Related
for example look at this array, I want to remove those objects which the the value of their "age" is the same.
var array =[
{age:21,name:"sam",class="C"},
{age:24,name:"david",class="f"},
{age:45,name:"zack",class="f"},
{age:21,name:"jeff",class="g"},
{age:21,name:"marco",class="a"},
{age:26,name:"john",class="d"},
];
I want to get this result:
[
{age:21,name:"sam",class="C"},
{age:24,name:"david",class="f"},
{age:45,name:"zack",class="f"},
{age:26,name:"john",class="d"},
];
You can use reduce
var array = [
{age:21,name:"sam",class:"C"},
{age:24,name:"david",class:"f"},
{age:45,name:"zack",class:"f"},
{age:21,name:"jeff",class:"g"},
{age:21,name:"marco",class:"a"},
{age:26,name:"john",class:"d"}
];
let result = array.reduce((a,v) => {
let i = a.findIndex(person => person.age === v.age);
if(i !== -1){
return a;
}
return [...a, {...v}];
},[]);
console.log(result);
You can do this
var array =[
{age:21,name:"sam",class:"C"},
{age:24,name:"david",class:"f"},
{age:45,name:"zack",class:"f"},
{age:21,name:"jeff",class:"g"},
{age:21,name:"marco",class:"a"},
{age:26,name:"john",class:"d"},
];
var res = array.filter((i, index, self) => self.findIndex(k => k.age === i.age)==index);
console.log(res);
//Another clever way with lesser complexity :)
var res = array.reduce((a,v)=>{
if(!a[v.age]){
a[v.age] = v
};
return a
},{})
console.log(Object.values(res))
Let's say I have an array which I filter by calling myItems.filter(filterFunction1) and get some items from it.
Then I want to run another filtering function filterFunction2 against the remaining items which were not selected by filterFunction1.
Is that possible to get the remaining items that were left out after calling a filtering function?
You'd have to rerun the filter with an inverted predicate, which seems wasteful. You should reduce the items instead and bin them into one of two bins:
const result = arr.reduce((res, item) => {
res[predicate(item) ? 'a' : 'b'].push(item);
return res;
}, { a: [], b: [] });
predicate here is the callback you'd give to filter.
Unfortunately, there is no one-step solution based on filter. Still the solution is a simple one-liner:
Here's an example
const arr = [ 1,2,3,4,5,6,7,8 ];
const filtered = arr.filter(x=>!!(x%2))
const remaining = arr.filter(x=>!filtered.includes(x))
console.log(filtered, remaining);
You could map an array of flags and then filter by the flags values.
const cond = v => !(v % 2);
var array = [1, 2, 3, 4, 5],
flags = array.map(cond),
result1 = array.filter((_, i) => flags[i]),
result2 = array.filter((_, i) => !flags[i]);
console.log(result1);
console.log(result2);
You can achieve that using Array.reduce.
const myItems = [...];
const { result1, result2 } = myItems.reduce(
(result, item) => {
if (filterFunc1(item)) {
result.result1.push(item);
} else if (filterFunc2(item)) {
result.result2.push(item);
}
return result;
},
{ result1: [], result2: [] },
);
If you don't want to use reduce, you may want to iterate the array once and acquire the filtered and unfiltered items in a single shot, using a plain efficient for..of loop:
function filterAndDiscriminate(arr, filterCallback) {
const res = [[],[]];
for (var item of arr) {
res[~~filterCallback(item)].push(item);
}
return res;
}
const [unfiltered, filtered] = filterAndDiscriminate([1,2,3,4,5], i => i <= 3);
console.log(filtered, unfiltered);
There's a way more simple and readable way to do this:
const array1 = []
const array2 = []
itemsToFilter.forEach(item => item.condition === met ? array1.push(challenge) : array2.push(challenge))
I have below two arrays:
array1 = [{
"type":"test",
"name":"name1"},
{
"type":"dev",
"name":"name2"}]
array2=[{
"type":"test",
"name":"name3"},
{
"type":"dev",
"name":"name4"},
{
"type":"prod",
"name":"name5"}]
I want to group two arrays with "type" and create a new array something like this:
finalArray=[{
"type":"test",
"info":[{
"type":"test",
"name":"name1"}],
[{
"type":"test",
"name":"name3"
}]},
{
"type":"dev",
"info":[{
"type":"dev",
"name":"name2"}],
[{
"type":"dev",
"name":"name4"}]},
{
"type":"prod",
"info":[],
[{
"type":"prod",
"name":"name5"}]
}]
Is there anyway that I can achieve this using javascript, angularjs2, lodash, jquery. I am able to group and create new object as mentioned in using lodash .groupBy. how to add your own keys for grouped output?
