Try this code
ArrayList<String> list = new ArrayList<String>();
JSONArray jsonArray = (JSONArray)jsonObject;
if (jsonArray != null) {
int len = jsonArray.length();
for (int i=0;i<len;i++){
list.add(jsonArray.get(i).toString());
}
}
//Remove the element from arraylist
list.remove(position);
//Recreate JSON Array
JSONArray jsArray = new JSONArray(list);
Edit:
Using ArrayList will add "\" to the key and values. So, use JSONArray itself
JSONArray list = new JSONArray();
JSONArray jsonArray = new JSONArray(jsonstring);
int len = jsonArray.length();
if (jsonArray != null) {
for (int i=0;i<len;i++)
{
//Excluding the item at position
if (i != position)
{
list.put(jsonArray.get(i));
}
}
}
Answer from Vinothkumar Arputharaj on Stack OverflowTry this code
ArrayList<String> list = new ArrayList<String>();
JSONArray jsonArray = (JSONArray)jsonObject;
if (jsonArray != null) {
int len = jsonArray.length();
for (int i=0;i<len;i++){
list.add(jsonArray.get(i).toString());
}
}
//Remove the element from arraylist
list.remove(position);
//Recreate JSON Array
JSONArray jsArray = new JSONArray(list);
Edit:
Using ArrayList will add "\" to the key and values. So, use JSONArray itself
JSONArray list = new JSONArray();
JSONArray jsonArray = new JSONArray(jsonstring);
int len = jsonArray.length();
if (jsonArray != null) {
for (int i=0;i<len;i++)
{
//Excluding the item at position
if (i != position)
{
list.put(jsonArray.get(i));
}
}
}
In case if someone returns with the same question for Android platform, you cannot use the inbuilt remove() method if you are targeting for Android API-18 or less. The remove() method is added on API level 19. Thus, the best possible thing to do is to extend the JSONArray to create a compatible override for the remove() method.
public class MJSONArray extends JSONArray {
@Override
public Object remove(int index) {
JSONArray output = new JSONArray();
int len = this.length();
for (int i = 0; i < len; i++) {
if (i != index) {
try {
output.put(this.get(i));
} catch (JSONException e) {
throw new RuntimeException(e);
}
}
}
return output;
//return this; If you need the input array in case of a failed attempt to remove an item.
}
}
EDIT As Daniel pointed out, handling an error silently is bad style. Code improved.
var json = { ... };
var key = "foo";
delete json[key]; // Removes json.foo from the dictionary.
You can use splice to remove elements from an array.
Do NOT have trailing commas in your OBJECT (JSON is a string notation)
UPDATE: you need to use array.splice and not delete if you want to remove items from the array in the object. Alternatively filter the array for undefined after removing
var data = {
"result": [{
"FirstName": "Test1",
"LastName": "User"
}, {
"FirstName": "user",
"LastName": "user"
}]
}
console.log(data.result);
console.log("------------ deleting -------------");
delete data.result[1];
console.log(data.result); // note the "undefined" in the array.
data = {
"result": [{
"FirstName": "Test1",
"LastName": "User"
}, {
"FirstName": "user",
"LastName": "user"
}]
}
console.log(data.result);
console.log("------------ slicing -------------");
var deletedItem = data.result.splice(1,1);
console.log(data.result); // here no problem with undefined.
var json = { ... };
var key = "foo";
delete json[key]; // Removes json.foo from the dictionary.
You can use splice to remove elements from an array.
Do NOT have trailing commas in your OBJECT (JSON is a string notation)
UPDATE: you need to use array.splice and not delete if you want to remove items from the array in the object. Alternatively filter the array for undefined after removing
var data = {
"result": [{
"FirstName": "Test1",
"LastName": "User"
}, {
"FirstName": "user",
"LastName": "user"
}]
}
console.log(data.result);
console.log("------------ deleting -------------");
delete data.result[1];
console.log(data.result); // note the "undefined" in the array.
data = {
"result": [{
"FirstName": "Test1",
"LastName": "User"
}, {
"FirstName": "user",
"LastName": "user"
}]
}
console.log(data.result);
console.log("------------ slicing -------------");
var deletedItem = data.result.splice(1,1);
console.log(data.result); // here no problem with undefined.
