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JS數據轉成父子節點數據

JS數據轉成父子節點數據

慕絲7291255 2019-05-19 21:00:40
原始的數據是平鋪的比如originalData:[{company:'AAA',department:'AAA-D1',office:'AAA-D1-O1',pass:1,onhold:3},{company:'AAA',department:'AAA-D1',office:'AAA-D1-O2',pass:3,onhold:5},{company:'AAA',department:'AAA-D2',office:'AAA-D2-O1',pass:2,onhold:7},{company:'BBB',department:'BBB-D1',office:'BBB-D1-O1',pass:1,onhold:3},{company:'BBB',department:'BBB-D2',office:'BBB-D2-O1',pass:4,onhold:3},{company:'BBB',department:'BBB-D3',office:'BBB-D3-O1',pass:1,onhold:3},........]希望轉換成treeData:[{label:'AAA',department:'AAA-D1',office:'AAA-D1-O1',pass:6,onhold:15,//companylevelchildren:[//departmentlevel{label:'AAA-D1',pass:4,onhold:8,children:[//officelevel{label:'AAA-D1-O1',pass:1,onhold:3},{label:'AAA-D1-O2',pass:1,onhold:3},]},{label:'AAA-D2',pass:2,onhold:7,children:[{label:'AAA-D2-O1',pass:1,onhold:3},]},]},......]根據指定的列轉換成父子結構的數據比如上面的是根據['company','department','office']不知道怎么寫了請大神指導:(:(:(:(:(
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2 回答

?
猛跑小豬

TA貢獻1858條經驗 獲得超8個贊

答完才發現你要的是能根據指定列進行映射...
那我再想想，下面的當參考吧ORZ...
varmap=originalData.reduce((p,c)=>{
let{company,department,office,pass,onhold}=c
if(!p[company])p[company]={}
if(!p[company][department])p[company][department]={}
p[company][department][office]={pass,onhold}
returnp
},{})
varreduce=(obj,key)=>obj.children.reduce((p,c)=>[p+=c[key],p][1],0)
varresult=Object.keys(map).map(cKey=>{
letcompany={label:cKey}
company.children=Object.keys(map[cKey]).map(dKey=>{
letdepartment={label:dKey}
department.children=Object.keys(map[cKey][dKey]).map(oKey=>{
letoffice=map[cKey][dKey][oKey]
return{label:oKey,pass:office.pass,onhold:office.onhold}
})
['onhold','pass'].forEach(key=>department[key]=reduce(department,key))
returndepartment
})
['onhold','pass'].forEach(key=>company[key]=reduce(company,key))
returncompany
})
輸出:
[
{
"label":"AAA","onhold":15,"pass":6
"children":[{"label":"AAA-D1","onhold":8,"pass":4
"children":[{"label":"AAA-D1-O1","pass":1,"onhold":3},
{"label":"AAA-D1-O2","pass":3,"onhold":5}]
},{
"label":"AAA-D2","onhold":7,"pass":2
"children":[{"label":"AAA-D2-O1","pass":2,"onhold":7}]
}]
},{
"label":"BBB","onhold":9,"pass":6
"children":[{"label":"BBB-D1","onhold":3,"pass":1
"children":[{"label":"BBB-D1-O1","pass":1,"onhold":3}]
},{
"label":"BBB-D2","onhold":3,"pass":4
"children":[{"label":"BBB-D2-O1","pass":4,"onhold":3}]
},{
"label":"BBB-D3","onhold":3,"pass":1
"children":[{"label":"BBB-D3-O1","pass":1,"onhold":3}]
}]
}]
思路上是先把原始數組做一次樹級映射，然后再從樹生成各層次信息，這樣會比較好理解一些。
你題目描述中有一些奇怪的地方：companylevel里的department和office的取值標準不清楚，所以就沒寫，不過這個要加也就是在生成result那一步，創建company時加入節點信息即可，不會很難。
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?
陪伴而非守候

TA貢獻1757條經驗 獲得超8個贊

你的輸出格式應該有問題，起碼label:'AAA',department:'AAA-D1',office:'AAA-D1-O1',pass:6,onhold:15,//companylevel是不合理的，最多是label:'AAA',pass:6,onhold:15,//companylevel吧如果數據能夠保證全是這樣的結構（只有如上的3級），其實還是比較好處理的。
varmoriginalData=[
{company:'AAA',department:'AAA-D1',office:'AAA-D1-O1',pass:1,onhold:3},
{company:'AAA',department:'AAA-D1',office:'AAA-D1-O2',pass:3,onhold:5},
{company:'AAA',department:'AAA-D2',office:'AAA-D2-O1',pass:2,onhold:7},
{company:'BBB',department:'BBB-D1',office:'BBB-D1-O1',pass:1,onhold:3},
{company:'BBB',department:'BBB-D2',office:'BBB-D2-O1',pass:4,onhold:3},
{company:'BBB',department:'BBB-D3',office:'BBB-D3-O1',pass:1,onhold:3}
];
functiona2o(originalData){
varoutData=[];
varoutObj={};
for(vari=0;ivarcompany=originalData[i].company;
vardepartment=originalData[i].department;
varoffice={label:originalData[i].office,pass:originalData[i].pass,onhold:originalData[i].onhold};
if(outObj[company]===undefined){
outObj[company]={childrenKey:[],pass:0,onhold:0};
}
if(outObj[company][department]===undefined){
outObj[company][department]={children:[],pass:0,onhold:0};
outObj[company].childrenKey.push(department)
}
outObj[company][department].children.push(office);
outObj[company][department].pass=outObj[company][department].pass+office.pass;
outObj[company].pass=outObj[company].pass+office.pass;
outObj[company][department].onhold=outObj[company][department].onhold+office.onhold;
outObj[company].onhold=outObj[company].onhold+office.onhold;
}
for(varcominoutObj){
vartmpA={label:com,pass:outObj[com].pass,onhold:outObj[com].onhold,children:[]};
for(varj=0;jvartD=outObj[com][outObj[com].childrenKey[j]];
tmpA.children.push({label:outObj[com].childrenKey[j],
pass:tD.pass,
onhold:tD.onhold,
children:tD.children});
}
outData.push(tmpA);
}
returnoutData;
}
vartreeDate=a2o(moriginalData)
console.log(treeDate);
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