javascript 下划线 where 过滤器
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Underscore where filter
提问by Croftc
I'm trying to use Underscore's whereto filter an array of objects, but I can't seem to figure out how to get it to filter multiple values for the same key. For example:
我正在尝试使用 Underscorewhere来过滤一组对象，但我似乎无法弄清楚如何让它过滤同一键的多个值。例如：
var itemsArr = [{name:"foo", color:"red"}, {name:"bar", color:"blue"}, {name:"bob", color:"yellow"}];
I'm trying to get it to return an array off all items with color redAND all items with color blue... Is this possible? I've tried
我试图让它从所有带有颜色的red项目和所有带有颜色的项目中返回一个数组blue......这可能吗？我试过了
tempArr = _.where(itemsArr, {color:["red", "blue"]});
but that didn't work. If I have to just use _.filterand write out my own predicate I will, but I was just wondering if anyone else had tried to do this and found a solution.
但这没有用。如果我必须只使用_.filter并写出我自己的谓词，我会，但我只是想知道是否有人尝试这样做并找到了解决方案。
回答by Praveen Prasannan
回答by Phil C
回答by Jeremy Schultz
Can you run _.where twice and concat the two resulting arrays?
你能运行 _.where 两次并连接两个结果数组吗？
tempArr = _.where(itemsArr, {color: "red"}).concat(_.where(itemsArr, {color: "blue"}));
回答by Rubel hasan
Here is the solution
这是解决方案
var itemsArr = [{name:"foo", color:"red"}, {name:"bar", color:"blue"}, {name:"bob", color:"red"}];
var filter = ["red","blue"];
var data = {};
_.each(filter, function (item) {
data[item] = true;
});
var returnData = _.filter(itemsArr, function (item) {
return data[item.color];
});
console.log(returnData)
回答by gwg
According to the documentation:
根据文档：
Looks through each value in the list, returning an array of all the values that contain all of the key-value pairslisted in properties.
查看列表中的每个值，返回包含属性中列出的所有键值对的所有值的数组。
You can also look at the source codeif you'd like. Under the hood, whereuses isMatch:
如果您愿意，也可以查看源代码。在引擎盖下，where使用isMatch：
_.isMatch = function(object, attrs) {
var keys = _.keys(attrs), length = keys.length;
if (object == null) return !length;
var obj = Object(object);
for (var i = 0; i < length; i++) {
var key = keys[i];
if (attrs[key] !== obj[key] || !(key in obj)) return false;
}
return true;
};
Since you cannot have two identical keys, you cannot do what you would like. I would use filter, as you suggest.
由于您不能拥有两个相同的密钥，因此您无法做自己想做的事。我会filter按照你的建议使用。