But only thing is always I want to push the data from second array in index=1 of "info" and first one to index=0. If any of the array does not have a "type" then the "info" array should have empty/null values.
use _.mapValues to iterate object values with key accessing
var res = _.chain(array1)
.concat(array2)
.groupBy('type')
.mapValues(function(val, key) {
return {
type: key,
info: val
};
})
.values()
.value();
It's possible to achieve the result you want in javascript, or using helper like lodash.
The last part of your question is hard to understand. If an array doesn't have "type", how would you group them. Please provide clearer explanation or modify your expected input and output.
[Updated]
Thanks for your explanation. This is the solution using plain javascript.
// get uniques type from two arrays
const uniqueTypes = new Set(array1
.concat(array2)
.map(x => x.type));
// loop the types, find item in both array
// group it
let result = Array.from(uniqueTypes).reduce((acc, curr) => {
const item1 = array1.find(x => x.type === curr);
const item2 = array2.find(x => x.type === curr);
const info1 = item1 ? [item1] : [];
const info2 = item2 ? [item2] : [];
acc = acc.concat({ type: curr, info: [info1, info2] });
return acc;
}, []);
console.log(result);
jsbin here: https://jsbin.com/mobezogaso/edit?js,console
Here's a working solution :). Hope it helps!
var array1 = [
{
"type":"test",
"name":"name1"
},
{
"type":"dev",
"name":"name2"
}
]
var array2 = [
{
"type":"test",
"name":"name3"
},
{
"type":"dev",
"name":"name4"
},
{
"type":"prod",
"name":"name5"
}
]
var newArray = array1.concat(array2);
var arr1 = [];
var arr2 = [];
var arr3 = [];
var arrTypes = [];
var finalArray = [];
var someArray = [];
for(var i in newArray)
{
if (arrTypes.indexOf(newArray[i].type) === -1){
arrTypes.push(newArray[i].type);
}
if(newArray[i].type === "test"){
arr1.push(newArray[i]);
}
else if(newArray[i].type === "dev"){
arr2.push(newArray[i]);
}
else if(newArray[i].type === "prod"){
arr3.push(newArray[i]);
}
}
someArray.push(arr1);
someArray.push(arr2);
someArray.push(arr3);
for(var j = 0; j < someArray.length; j++){
finalArray.push({
"type": arrTypes[j],
"info": someArray[j]
});
}
console.log(finalArray);
And a short (unreadable?) ES6 solution:
Concat the arrays
Reduce the array into a Map object, with the type as the key
Get the entries iterator - key (type) - value (array of objects)
Use spread to convert the entry iterator to an array
Array#Map the array of entries to the type/info objects
const array1 = [{"type":"test","name":"name1"},{"type":"dev","name":"name2"}];
const array2=[{"type":"test","name":"name3"},{"type":"dev","name":"name4"},{"type":"prod","name":"name5"}];
const result = [...array1.concat(array2).reduce((r, o) => {
r.has(o.type) ? r.get(o.type).push(o) : r.set(o.type, [o]);
return r;
}, new Map).entries()]
.map(([type, info]) => ({
type, info
}));
console.log(result);
I want to check if an array contains "role". If it does, I want to move the "role" to the front of the array.
var data= ["email","role","type","name"];
if ("role" in data) data.remove(data.indexOf("role")); data.unshift("role")
data;
Here, I got the result:
["role", "email", "role", "type", "name"]
How can I fix this?
