[英]Remove an item from an array by value
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。
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?
这是我的 jsFiddle: http : //jsfiddle.net/zafrX/
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;
}
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。
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:
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;
}
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');
您可以使用splice
或自己运行删除。 下面是一个例子:
for (var i = 0; i < items.length; i ++) {
if (items[i] == "animal") {
items.splice(i, 1);
break;
}
}
有一个简单的方法!
myItems.splice(myItems.indexOf(myItems.find(row => row.id == id)), 1);
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);
你应该这样做(确保你有正确的语法......你不能有带属性的数组,但对象在{}
,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]
}
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)。
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);
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: 按值从数组中删除项目
-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;
}
ES6 解决方案
persons.splice(persons.findIndex((pm) => pm.id === personToDelete.id), 1);
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