"4"
],
"id": 2
},
…
]
}
And I imported it this way (I've been correctly using it):
var JsonData = require('./JsonData.json');
I want to get a name (galaxy, iPhone...) when temp = "4", from both phone and computer data.
How can I get them?
json react-native
edited Nov 22 at 12:16
c-chavez
2,11521630
asked Nov 22 at 11:57
ello
154
add a comment |
up vote
1
down vote
favorite
1
up vote
1
down vote
favorite
1
1
This is my local JSON file (JsonData.json):
{
"Phone": [
{
"num": "phone1",
"name": "galaxy",
"temp": [
"5",
"6"
],
"id": 1
},
{
"num": "phone2",
"name": "iPhone",
"temp": [
"4",
"5",
"6"
],
"id": 2
},
…
],
"Computer": [
{
"num": "computer1",
"name": "Mac",
"temp": [
"1"
],
"id": 1
},
{
"num": "computer2",
"name": "Samsung",
"temp": [
"4"
],
"id": 2
},
…
]
}
And I imported it this way (I've been correctly using it):
var JsonData = require('./JsonData.json');
I want to get a name (galaxy, iPhone...) when temp = "4", from both phone and computer data.
How can I get them?
json react-native
edited Nov 22 at 12:16
c-chavez
2,11521630
asked Nov 22 at 11:57
ello
154
This is my local JSON file (JsonData.json):
{
"Phone": [
{
"num": "phone1",
"name": "galaxy",
"temp": [
"5",
"6"
],
"id": 1
},
{
"num": "phone2",
"name": "iPhone",
"temp": [
"4",
"5",
"6"
],
"id": 2
},
…
],
"Computer": [
{
"num": "computer1",
"name": "Mac",
"temp": [
"1"
],
"id": 1
},
{
"num": "computer2",
"name": "Samsung",
"temp": [
"4"
],
"id": 2
},
…
]
}
And I imported it this way (I've been correctly using it):
var JsonData = require('./JsonData.json');
I want to get a name (galaxy, iPhone...) when temp = "4", from both phone and computer data.
How can I get them?
json react-native
json react-native
edited Nov 22 at 12:16
c-chavez
2,11521630
asked Nov 22 at 11:57
ello
154
edited Nov 22 at 12:16
c-chavez
2,11521630
asked Nov 22 at 11:57
ello
154
edited Nov 22 at 12:16
c-chavez
2,11521630
edited Nov 22 at 12:16
c-chavez
2,11521630
edited Nov 22 at 12:16
c-chavez
2,11521630
2,11521630
asked Nov 22 at 11:57
ello
154
asked Nov 22 at 11:57
ello
154
asked Nov 22 at 11:57
ello
154
154
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
up vote
0
down vote
accepted
Try this:
const newArray = [...JsonData["Phone"], ...JsonData["Computer"]];
const result = newArray.filter(item => {
if(item.temp.indexOf("4") !== -1) return item.name;
});
The result will be: ["iPhone", "Samsung"].
answered Nov 22 at 22:56
Diamant
10816
Thank you! By the Way, I tried to print out the result like this, <Text style={styles.title}>{result}</Text> But ran into this error. "Invariant Violation: Objects are not valid as a React child(found:object with keys)". Do you know how to solve it?
– ello
Nov 23 at 2:55
I solved it using the flat list!
– ello
Nov 23 at 6:26
add a comment |
up vote
0
down vote
This code will help you to parse the JSON
function loopObj(value, array){
let keyList = Object.keys(array)
var namesArray =
for (var i = 0; i < keys.length; i++) {
let arrayVal = a[keys[i]];
for (var j = 0; j < arrayVal.length; j++) {
var innerArray = arrayVal[j]
if (innerArray.temp.indexOf(value) >= 0) {
namesArray.push(innerArray.name)
}
}
}
return(namesArray)
}
loopObj("4", a) // will give you the array ["iPhone", "Samsung"]
answered Nov 22 at 13:21
Manju Basha
302319
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
Try this:
const newArray = [...JsonData["Phone"], ...JsonData["Computer"]];
const result = newArray.filter(item => {
if(item.temp.indexOf("4") !== -1) return item.name;
});
The result will be: ["iPhone", "Samsung"].
answered Nov 22 at 22:56
Diamant
10816
Thank you! By the Way, I tried to print out the result like this, <Text style={styles.title}>{result}</Text> But ran into this error. "Invariant Violation: Objects are not valid as a React child(found:object with keys)". Do you know how to solve it?
– ello
Nov 23 at 2:55
I solved it using the flat list!
– ello
Nov 23 at 6:26
add a comment |
up vote
0
down vote
accepted
Try this:
const newArray = [...JsonData["Phone"], ...JsonData["Computer"]];
const result = newArray.filter(item => {
if(item.temp.indexOf("4") !== -1) return item.name;
});
The result will be: ["iPhone", "Samsung"].
answered Nov 22 at 22:56
Diamant
10816
Thank you! By the Way, I tried to print out the result like this, <Text style={styles.title}>{result}</Text> But ran into this error. "Invariant Violation: Objects are not valid as a React child(found:object with keys)". Do you know how to solve it?
