我正在尝试格式化一堆数据,我能够以下面的格式检索。 (我还把一个可读的版本注释掉了) 我希望以OUTPUT(见底部)显示的方式将每一行放在 json对象 中。但它只保留最后一项的细节,而我希望每个项目都保持在正确的MGId和FGId值之下。
MGId MGName FGId FGName itemId itemName
6272440:饮料:6272963:焦炭:6274904:百事可乐@@@
我使用jsonObj [MGId] [FGId] [MIId],我期待这个创建单独的[MIId] jsonObj [MGId] [FGId] ..显然我错过了一些东西,或者它不能以这种方式工作。 任何人都可以告诉我缺少什么或任何替代方式欢迎。谢谢!
/*
DATA READABLE FORMAT
6272440:Beverage:6272963:Coke:6274904:pepsi@@@
6272440:Beverage:6272963:coke:6274494:coco cola 1 lt@@@
6272440:Beverage:6272951:Milk:6274300:Skim 1 lt@@@
6272440:Beverage:6272951:Milk:6274130:Full Cream@@@
6272437:Food:6272919:Breakfast Menu:6274947:Bacon and Egg@@@
6272437:Food:6272915:Lunch Menu:6274766:Chicken Burger@@@
6272437:Food:6272915:Lunch Menu:6274922:Vegi Sandwich@@@
6272437:Food:6272915:Lunch Menu:6274900:Garlic Bread";
*/
var sampleData = "6272440:Beverage:6272963:Coke:6274904:pepsi@@@6272440:Beverage:6272963:coke:6274494:coco cola 1 lt@@@6272440:Beverage:6272951:Milk:6274300:Skim 1 lt@@@6272440:Beverage:6272951:Milk:6274130:Full Cream@@@6272437:Food:6272919:Breakfast Menu:6274947:Bacon and Egg@@@6272437:Food:6272915:Lunch Menu:6274766:Chicken Burger@@@6272437:Food:6272915:Lunch Menu:6274922:Vegi Sandwich@@@6272437:Food:6272915:Lunch Menu:6274900:Garlic Bread";
var elements = sampleData.split("@@@");
var jsonObj = {};
var menuItemsArr = [];
var previousMGId="";
var previousFGId="";
for(var i=0; i<elements.length; i++) {
var nextMenuPack = elements[i];
var nextMenuPackInnerHtml = nextMenuPack; //nextMenuPack.innerHTML;
var nextMenuPackArr = nextMenuPackInnerHtml.split(":");
var MGId = nextMenuPackArr[0];
var MGName = nextMenuPackArr[1];
var FGId = nextMenuPackArr[2];
var FGName = nextMenuPackArr[3];
var MIId = nextMenuPackArr[4];
var MIName = nextMenuPackArr[5];
console.log(MGName +"-"+FGName+"-"+MIName);
jsonObj[MGId] = {} ;
jsonObj[MGId][FGId] = {};
//var jsonObj = {};
jsonObj[MGId][FGId][MIId] = {
"MGName" : MGName,
"MGId": MIId,
"FGName": FGName,
"FGId": FGId,
"MIName" : MIName,
"MIId": MIId
};
}
document.write('<pre>'+JSON.stringify(jsonObj, null, "\t")+'</pre>');
输出
{
"6272440": { //major group id
"6272951": { //family group id
"6274130": { // menuitem id - JUST LAST ONE, OTHERS OVERWRITTEN!
"MGName": "Beverage",
"MGId": "6274130",
"FGName": "Milk",
"FGId": "6272951",
"MIName": "Full Cream",
"MIId": "6274130"
}
}
},
"6272437": {
"6272915": {
"6274900": {// menuitem id - JUST LAST ONE, OTHERS OVERWRITTEN!
"MGName": "Food",
"MGId": "6274900",
"FGName": "Lunch Menu",
"FGId": "6272915",
"MIName": "Garlic Bread",
"MIId": "6274900"
}
}
}
}
答案 0 :(得分:0)
只有初始化jsonObj[MGId]
和jsonObj[MGId][FGId]
(如果它们尚未存在)。
jsonObj[MGId] = jsonObj[MGId] || {} ;
jsonObj[MGId][FGId] = jsonObj[MGId][FGId] || {};
//var jsonObj = {};
jsonObj[MGId][FGId][MIId] = {
"MGName" : MGName,
"MGId": MIId,
"FGName": FGName,
"FGId": FGId,
"MIName" : MIName,
"MIId": MIId
};