首頁猿問 JS數(shù)據(jù)轉(zhuǎn)成父子節(jié)點數(shù)據(jù)

JS數(shù)據(jù)轉(zhuǎn)成父子節(jié)點數(shù)據(jù)

JavaScript

慕桂英3389331 2019-05-14 11:41:20

原始的數(shù)據(jù)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(zhuǎn)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(jù)指定的列轉(zhuǎn)換成父子結(jié)構(gòu)的數(shù)據(jù)比如上面的是根據(jù)['company','department','office']不知道怎么寫了請大神指導(dǎo):(:(:(:(:(

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qq_笑_17

TA貢獻1818條經(jīng)驗獲得超7個贊

答完才發(fā)現(xiàn)你要的是能根據(jù)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}=cif(!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(shù)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(chuàng)建company時加入節(jié)點信息即可，不會很難。

反對回復(fù) 2019-05-14

至尊寶的傳說

TA貢獻1789條經(jīng)驗獲得超10個贊

你的輸出格式應(yīng)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(shù)據(jù)能夠保證全是這樣的結(jié)構(gòu)（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);