我有以下对象:
[{
"id": 2,
"price": 2000,
"name": "Mr Robot T1",
"image": "http://placehold.it/270x335"
}, {
"id": 1,
"price": 1000,
"name": "Mr Robot T2",
"image": "http://placehold.it/270x335"
}]
我想要的是删除第一项(id = 1),结果是:
[{
"id": 2,
"price": 2000,
"name": "Mr Robot T1",
"image": "http://placehold.it/270x335"
}]
它可以吗?
答案 0 :(得分:0)
类似下面的代码?
var array = [{
"id": 2,
"price": 2000,
"name": "Mr Robot T1",
"image": "http://placehold.it/270x335"
}, {
"id": 1,
"price": 1000,
"name": "Mr Robot T2",
"image": "http://placehold.it/270x335"
}];
var newArray = [];
array.forEach(function(item){
if (item.id != 1) {
newArray.push(item);
}
});
基本上,这将遍历您的数组,并且所有没有id = 1
的元素都会被推送到变量newArray
。
修改强>
要删除该项目,您始终可以splice。如下所示。
var array = [{
"id": 2,
"price": 2000,
"name": "Mr Robot T1",
"image": "http://placehold.it/270x335"
}, {
"id": 1,
"price": 1000,
"name": "Mr Robot T2",
"image": "http://placehold.it/270x335"
}];
array.forEach(function(item, index){
if (item.id == 1) {
array.splice(index, 1);
}
});