– ello
Nov 23 at 2:55
I solved it using the flat list!
– ello
Nov 23 at 6:26
add a comment |
up vote
0
down vote
accepted
up vote
0
down vote
accepted
Try this:
const newArray = [...JsonData["Phone"], ...JsonData["Computer"]];
const result = newArray.filter(item => {
if(item.temp.indexOf("4") !== -1) return item.name;
});
The result will be: ["iPhone", "Samsung"].
answered Nov 22 at 22:56
Diamant
10816
Try this:
const newArray = [...JsonData["Phone"], ...JsonData["Computer"]];
const result = newArray.filter(item => {
if(item.temp.indexOf("4") !== -1) return item.name;
});
The result will be: ["iPhone", "Samsung"].
answered Nov 22 at 22:56
Diamant
10816
answered Nov 22 at 22:56
Diamant
10816
answered Nov 22 at 22:56
Diamant
10816
answered Nov 22 at 22:56
Diamant
10816
10816
Thank you! By the Way, I tried to print out the result like this, <Text style={styles.title}>{result}</Text> But ran into this error. "Invariant Violation: Objects are not valid as a React child(found:object with keys)". Do you know how to solve it?
– ello
Nov 23 at 2:55
I solved it using the flat list!
– ello
Nov 23 at 6:26
add a comment |
Thank you! By the Way, I tried to print out the result like this, <Text style={styles.title}>{result}</Text> But ran into this error. "Invariant Violation: Objects are not valid as a React child(found:object with keys)". Do you know how to solve it?
– ello
Nov 23 at 2:55
I solved it using the flat list!
– ello
Nov 23 at 6:26
Thank you! By the Way, I tried to print out the result like this, <Text style={styles.title}>{result}</Text> But ran into this error. "Invariant Violation: Objects are not valid as a React child(found:object with keys)". Do you know how to solve it?
– ello
Nov 23 at 2:55
Thank you! By the Way, I tried to print out the result like this, <Text style={styles.title}>{result}</Text> But ran into this error. "Invariant Violation: Objects are not valid as a React child(found:object with keys)". Do you know how to solve it?
– ello
Nov 23 at 2:55
I solved it using the flat list!
– ello
Nov 23 at 6:26
I solved it using the flat list!
– ello
Nov 23 at 6:26
add a comment |
up vote
0
down vote
This code will help you to parse the JSON
function loopObj(value, array){
let keyList = Object.keys(array)
var namesArray =
for (var i = 0; i < keys.length; i++) {
let arrayVal = a[keys[i]];
for (var j = 0; j < arrayVal.length; j++) {
var innerArray = arrayVal[j]
if (innerArray.temp.indexOf(value) >= 0) {
namesArray.push(innerArray.name)
}
}
}
return(namesArray)
}
loopObj("4", a) // will give you the array ["iPhone", "Samsung"]
answered Nov 22 at 13:21
Manju Basha
302319
add a comment |
up vote
0
down vote
This code will help you to parse the JSON
function loopObj(value, array){
let keyList = Object.keys(array)
var namesArray =
for (var i = 0; i < keys.length; i++) {
let arrayVal = a[keys[i]];
for (var j = 0; j < arrayVal.length; j++) {
var innerArray = arrayVal[j]
if (innerArray.temp.indexOf(value) >= 0) {
namesArray.push(innerArray.name)
}
}
}
return(namesArray)
}
loopObj("4", a) // will give you the array ["iPhone", "Samsung"]
answered Nov 22 at 13:21
Manju Basha
302319
add a comment |
up vote
0
down vote
up vote
0
down vote
This code will help you to parse the JSON
function loopObj(value, array){
let keyList = Object.keys(array)
var namesArray =
for (var i = 0; i < keys.length; i++) {
let arrayVal = a[keys[i]];
for (var j = 0; j < arrayVal.length; j++) {
var innerArray = arrayVal[j]
if (innerArray.temp.indexOf(value) >= 0) {
namesArray.push(innerArray.name)
}
}
}
return(namesArray)
}
loopObj("4", a) // will give you the array ["iPhone", "Samsung"]
answered Nov 22 at 13:21
Manju Basha
302319
This code will help you to parse the JSON
function loopObj(value, array){
let keyList = Object.keys(array)
var namesArray =
for (var i = 0; i < keys.length; i++) {
let arrayVal = a[keys[i]];
for (var j = 0; j < arrayVal.length; j++) {
var innerArray = arrayVal[j]
if (innerArray.temp.indexOf(value) >= 0) {
namesArray.push(innerArray.name)
}
}
}
return(namesArray)
}
loopObj("4", a) // will give you the array ["iPhone", "Samsung"]
answered Nov 22 at 13:21
Manju Basha
302319
answered Nov 22 at 13:21
Manju Basha
302319
answered Nov 22 at 13:21
Manju Basha
302319
answered Nov 22 at 13:21
Manju Basha
302319
302319
add a comment |
add a comment |
draft saved
draft discarded
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.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- 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.
draft saved
draft discarded
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%2f53430513%2freact-native-how-to-filter-json%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