You can sort the array and specify that the value "role" comes before all other values, and that all other values are equal:
var first = "role";
data.sort(function(x,y){ return x == first ? -1 : y == first ? 1 : 0; });
Demo: http://jsfiddle.net/Guffa/7ST24/
The cleanest solution in ES6 in my opinion:
let data = ["email","role","type","name"];
data = data.filter(item => item !== "role");
data.unshift("role");
let data = [0, 1, 2, 3, 4, 5];
let index = 3;
data.unshift(data.splice(index, 1)[0]);
// data = [3, 0, 1, 2, 4, 5]
My first thought would be:
var data= ["email","role","type","name"];
// if it's not there, or is already the first element (of index 0)
// then there's no point going further:
if (data.indexOf('role') > 0) {
// find the current index of 'role':
var index = data.indexOf('role');
// using splice to remove elements from the array, starting at
// the identified index, and affecting 1 element(s):
data.splice(index,1);
// putting the 'role' string back in the array:
data.unshift('role');
}
console.log(data);
To revise, and tidy up a little:
if (data.indexOf('role') > 0) {
data.splice(data.indexOf('role'), 1);
data.unshift('role');
}
References:
Array.indexOf().
Array.prototype.splice().
Array.unshift().
Here is an immutable solution if needed :
const newData = [
data.find(item => item === 'role'),
...data.filter(item => item !== 'role'),
],
If you don't want to alter the existing array, you can use ES6 destructuring with the filter method to create a new copy while maintaining the order of the other items.
const data = ["email", "role", "type", "name"];
const newData = ['role', ...data.filter(item => item !== 'role')];
If you have an array of objects you could shift the start-index with splice and push. Splice replaces the original array with the part of the array starting from the desired index and returns the part it removes (the stuff before the index) which you push.
let friends = [{
id: 1,
name: "Sam",
},
{
id: 2,
name: "Steven",
},
{
id: 3,
name: "Tom",
},
{
id: 4,
name: "Nora",
},
{
id: 5,
name: "Jessy",
}
];
const tomsIndex = friends.findIndex(friend => friend.name == 'Tom');
friends.push(...friends.splice(0, tomsIndex));
console.log(friends);
To check whether an item exists in an array you should to use .includes() instead of in (as already noted here, in is for properties in objects).
This function does what you are looking for:
(removes the item from the position it is in and reads
in front)
data = ["email","role","type","name"];
moveToFirst("role", data);
function moveToFirst( stringToMove, arrayIn ){
if ( arrayIn.includes(stringToMove) ){
let currentIndex = arrayIn.indexOf(stringToMove);
arrayIn.splice(currentIndex, 1);
arrayIn.unshift(stringToMove);
}
}
console.log(data);
Similar to #Tandroid's answer but a more general solution:
const putItemsFirst = ({ findFunction, array }) => [
...array.filter(findFunction),
...array.filter(item => !findFunction(item)),
];
Can be used like this
putItemsFirst({
array: ["email","role","type","name"],
findFunction: item => item === 'role',
})
Something similar to this is what I ended up using,
I would go with this ES6 solution. It doesn't mutate the original array(considering it's not nested), doesn't traverse through the array(filter) and you're not just limited to 0th index for shifting the array item.
const moveArrayItem = (array, fromIndex, toIndex) => {
const arr = [...array];
arr.splice(toIndex, 0, ...arr.splice(fromIndex, 1));
return arr;
}
const arr = ["a", "b", "c", "d", "e", "f", "g"];
console.log(moveArrayItem(arr, 4, 0))
// [ 'e', 'a', 'b', 'c', 'd', 'f', 'g' ]
The most readable way in my opinion.
array.sort((a, b) => (a === value && -1) || (b === value && 1) || 0)
var data= ["email","role","type","name"];
data.splice(data.indexOf("role"), 1);
data.unshift('role');
You could take the delta of the check with the wanted value at top.
var data = ["email", "role", "type", "name"];
data.sort((a, b) => (b === 'role') - (a === 'role'));
console.log(data);
A reusable ES6/Typescript solution:
const moveToStart = <T>(array: T[], predicate: (item: T) => boolean): T[] => {
return array.sort((a, b) => {
if (predicate(a)) return -1;
if (predicate(b)) return 1;
return 0;
});
};
const data = ["email", "role", "type", "name"];
const result = moveToStart(data, (item) => item === "role"))
the in operator is about properties, not about items in arrays. See How do I check if an array includes an object in JavaScript? for what to use else.
