按值从数组中删除项目

时间:2012-01-15 21:42:03

标签: javascript jquery arrays each

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/

9 个答案:

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

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)

哇,这么多的想法,但仍然不是我想要的xD

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);