我有一个像这样的数组:
0: "SuSPENSE"
1: "Subcontractor Expense"
2: "Data Entry"
3: "Design"
4: "Programming"
5: "Subcontractor Expense - Other"
6: "Total Subcontractor Expense"
7: "Technology-Communication"
8: "Licenses"
9: "Domain Hosting Fee"
10: "Internet"
11: "Internet Servers"
12: "Internet - Other"
13: "Total Internet"
14: "Telephone"
15: "Call Center"
16: "Cellular Phones"
17: "Fax"
18: "Telephone - Other"
19: "Total Telephone"
20: "Hardware/Software"
21: "Computer Software"
22: "Hardware/Software - Other"
23: "Total Hardware/Software"
24: "Technology-Communication - Other"
25: "Total Technology-Communication"
这是类别和子类别的列表。 例如，“分包商费用”是一个子类别，以“分包商费用总额”结尾。 “internet”和“total internet”相同。 模板总是相同的，category以“（name）”开头，以“total（name）”结尾。 但是每个类别可以有很多级别的子类别，就像一棵树。 我试图使用递归函数将这个数组解析为类似JSON的数组或多维数组，但我从来不知道最大深度是多少。 我尝试使用js执行以下操作:
var parsedArray = [];
var items = getLineItems("Expense", "Total Expense"); //This is a function to return the mentioned array
var newArray = parseArray(items, "Expense");
function parseArray(items, category){
var lineItems = [];
for(var i = 0; i < items.length; i++){
var inArray = $.inArray("Total " + items[i], items);
if(inArray !== -1){
parseArray(getLineItems(items[i], "Total " + items[i]), items[i]);
}
else {
lineItems.push(items[i]);
}
}
parsedArray[category] = lineItems;
}
但是这个递归函数永远不会深入到2级以上。 有可能生成这样的东西吗？
"SuSPENSE"
"Subcontractor Expense"
"Data Entry"
"Design"
"Programming"
"Subcontractor Expense - Other"
"Technology-Communication"
"Licenses"
"Domain Hosting Fee"
"Internet"
"Internet Servers"
"Internet - Other"
"Telephone"
"Call Center"
"Cellular Phones"
"Fax"
"Telephone - Other"
"Hardware/Software"
"Computer Software"
"Hardware/Software - Other"
"Technology-Communication - Other"
您可以通过检查当前元素是否具有以单词total开头，然后以当前元素的文本继续的相应元素来完成此操作，如果是，则递增级别。 当当前元素以word total开头时，则递减级别。
null
const data = {"0":"SuSPENSE","1":"Subcontractor Expense","2":"Data Entry","3":"Design","4":"Programming","5":"Subcontractor Expense - Other","6":"Total Subcontractor Expense","7":"Technology-Communication","8":"Licenses","9":"Domain Hosting Fee","10":"Internet","11":"Internet Servers","12":"Internet - Other","13":"Total Internet","14":"Telephone","15":"Call Center","16":"Cellular Phones","17":"Fax","18":"Telephone - Other","19":"Total Telephone","20":"Hardware/Software","21":"Computer Software","22":"Hardware/Software - Other","23":"Total Hardware/Software","24":"Technology-Communication - Other","25":"Total Technology-Communication"}
function toNested(data) {
const result = [];
const levels = [result]
let level = 0;
const checkChildren = (string, data) => {
return data.some(e => e === `Total ${string}`)
}
data.forEach((e, i) => {
const object = { name: e, children: []}
levels[level + 1] = object.children;
if (e.startsWith('Total')) level--;
else levels[level].push(object);
if (checkChildren(e, data.slice(i))) level++;
})
return result;
}
const result = toNested(Object.values(data));
console.log(result)
我还在猜测你的输出格式。 下面是一个递归解决方案，它给出了我在评论中询问的格式:
[
{name: "SuSPENSE"},
{name: "Subcontractor Expense", children: [
{name: "Data Entry"},
{name: "Design"},
{name: "Programming"},
{name: "Subcontractor Expense - Other"}
]},
{name: "Technology-Communication", children: [
//...
]}
]
null
const restructure = (
[s = undefined, ...ss],
index = s == undefined ? -1 : ss .indexOf ('Total ' + s)
) =>
s == undefined
? []
: index > -1
? [
{name: s, children: restructure (ss .slice (0, index))},
... restructure (ss .slice (index + 1))
]
: [{name: s}, ... restructure (ss)]
const data = ["SuSPENSE", "Subcontractor Expense", "Data Entry", "Design", "Programming", "Subcontractor Expense - Other", "Total Subcontractor Expense", "Technology-Communication", "Licenses", "Domain Hosting Fee", "Internet", "Internet Servers", "Internet - Other", "Total Internet", "Telephone", "Call Center", "Cellular Phones", "Fax", "Telephone - Other", "Total Telephone", "Hardware/Software", "Computer Software", "Hardware/Software - Other", "Total Hardware/Software", "Technology-Communication - Other", "Total Technology-Communication"]
console .log (
restructure (data)
).as-console-wrapper {min-height: 100% !important; top: 0}