You're missing braces around the two (!) statements in your if-block
I'm not sure whether that .remove() function you're using does take an index of an item.
Using lodash _.sortBy. If the item is role, it will be sorted first, otherwise second. This works fine too if there is no role
var data = ["email", "role", "type", "name"];
var sorted = _.sortBy(data, function(item) {
return item === 'role' ? 0 : 1;
});
console.log(sorted);
<script src="https://cdnjs.cloudflare.com/ajax/libs/lodash.js/4.17.4/lodash.min.js"></script>
Just wanted to drop this on here since according to other comments Guffa's answer seems to be gaining traction, the final tertiary - which was one of the negative comments on that answer is unnecessary. Also using arrow functions makes it seem much cleaner.
Also, it is easily expandable to handling Arrays of objects.
const first = "role";
data.sort((x, y) => first === x ? -1 : first === y)
I believe this should also handle the worry of the rest of the array being affected. When the sort function returns a number less than 0 (first === x), the element will move toward the start of the Array, when it returns 0 (first !== y), there will be no movement, and when a number greater than 0 (first === y), x will move toward the end of the Array, all in relation to x and y. Therefore, when neither x or y are equivalent to the desired first element (or it's identifier in the case of sorting objects), there will be no movement of the two in relation to each other.
For an object:
const unsorted = [{'id': 'test'}, {'id': 'something'}, {'id': 'else'}];
const first = 'something';
const sorted = unsorted.sort((x,y) => x['id'] === first ? -1 : y['id'] === first);
My solution is a bit different as it mutates original array instead of creating a new one.
It will move given item to start of the array and move item that was previously at start in the place of requested item.
function moveElementToStart<T>(items: T[], item: T) {
const itemIndex = items.indexOf(item);
// Item is not found or it is already on start
if (itemIndex === -1 || itemIndex === 0) return;
// Get item that is currently at start
const currentItemAtStart = items[0];
// Swap this item position with item we want to put on start
items[0] = item;
items[itemIndex] = currentItemAtStart;
}
Generalized one-liners:
const data = ["a", "b", "c", "d", "e", "f"];
const [from, take] = [3, 2];
data.unshift(...data.splice(from, take));
// alternatively
data = [...data.splice(from, take), ...data];
// ["d", "e", "a", "b", "c", "f"]
const moveToFront = (arr, queryStr) =>
arr.reduce((acc, curr) => {
if (queryStr === curr) {
return [curr, ...acc];
}
return [...acc, curr];
}, []);
const data = ['email', 'role', 'type', 'name'];
console.log(moveToFront(data, 'role'))
const moveTargetToBeginningOfArray = (arr, target) => {
// loop through array
for (let i = 0; i < arr.length; i++){
// if current indexed element is the target
if(arr[i] === target){
// remove that target element
arr.splice(i, 1)
// then add a target element to the beginning of the array
arr.unshift(target)
}
}
return arr;
};
// quick sanity check, before and after both are correct
const arrayOfStrings = ["email", "role", "type", "name", "role", "role"];
console.log('before:', arrayOfStrings)
console.log('after:', moveTargetToBeginningOfArray(arrayOfStrings, "role"))
// this would also work for numbers
var arrayOfNumbers = [2,4,0,3,0,1,0]
console.log('before:', arrayOfNumbers)
console.log('after:', moveTargetToBeginningOfArray(arrayOfNumbers, 0))
function unshiftFrom(arr, index) {
if (index > -1 && index < arr.length) { // validate index
var [itemToMove] = arr.splice(index, 1)
arr.unshift(itemToMove)
}
return arr // optional
}
//we can do this from scratch
let tempList=["person1","person2","person3"];
let result=[];
//suppose i need to move "person2" to first place
let movableValue=null;
let query="person2"; //here you could use any type of query based on your problem
tempList.map((e)=>{
if(e!==query){
result.push(e);
}else if(e===query){
movableValue=e;
}
})
if(movableValue!==null){
result.unshift(movableValue);
}
console.log(result)
)
var i = -1;
while (i < data.length) {
if (data[i] === "role") {
data.splice(i, 1);
break;
}
i++;
}
data.unshift("role");
indexOf only has limited browser support, not being recognized by IE7-8. So I wouldn't use it if I were you, even at the expense of a few lines' worth of code conciseness. You also want to put a semicolon at the end of the "unshift" statement. splice()'s first argument specifies the index to start removing elements, and the second argument specifies the number of arguments to remove.
