Javascript 按值从数组中删除项目
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Remove an item from an array by value
提问by Alex Guerin
I have an array of items like:
我有一系列项目，例如：
var items = [id: "animal", type: "cat", cute: "yes"]
And I'm trying to remove any items that match the ID given. In this case; animal
我正在尝试删除与给定 ID 匹配的所有项目。在这种情况下;animal
I'm stuck! I can get it to work easily by having a more simpler array but this is not what I need... I also need to remove the item by value as I don't want the hassle of referring to items by their index.
我被卡住了！通过使用更简单的数组，我可以让它轻松工作，但这不是我需要的......我还需要按值删除项目，因为我不希望通过索引引用项目的麻烦。
Is there a jQuery method I could use where I don't need to iterate through the items array, rather specify a selector?
有没有我可以使用的 jQuery 方法，我不需要遍历 items 数组，而是指定一个选择器？
Here is my jsFiddle: http://jsfiddle.net/zafrX/
这是我的 jsFiddle：http: //jsfiddle.net/zafrX/
回答by Adam Rackis
I'm not sure how much of a hassle it is to refer to array items by index. The standard way to remove array items is with the splice method
我不确定按索引引用数组项有多少麻烦。删除数组项的标准方法是使用splice 方法
for (var i = 0; i < items.length; i++)
if (items[i] === "animal") {
items.splice(i, 1);
break;
}
And of course you can generalize this into a helper function so you don't have to duplicate this everywhere.
当然，您可以将其概括为辅助函数，这样您就不必在任何地方复制它。
EDIT
编辑
I just noticed this incorrect syntax:
我刚刚注意到这个不正确的语法：
var items = [id: "animal", type: "cat", cute: "yes"]
Did you want something like this:
你想要这样的东西：
var items = [ {id: "animal", type: "cat", cute: "yes"}, {id: "mouse", type: "rodent", cute: "no"}];
That would change the removal code to this:
这会将删除代码更改为：
for (var i = 0; i < items.length; i++)
if (items[i].id && items[i].id === "animal") {
items.splice(i, 1);
break;
}
回答by Chtiwi Malek
No need for jQuery or any third party lib for this, now we can use the new ES5 filter :
不需要 jQuery 或任何第三方库，现在我们可以使用新的 ES5 过滤器：
let myArray = [{ id : 'a1', name : 'Rabbit'}, { id : 'a2', name : 'Cat'}];
myArray = myArray.filter(i => i.id !== 'a1');
回答by Milad Naseri
You can either use spliceor run a delete yourself. Here's an example:
您可以splice自己使用或运行删除。下面是一个例子：
for (var i = 0; i < items.length; i ++) {
if (items[i] == "animal") {
items.splice(i, 1);
break;
}
}
回答by Playnox
There's a simple way!
有一个简单的方法！
myItems.splice(myItems.indexOf(myItems.find(row => row.id == id)), 1);
Demo below:
演示如下：
// define function
function delete_by_id(id) {
var myItems = [{
id: 1,
type: "cat",
cute: "yes"
}, {
id: 2,
type: "rat",
cute: "yes"
}, {
id: 3,
type: "mouse",
cute: "yes"
}];
// before
console.log(myItems);
myItems.splice(myItems.indexOf(myItems.find(item => item.id == id)), 1);
// after
console.log(myItems);
}
// call function
delete_by_id(1);回答by Aleksandar Vucetic
You should do it like this (make sure that you have the right syntax...you cannot have array with properties, but object inside {}and then you can iterate by keys and delete unwanted key):
你应该这样做（确保你有正确的语法......你不能有带有属性的数组，但里面有对象{}，然后你可以通过键进行迭代并删除不需要的键）：
var items = {id: "animal", type: "cat", cute: "yes"}
var removeItem = "animal"; // or with the ID matching animal...
for(var p in items){
if(items[p] === removeItem)
delete items[p]
}
And to answer you question, you cannot apply jquery selectors to javascript objects. The best you can do to avoid forloop is to use $.each(which is a loop written in a more "functional" way).
