最新消息:雨落星辰是一个专注网站SEO优化、网站SEO诊断、搜索引擎研究、网络营销推广、网站策划运营及站长类的自媒体原创博客

javascript - delete data from JSON object - Stack Overflow

programmeradmin4浏览0评论

I've following output code

I want to remove position from array. When I try delete $products.position; It is not deleting. But When I try delete $products[0].position; It is deleting only from first.

I've following output code

I want to remove position from array. When I try delete $products.position; It is not deleting. But When I try delete $products[0].position; It is deleting only from first.

Share Improve this question asked Jun 14, 2017 at 6:00 w3debuggerw3debugger 2,10219 silver badges24 bronze badges 2
  • 3 As you already know $products is an array, so iterate and use delete operator. You should search "How to iterate an array?" – Satpal Commented Jun 14, 2017 at 6:01
  • luckily this isn't JSON, but a palin ol' javascript Object - doing this with JSON would be painful – Jaromanda X Commented Jun 14, 2017 at 6:07
Add a ment  | 

5 Answers 5

Reset to default 5

Loop through the elements of the array and delete position from each.

for (var i = 0; i < $products.length; i++) {
   delete $products[i].position;
}

I would use map instead of forEach.

const products = [{ foo: 'bar', position: 1 }, { foo: 'baz', position: 2}, { doo: 'daz' }];
console.log(products.map(obj => delete obj.position && obj));
.as-console-wrapper { max-height: 100% !important; top: 0; }

Although it is probably less readable it makes more sense to me because you want to generate a new array applying a delete function to each element.

try this

$products.forEach(function(item){ 
    delete item.position; 
});

There are multiple methods to remove an JSON object from particular position from an array, but I will remended to use splice method since its paratively faster then other methods.

products.splice(position, 1);

The first parameter indicates the position to start from and the second indicates the number of objects you want to splice. Since you only want to remove one object, so it should be written 1(one). Position here starts from 0(zero) so insert the position accordingly.

This one works for me.

for(var i in $products){
    delete $products[i].position;
}
发布评论

评论列表(0)

  1. 暂无评论