使用多个键/值对查找JSON对象，然后更新该Object的其他属性
特定
sessionStorage.cart = "[ {"id":121,"name":"Pants","number":1,"specification":""}, {"id":121,"name":"Pants","number":2,"specification":""}, {"id":121,"name":"Pants","number":3,"specification":""} ]" 我想编写一个函数，找到id为121，Pants名称，数量为2的对象，以便我可以更新该对象的规范。 所以我会传递id，名称，数字和所需的新规范值，并获得以下输出：
sessionStorage.cart = "[ {"id":121,"name":"Pants","number":1,"specification":""}, {"id":121,"name":"Pants","number":2,"specification":"new value"}, {"id":121,"name":"Pants","number":3,"specification":""} ]"
我真的在努力思考这个……欢迎指导！
用这个：
var cart = [ { "id": 121, "name": "Pants", "number": 1, "specification": "" }, { "id": 121, "name": "Pants", "number": 2, "specification": "" }, { "id": 121, "name": "Pants", "number": 3, "specification": "" } ]; cart.forEach(function(entry) { if (entry.id == 121 && entry.name == 'Pants' && entry.number == 2) { entry.specification = 'new value'; } }); console.log(cart);
使用Array.prototype.find的简单解决方案
sessionStorage.cart.find(e => e.id === 121 && e.name === 'Pants' && e.number === 2 ).specification = 'new value'; console.log(JSON.stringify(sessionStorage.cart, null, ' '));
产量
[ { "id": 121, "name": "Pants", "number": 1, "specification": "" }, { "id": 121, "name": "Pants", "number": 2, "specification": "new value" }, { "id": 121, "name": "Pants", "number": 3, "specification": "" } ]
注意：这需要ES6。
您可能想尝试以下代码：
function update(oldValues, newValues, keyToCompare) { var keyToCompareLen = keyToCompare.length; var result = []; oldValues.forEach(function(oldValue){ newValues.forEach(function(newValue){ var cnt = 0; keyToCompare.forEach(function(key){ if(newValue[key] == oldValue[key]) { ++cnt; } }); if(cnt == keyToCompareLen) { oldValue = $.extend(true, {}, oldValue, newValue); } }); result.push(oldValue); }); return result; } var result = update([ {"id":121,"name":"Pants","number":1,"specification":""}, {"id":121,"name":"Pants","number":2,"specification":""}, {"id":121,"name":"Pants","number":3,"specification":""} ], [ {"id":121,"name":"Pants","number":2,"specification":"test"} ], [ "id", "name", "number" ]); console.log(result);
注意：您需要包含jquery库。 你也可以在这里运行代码https://fiddle.jshell.net/b17fx6qk/1/
迭代数组可能效率低下。 为了提高性能，您可以对所有索引值进行字符串化，并使用它们将对象存储在散列映射中以便快速查找。 就像是：
function cartItem() { this.id; this.name; this.number; this.specification; } CartItem.hash = function(id, name, number) { return [id, name, number].join('|'); }; cartItem.prototype.toHash = function() { return CartItem.hash(this.id, this.name, this.number); }; var cartMap = cart.reduce(function(map, item) { var key = item.toHash(); map[key] = item; }, { });
然后你可以（很快）通过哈希查看项目：
cartMap[CartItem.hash(id, name, number)];
试试这个。 只需匹配和更新
$(document).ready(function() { var o = []; o.id = 121; o.name = "Pants"; o.number = 2; updateVal(o); }); function updateVal(o) { var cart = [{ "id": 121, "name": "Pants", "number": 1, "specification": "" }, { "id": 121, "name": "Pants", "number": 2, "specification": "" }, { "id": 121, "name": "Pants", "number": 3, "specification": "" }]; $.each(cart, function(a, b) { if (b.id == o.id && b.name == o.name && b.number == o.number) { b.specification = "new value"; } }); alert("value updated. specification:" + cart[1].specification); }