// Assuming this is your fetched data
const fetchMethodJsonArray = [{
"val": "One"
}, {
"val": "Two"
}, {
"val": "Three"
}];
var setValue = fetchMethodJsonArray;
const dataRemoved = setValue.filter((el) => {
return el.val !== "One";
});
console.log(dataRemoved);
Answer from what i got..
jsonArray.splice(jsonArray.indexOf('string_to_search'));
It will delete the found item and return remaining array.
You should convert it first, remove the element and then re-encode.
$json = json_decode($status, true); //return an array
foreach($json as $key => $value) {
if($value['value'] == 'Dispatched') {
unset($json[$key]);
}
}
$status = json_encode($json);
Since you have boolean true values in the array (that will match a type juggled true value such as string "Dispatched"), you need to pass true as the third parameter to in_array() for strict comparison.
Assuming you have run json_decode() and passed true for an array, just use strict comparison in in_array():
if (in_array('Dispatched', $value, true)) {
unset($status[$key]);
}
In this case, knowing the key I personally would use:
if ($value['value'] === 'Dispatched') {
unset($status[$key]);
}
delete operator is used to remove an object property.
delete operator does not returns the new object, only returns a boolean: true or false.
In the other hand, after interpreter executes var updatedjsonobj = delete myjsonobj['otherIndustry']; , updatedjsonobj variable will store a boolean
value.
How to remove Json object specific key and its value ?
You just need to know the property name in order to delete it from the object's properties.
delete myjsonobj['otherIndustry'];
let myjsonobj = {
"employeeid": "160915848",
"firstName": "tet",
"lastName": "test",
"email": "[email protected]",
"country": "Brasil",
"currentIndustry": "aaaaaaaaaaaaa",
"otherIndustry": "aaaaaaaaaaaaa",
"currentOrganization": "test",
"salary": "1234567"
}
delete myjsonobj['otherIndustry'];
console.log(myjsonobj);
If you want to remove a key when you know the value you can use Object.keys function which returns an array of a given object's own enumerable properties.
let value="test";
let myjsonobj = {
"employeeid": "160915848",
"firstName": "tet",
"lastName": "test",
"email": "[email protected]",
"country": "Brasil",
"currentIndustry": "aaaaaaaaaaaaa",
"otherIndustry": "aaaaaaaaaaaaa",
"currentOrganization": "test",
"salary": "1234567"
}
Object.keys(myjsonobj).forEach(function(key){
if (myjsonobj[key] === value) {
delete myjsonobj[key];
}
});
console.log(myjsonobj);
There are several ways to do this, lets see them one by one:
- delete method: The most common way
const myObject = {
"employeeid": "160915848",
"firstName": "tet",
"lastName": "test",
"email": "[email protected]",
"country": "Brasil",
"currentIndustry": "aaaaaaaaaaaaa",
"otherIndustry": "aaaaaaaaaaaaa",
"currentOrganization": "test",
"salary": "1234567"
};
delete myObject['currentIndustry'];
// OR delete myObject.currentIndustry;
console.log(myObject);
- By making key value undefined: Alternate & a faster way:
let myObject = {
"employeeid": "160915848",
"firstName": "tet",
"lastName": "test",
"email": "[email protected]",
"country": "Brasil",
"currentIndustry": "aaaaaaaaaaaaa",
"otherIndustry": "aaaaaaaaaaaaa",
"currentOrganization": "test",
"salary": "1234567"
};
myObject.currentIndustry = undefined;
myObject = JSON.parse(JSON.stringify(myObject));
console.log(myObject);
- With es6 spread Operator:
const myObject = {
"employeeid": "160915848",
"firstName": "tet",
"lastName": "test",
"email": "[email protected]",
"country": "Brasil",
"currentIndustry": "aaaaaaaaaaaaa",
"otherIndustry": "aaaaaaaaaaaaa",
"currentOrganization": "test",
"salary": "1234567"
};
const {currentIndustry, ...filteredObject} = myObject;
console.log(filteredObject);
Or if you can use omit() of underscore js library:
const filteredObject = _.omit(currentIndustry, 'myObject');
console.log(filteredObject);
When to use what??
If you don't wanna create a new filtered object, simply go for either option 1 or 2. Make sure you define your object with let while going with the second option as we are overriding the values. Or else you can use any of them.
hope this helps :)