为了回答您的问题，您不能将 jquery 选择器应用于 javascript 对象。避免for循环的最佳方法是使用$.each（这是一个以更“功能”的方式编写的循环）。
回答by Jeremy D
By using object notation : http://jsfiddle.net/jrm2k6/zafrX/2/
通过使用对象表示法：http: //jsfiddle.net/jrm2k6/zafrX/2/
var animal1 = {id: "animal", type: "cat", cute: "yes"}
var car2 = {id: "car", type: "pick-up", cute: "no"}
var animal3 = {id: "animal", type: "dog", cute: "yes"}
var removeItem = "animal"; // or with the ID matching animal...
var array_items = []
array_items.push(animal1);
array_items.push(car2);
array_items.push(animal3);
for(var i=0;i<array_items.length;i++){
if(array_items[i].id == removeItem){
array_items.splice(i,1);
}
}
//alert(array_items.length);
回答by M?hre
Wow, so many ideas but still not what I wanted xD
哇，这么多想法，但仍然不是我想要的 xD
This will remove ALL entries of the given value and return the removed value:
这将删除给定值的所有条目并返回删除的值：
function removeOfArray(val, arr){
var idx;
var ret;
while ((idx = arr.indexOf(val)) > -1){
arr.splice(idx, 1);
ret = val;
}
return ret;
}
Also I found other solutions here: Remove item from array by value
我还在这里找到了其他解决方案： 按值从数组中删除项目
回答by R Maharajan
-parray : list of array of object
-pstring :value to remove from the array
-ptag :using which tag we
function removeFromArr (parray,ptag,pstring){
var b =[];
var count = 0;
for (var i =0;i<parray.length;i++){
if(pstring != parray[i][ptag]){
b[count] = parray[i];
count++;
}
}
return b;
}
var lobj = [ {
"SCHEME_CODE": "MIP65",
"YEARS": "1",
"CURRENCY": "GBP",
"MAX_AMT": 200000,
"MIN_AMT": 1000,
"AER_IR": "1.80",
"FREQUENCY": "Monthly",
"CUST_TYPE": "RETAIL",
"GROSS_IR": "1.79"
},
{
"SCHEME_CODE": "MIP65",
"YEARS": "2",
"CURRENCY": "GBP",
"MAX_AMT": 200000,
"MIN_AMT": 1000,
"AER_IR": "1.98",
"FREQUENCY": "Monthly",
"CUST_TYPE": "RETAIL",
"GROSS_IR": "1.96"
},
{
"SCHEME_CODE": "MIP65",
"YEARS": "3",
"CURRENCY": "GBP",
"MAX_AMT": 200000,
"MIN_AMT": 1000,
"AER_IR": "2.05",
"FREQUENCY": "Monthly",
"CUST_TYPE": "RETAIL",
"GROSS_IR": "2.03"
},
{
"SCHEME_CODE": "MIP65",
"YEARS": "5",
"CURRENCY": "GBP",
"MAX_AMT": 200000,
"MIN_AMT": 1000,
"AER_IR": "2.26",
"FREQUENCY": "Monthly",
"CUST_TYPE": "RETAIL",
"GROSS_IR": "2.24"
},
{
"SCHEME_CODE": "QIP65",
"YEARS": "1",
"CURRENCY": "GBP",
"MAX_AMT": 200000,
"MIN_AMT": 1000,
"AER_IR": "1.80",
"FREQUENCY": "Quarterly",
"CUST_TYPE": "RETAIL",
"GROSS_IR": "1.79"
},
{
"SCHEME_CODE": "QIP65",
"YEARS": "2",
"CURRENCY": "GBP",
"MAX_AMT": 200000,
"MIN_AMT": 1000,
"AER_IR": "1.98",
"FREQUENCY": "Quarterly",
"CUST_TYPE": "RETAIL",
"GROSS_IR": "1.97"
},
]
function myFunction(){
var final = removeFromArr(lobj,"SCHEME_CODE","MIP65");
console.log(final);
}<html>
<button onclick="myFunction()">Click me</button>
</html>are going to remove from the objects
将从对象中删除
function removeFromArr(parray, pstring, ptag) {
var farr = [];
var count = 0;
for (var i = 0; i < pa.length; i++) {
if (pstring != pa[i][ptag]) {
farr[count] = pa[i];
count++;
}
}
return farr;
}