data.unshift(data.splice(data.indexOf('role'), 1)[0])
data.indexOf('role') will find the index of 'role' in the array and then the original array is spliced to remove the 'role' element, which is added to the beginning of the array using unshift
var data= ["email","role","type","name"];
if ("role" in data) data.splice(data.indexOf("role"),1); data.unshift("role");
data;
Each item of this array is some number:
var items = Array(523,3452,334,31, ...5346);
How to replace some item with a new one?
For example, we want to replace 3452 with 1010, how would we do this?
var index = items.indexOf(3452);
if (index !== -1) {
items[index] = 1010;
}
Also it is recommend you not use the constructor method to initialize your arrays. Instead, use the literal syntax:
var items = [523, 3452, 334, 31, 5346];
You can also use the ~ operator if you are into terse JavaScript and want to shorten the -1 comparison:
var index = items.indexOf(3452);
if (~index) {
items[index] = 1010;
}
Sometimes I even like to write a contains function to abstract this check and make it easier to understand what's going on. What's awesome is this works on arrays and strings both:
var contains = function (haystack, needle) {
return !!~haystack.indexOf(needle);
};
// can be used like so now:
if (contains(items, 3452)) {
// do something else...
}
Starting with ES6/ES2015 for strings, and proposed for ES2016 for arrays, you can more easily determine if a source contains another value:
if (haystack.includes(needle)) {
// do your thing
}
The Array.indexOf() method will replace the first instance. To get every instance use Array.map():
a = a.map(function(item) { return item == 3452 ? 1010 : item; });
Of course, that creates a new array. If you want to do it in place, use Array.forEach():
a.forEach(function(item, i) { if (item == 3452) a[i] = 1010; });
Answer from #gilly3 is great.
Replace object in an array, keeping the array order unchanged
I prefer the following way to update the new updated record into my array of records when I get data from the server. It keeps the order intact and quite straight forward one liner.
users = users.map(u => u.id !== editedUser.id ? u : editedUser);
var users = [
{id: 1, firstname: 'John', lastname: 'Ken'},
{id: 2, firstname: 'Robin', lastname: 'Hood'},
{id: 3, firstname: 'William', lastname: 'Cook'}
];
var editedUser = {id: 2, firstname: 'Michael', lastname: 'Angelo'};
users = users.map(u => u.id !== editedUser.id ? u : editedUser);
console.log('users -> ', users);
My suggested solution would be:
items.splice(1, 1, 1010);
The splice operation will start at index 1, remove 1 item in the array (i.e. 3452), and will replace it with the new item 1010.
