我有一系列项目,如:
var items = [id: "animal", type: "cat", cute: "yes"]
我正在尝试删除与给定ID相匹配的任何项目。在这种情况下; animal
我被困住了!我可以通过使用更简单的数组来轻松地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。
我是否可以使用jQuery方法,我不需要遍历items数组,而是指定一个选择器?
这是我的jsFiddle:http://jsfiddle.net/zafrX/
答案 0 :(得分:21)
我不确定通过索引引用数组项有多麻烦。删除数组项的标准方法是使用splice method
for (var i = 0; i < items.length; i++)
if (items[i] === "animal") {
items.splice(i, 1);
break;
}
当然,您可以将其概括为辅助功能,因此您无需在任何地方复制此功能。
修改强>
我刚注意到这种错误的语法:
var items = [id: "animal", type: "cat", cute: "yes"]
你想要这样的东西:
var items = [ {id: "animal", type: "cat", cute: "yes"}, {id: "mouse", type: "rodent", cute: "no"}];
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;
}
答案 1 :(得分:7)
不需要jQuery或任何第三方lib,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');
答案 2 :(得分:6)
您可以使用splice
或自行运行删除。这是一个例子:
for (var i = 0; i < items.length; i ++) {
if (items[i] == "animal") {
items.splice(i, 1);
break;
}
}
答案 3 :(得分:2)
你应该这样做(确保你有正确的语法......你不能拥有属性的数组,但是{}
内的对象然后你可以按键迭代并删除不需要的键): / p>
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]
}
要回答你的问题,你不能将jquery选择器应用于javascript对象。避免for
循环的最佳方法是使用$.each(这是一种以更“功能”方式编写的循环)。
答案 4 :(得分:2)
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);
答案 5 :(得分:1)
有一个简单的方法!
myItems.splice(myItems.indexOf(myItems.find(row => row.id == id)), 1);
下面的演示:
// 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);
答案 6 :(得分:0)
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;
}
我也在这里找到了其他解决方案: Remove item from array by value
答案 7 :(得分:0)
-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>
将要从对象中删除
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;
}
答案 8 :(得分:0)
ES6解决方案
persons.splice(persons.findIndex((pm) => pm.id === personToDelete.id), 1);