按值从数组中删除一个项目

Question

我有一系列项目，例如：

var items = [id: "animal", type: "cat", cute: "yes"]

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

我被困住了！我可以通过拥有一个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。

有没有一种 jQuery 方法我可以在不需要遍历 items 数组的地方使用，而是指定一个选择器？

这是我的 jsFiddle：http://jsfiddle.net/zafrX/

Answer 1

我不确定按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，您可以将其概括为一个辅助函数，这样您就不必在任何地方重复它。

---

编辑

我刚刚注意到这个不正确的语法：

var items = [id: "animal", type: "cat", cute: "yes"]

你想要这样的东西吗：

 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;
    }

Answer 2

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');

Answer 3

您可以使用splice 或自己运行删除。这是一个例子：

for (var i = 0; i < items.length; i ++) {
    if (items[i] == "animal") { 
        items.splice(i, 1);
        break;
    }
}

Answer 4

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);

Answer 5

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，而是{}中的对象，然后你可以通过键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]
}

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 循环的最好方法是使用$.each（这是一个以更“功能性”的方式编写的循环）。

Answer 6

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);  

Answer 7

哇，这么多想法，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;
}

我还在这里找到了其他解决方案： Remove item from array by value

Answer 8

-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;
    }

Answer 9

ES6 解决方案

persons.splice(persons.findIndex((pm) => pm.id === personToDelete.id), 1);