Use indexOf to find an element.
var i = items.indexOf(3452);
items[i] = 1010;
First method
Best way in just one line to replace or update item of array
array.splice(array.indexOf(valueToReplace), 1, newValue)
Eg:
let items = ['JS', 'PHP', 'RUBY'];
let replacedItem = items.splice(items.indexOf('RUBY'), 1, 'PYTHON')
console.log(replacedItem) //['RUBY']
console.log(items) //['JS', 'PHP', 'PYTHON']
Second method
An other simple way to do the same operation is :
items[items.indexOf(oldValue)] = newValue
Easily accomplished with a for loop.
for (var i = 0; i < items.length; i++)
if (items[i] == 3452)
items[i] = 1010;
If using a complex object (or even a simple one) and you can use es6, Array.prototype.findIndex is a good one. For the OP's array, they could do,
const index = items.findIndex(x => x === 3452)
items[index] = 1010
For more complex objects, this really shines. For example,
const index =
items.findIndex(
x => x.jerseyNumber === 9 && x.school === 'Ohio State'
)
items[index].lastName = 'Utah'
items[index].firstName = 'Johnny'
You can edit any number of the list using indexes
for example :
items[0] = 5;
items[5] = 100;
ES6 way:
const items = Array(523, 3452, 334, 31, ...5346);
We wanna replace 3452 with 1010, solution:
const newItems = items.map(item => item === 3452 ? 1010 : item);
Surely, the question is for many years ago and for now I just prefer to use immutable solution, definitely, it is awesome for ReactJS.
For frequent usage I offer below function:
const itemReplacer = (array, oldItem, newItem) =>
array.map(item => item === oldItem ? newItem : item);
A functional approach to replacing an element of an array in javascript:
const replace = (array, index, ...items) => [...array.slice(0, index), ...items, ...array.slice(index + 1)];
The immutable way to replace the element in the list using ES6 spread operators and .slice method.
const arr = ['fir', 'next', 'third'], item = 'next'
const nextArr = [
...arr.slice(0, arr.indexOf(item)),
'second',
...arr.slice(arr.indexOf(item) + 1)
]
Verify that works
console.log(arr) // [ 'fir', 'next', 'third' ]
console.log(nextArr) // ['fir', 'second', 'third']
Replacement can be done in one line:
var items = Array(523, 3452, 334, 31, 5346);
items[items.map((e, i) => [i, e]).filter(e => e[1] == 3452)[0][0]] = 1010
console.log(items);
Or create a function to reuse:
Array.prototype.replace = function(t, v) {
if (this.indexOf(t)!= -1)
this[this.map((e, i) => [i, e]).filter(e => e[1] == t)[0][0]] = v;
};
//Check
var items = Array(523, 3452, 334, 31, 5346);
items.replace(3452, 1010);
console.log(items);
var items = Array(523,3452,334,31,5346);
If you know the value then use,
items[items.indexOf(334)] = 1010;
If you want to know that value is present or not, then use,
var point = items.indexOf(334);
if (point !== -1) {
items[point] = 1010;
}
If you know the place (position) then directly use,
items[--position] = 1010;
If you want replace few elements, and you know only starting position only means,
items.splice(2, 1, 1010, 1220);
for more about .splice
The easiest way is to use some libraries like underscorejs and map method.
var items = Array(523,3452,334,31,...5346);
_.map(items, function(num) {
return (num == 3452) ? 1010 : num;
});
=> [523, 1010, 334, 31, ...5346]
If you want a simple sugar sintax oneliner you can just:
(elements = elements.filter(element => element.id !== updatedElement.id)).push(updatedElement);
Like:
let elements = [ { id: 1, name: 'element one' }, { id: 2, name: 'element two'} ];
const updatedElement = { id: 1, name: 'updated element one' };
If you don't have id you could stringify the element like:
(elements = elements.filter(element => JSON.stringify(element) !== JSON.stringify(updatedElement))).push(updatedElement);
var index = Array.indexOf(Array value);
if (index > -1) {
Array.splice(index, 1);
}
from here you can delete a particular value from array and based on the same index
you can insert value in array .
Array.splice(index, 0, Array value);
Well if anyone is interresting on how to replace an object from its index in an array, here's a solution.
