我一直在尝试几种方法，如何在一个数组中找到一个对象，其中ID = var，如果找到，从数组中删除对象，并返回新的对象数组。
数据:
[
{"id":"88","name":"Lets go testing"},
{"id":"99","name":"Have fun boys and girls"},
{"id":"108","name":"You are awesome!"}
]
我能够搜索数组使用jQuery $grep;
var id = 88;
var result = $.grep(data, function(e){
return e.id == id;
});
但是，当id == 88时，如何删除整个对象，并返回如下数据?
数据:
[
{"id":"99", "name":"Have fun boys and girls"},
{"id":"108", "name":"You are awesome!"}
]
当前回答
const data = [
{"id":"88","name":"Lets go testing"},
{"id":"99","name":"Have fun boys and girls"},
{"id":"108","name":"You are awesome!"}
];
这里我们得到了id值为88的对象的索引
const index = data.findIndex(item => item.id === "88");
console.log(index); // 0
我们使用splice函数从数据数组中删除指定的对象
data.splice(index,1);
console.log(data); // [{"id":"99","name":"Have fun boys and girls"},{"id":"108","name":"You are awesome!"}]
其他回答
如果测试严格相等，请确保将对象id强制为整数:
var result = $.grep(data, function(e, i) {
return +e.id !== id;
});
Demo
var items = [
{"id":"88","name":"Lets go testing"},
{"id":"99","name":"Have fun boys and girls"},
{"id":"108","name":"You are awesome!"}
];
如果你用的是jQuery，那就用jQuery吧。Grep是这样的:
items = $.grep(items, function(item) {
return item.id !== '88';
});
// items => [{ id: "99" }, { id: "108" }]
使用ES5 Array.prototype.filter:
items = items.filter(function(item) {
return item.id !== '88';
});
// items => [{ id: "99" }, { id: "108" }]
const data = [
{"id":"88","name":"Lets go testing"},
{"id":"99","name":"Have fun boys and girls"},
{"id":"108","name":"You are awesome!"}
];
这里我们得到了id值为88的对象的索引
const index = data.findIndex(item => item.id === "88");
console.log(index); // 0
我们使用splice函数从数据数组中删除指定的对象
data.splice(index,1);
console.log(data); // [{"id":"99","name":"Have fun boys and girls"},{"id":"108","name":"You are awesome!"}]
ES6原生解决方案:
const pos = data.findIndex(el => el.id === ID_TO_REMOVE);
if (pos >= 0)
data.splice(pos, 1);
如果你确定元素在数组中:
data.splice(data.findIndex(el => el.id === ID_TO_REMOVE), 1);
原型:
Array.prototype.removeByProp = function(prop,val) {
const pos = this.findIndex(x => x[prop] === val);
if (pos >= 0)
return this.splice(pos, 1);
};
// usage:
ar.removeByProp('id', ID_TO_REMOVE);
http://jsfiddle.net/oriadam/72kgprw5/
注意:这将就地移除项目。如果你需要一个新的数组，使用过滤器在前面的答案中提到的。
也许你正在寻找$.grep()函数:
arr = [
{"id":"88","name":"Lets go testing"},
{"id":"99","name":"Have fun boys and girls"},
{"id":"108","name":"You are awesome!"}
];
id = 88;
arr = $.grep(arr, function(data, index) {
return data.id != id
});
