如何从数组中删除对象? 我希望从someArray中删除包含名称Kristian的对象。例如:
someArray = [{name:"Kristian", lines:"2,5,10"},
{name:"John", lines:"1,19,26,96"}];
我想实现:
someArray = [{name:"John", lines:"1,19,26,96"}];
如何从数组中删除对象? 我希望从someArray中删除包含名称Kristian的对象。例如:
someArray = [{name:"Kristian", lines:"2,5,10"},
{name:"John", lines:"1,19,26,96"}];
我想实现:
someArray = [{name:"John", lines:"1,19,26,96"}];
当前回答
你也可以尝试这样做:
var myArray = [{'name': 'test'}, {'name':'test2'}];
var myObject = {'name': 'test'};
myArray.splice(myArray.indexOf(myObject),1);
其他回答
你也可以用一些:
someArray = [{name:"Kristian", lines:"2,5,10"},
{name:"John", lines:"1,19,26,96"}];
someArray.some(item => {
if(item.name === "Kristian") // Case sensitive, will only remove first instance
someArray.splice(someArray.indexOf(item),1)
})
虽然这可能不适合这种情况,我发现前几天,如果你不需要改变数组的大小,你也可以使用delete关键字从数组中删除一个项目。
var myArray = [1,2,3];
delete myArray[1];
console.log(myArray[1]); //undefined
console.log(myArray.length); //3 - doesn't actually shrink the array down
你可以使用以下几种方法从数组中删除项:
//1
someArray.shift(); // first element removed
//2
someArray = someArray.slice(1); // first element removed
//3
someArray.splice(0, 1); // first element removed
//4
someArray.pop(); // last element removed
//5
someArray = someArray.slice(0, someArray.length - 1); // last element removed
//6
someArray.length = someArray.length - 1; // last element removed
如果你想移除x位置的元素,使用:
someArray.splice(x, 1);
Or
someArray = someArray.slice(0, x).concat(someArray.slice(-x));
回复@chill182的评论:您可以使用array从数组中删除一个或多个元素。过滤器或数组。拼接结合数组。findIndex(参见MDN)。
请看这个Stackblitz项目或下面的代码片段:
// non destructive filter > noJohn = John removed, but someArray will not change let someArray = getArray(); let noJohn = someArray.filter( el => el.name !== "John" ); log(`let noJohn = someArray.filter( el => el.name !== "John")`, `non destructive filter [noJohn] =`, format(noJohn)); log(`**someArray.length ${someArray.length}`); // destructive filter/reassign John removed > someArray2 = let someArray2 = getArray(); someArray2 = someArray2.filter( el => el.name !== "John" ); log("", `someArray2 = someArray2.filter( el => el.name !== "John" )`, `destructive filter/reassign John removed [someArray2] =`, format(someArray2)); log(`**someArray2.length after filter ${someArray2.length}`); // destructive splice /w findIndex Brian remains > someArray3 = let someArray3 = getArray(); someArray3.splice(someArray3.findIndex(v => v.name === "Kristian"), 1); someArray3.splice(someArray3.findIndex(v => v.name === "John"), 1); log("", `someArray3.splice(someArray3.findIndex(v => v.name === "Kristian"), 1),`, `destructive splice /w findIndex Brian remains [someArray3] =`, format(someArray3)); log(`**someArray3.length after splice ${someArray3.length}`); // if you're not sure about the contents of your array, // you should check the results of findIndex first let someArray4 = getArray(); const indx = someArray4.findIndex(v => v.name === "Michael"); someArray4.splice(indx, indx >= 0 ? 1 : 0); log("", `someArray4.splice(indx, indx >= 0 ? 1 : 0)`, `check findIndex result first [someArray4] = (nothing is removed)`, format(someArray4)); log(`**someArray4.length (should still be 3) ${someArray4.length}`); // -- helpers -- function format(obj) { return JSON.stringify(obj, null, " "); } function log(...txt) { document.querySelector("pre").textContent += `${txt.join("\n")}\n` } function getArray() { return [ {name: "Kristian", lines: "2,5,10"}, {name: "John", lines: "1,19,26,96"}, {name: "Brian", lines: "3,9,62,36"} ]; } <pre> **Results** </pre>
如果你想删除给定对象的所有出现(基于某些条件),那么使用for循环中的javascript splice方法。
因为删除对象会影响数组长度,所以请确保减少计数器一步,以保持长度检查不变。
var objArr=[{Name:"Alex", Age:62},
{Name:"Robert", Age:18},
{Name:"Prince", Age:28},
{Name:"Cesar", Age:38},
{Name:"Sam", Age:42},
{Name:"David", Age:52}
];
for(var i = 0;i < objArr.length; i ++)
{
if(objArr[i].Age > 20)
{
objArr.splice(i, 1);
i--; //re-adjust the counter.
}
}
上面的代码片段删除了年龄大于20的所有对象。
someArray = jQuery.grep(someArray , function (value) {
return value.name != 'Kristian';
});