Find the index of the object by its id:
const index = items.map(item => item.id).indexOf(objectId)
Replace the object using Object.assign() method:
Object.assign(items[index], newValue)
items[items.indexOf(3452)] = 1010
great for simple swaps. try the snippet below
const items = Array(523, 3452, 334, 31, 5346);
console.log(items)
items[items.indexOf(3452)] = 1010
console.log(items)
Here is the basic answer made into a reusable function:
function arrayFindReplace(array, findValue, replaceValue){
while(array.indexOf(findValue) !== -1){
let index = array.indexOf(findValue);
array[index] = replaceValue;
}
}
Here's a one liner. It assumes the item will be in the array.
var items = [523, 3452, 334, 31, 5346]
var replace = (arr, oldVal, newVal) => (arr[arr.indexOf(oldVal)] = newVal, arr)
console.log(replace(items, 3452, 1010))
const items = Array(1, 2, 3, 4, 5);
console.log(items)
items[items.indexOf(2)] = 1010
console.log(items)
First, rewrite your array like this:
var items = [523,3452,334,31,...5346];
Next, access the element in the array through its index number. The formula to determine the index number is: n-1
To replace the first item (n=1) in the array, write:
items[0] = Enter Your New Number;
In your example, the number 3452 is in the second position (n=2). So the formula to determine the index number is 2-1 = 1. So write the following code to replace 3452 with 1010:
items[1] = 1010;
I solved this problem using for loops and iterating through the original array and adding the positions of the matching arreas to another array and then looping through that array and changing it in the original array then return it, I used and arrow function but a regular function would work too.
var replace = (arr, replaceThis, WithThis) => {
if (!Array.isArray(arr)) throw new RangeError("Error");
var itemSpots = [];
for (var i = 0; i < arr.length; i++) {
if (arr[i] == replaceThis) itemSpots.push(i);
}
for (var i = 0; i < itemSpots.length; i++) {
arr[itemSpots[i]] = WithThis;
}
return arr;
};
presentPrompt(id,productqty) {
let alert = this.forgotCtrl.create({
title: 'Test',
inputs: [
{
name: 'pickqty',
placeholder: 'pick quantity'
},
{
name: 'state',
value: 'verified',
disabled:true,
placeholder: 'state',
}
],
buttons: [
{
text: 'Ok',
role: 'cancel',
handler: data => {
console.log('dataaaaname',data.pickqty);
console.log('dataaaapwd',data.state);
for (var i = 0; i < this.cottonLists.length; i++){
if (this.cottonLists[i].id == id){
this.cottonLists[i].real_stock = data.pickqty;
}
}
for (var i = 0; i < this.cottonLists.length; i++){
if (this.cottonLists[i].id == id){
this.cottonLists[i].state = 'verified';
}
}
//Log object to console again.
console.log("After update: ", this.cottonLists)
console.log('Ok clicked');
}
},
]
});
alert.present();
}
As per your requirement you can change fields and array names.
thats all. Enjoy your coding.
The easiest way is this.
var items = Array(523,3452,334,31, 5346);
var replaceWhat = 3452, replaceWith = 1010;
if ( ( i = items.indexOf(replaceWhat) ) >=0 ) items.splice(i, 1, replaceWith);
console.log(items);
>>> (5) [523, 1010, 334, 31, 5346]
When your array have many old item to replace new item, you can use this way:
function replaceArray(array, oldItem, newItem) {
for (let i = 0; i < array.length; i++) {
const index = array.indexOf(oldItem);
if (~index) {
array[index] = newItem;
}
}
return array
}
console.log(replaceArray([1, 2, 3, 2, 2, 8, 1, 9], 2, 5));
console.log(replaceArray([1, 2, 3, 2, 2, 8, 1, 9], 2, "Hi"));
let items = Array(523,3452,334,31, 5346);
items[0]=1010;
This will do the job
Array.prototype.replace = function(a, b) {
return this.map(item => item == a ? b : item)
}
Usage:
let items = ['hi', 'hi', 'hello', 'hi', 'hello', 'hello', 'hi']
console.log(items.replace('hello', 'hi'))
Output:
['hi', 'hi', 'hi', 'hi', 'hi', 'hi', 'hi']
The nice thing is, that EVERY array will have .replace() property.