cod*_*sed 8 javascript underscore.js lodash
我有以下对象records
:
{
"notes":[
{
"id":1,
"description":"hey",
"userId":2,
"replyToId":null,
"postId":2,
"parentId":null
},
{
"id":5,
"description":"hey test",
"userId":3,
"replyToId":null,
"postId":2,
"parentId":null
},
{
"id":2,
"description":"how are you",
"userId":null,
"replyToId":2,
"postId":2,
"parentId":null,
"user":null
}
]
}
Run Code Online (Sandbox Code Playgroud)
我想将其输出为:
2
object with id 1
object with id 2 (because replyToId value is same as userId
3
object with id 5
Run Code Online (Sandbox Code Playgroud)
所以基本上我想在同一组下考虑UserId和replyToId值.
我在lodash下构建了自己的mixin,将groupBy方法包装为:
mixin({
splitGroupBy: function(list, groupByIter){
if (_.isArray(groupByIter)) {
function groupBy(obj) {
return _.forEach(groupByIter, function (key){
if ( !!obj[key] ) return obj[key]
});
}
} else {
var groupBy = groupByIter;
}
debugger;
var groups = _.groupBy(list, groupBy);
return groups;
}
});
Run Code Online (Sandbox Code Playgroud)
电话看起来像这样:
_.splitGroupBy(data.notes,['userId', 'replyToId']);
Run Code Online (Sandbox Code Playgroud)
输出没有组.甚至当我试图用_.map
,而不是_.forEach
分裂没有正确发生.
使用下划线的解决方案:
var props = ['userId', 'replyToId'];
var notNull = _.negate(_.isNull);
var groups = _.groupBy(record.notes, function(note){
return _.find(_.pick(note, props), notNull);
});
Run Code Online (Sandbox Code Playgroud)
归档时间: |
|
查看次数: |
16773 次 |
最近记录: |