[
{
id:265,
leaveType:{
id:1,
name:'Casual Leave'
},
currentProject:'parlo,',
initialDate:'2017-07-06',
finalDate:'2017-07-06',
workingDays:1,
employee:{
id:211,
name:'appu'
},
actionItems:[
{
id:2,
name:'APPROVE'
},
{
id:3,
name:'REJECT'
},
{
id:4,
name:'CANCEL'
}
],
halfDay:false
},
{
id:277,
leaveType:{
id:1,
name:'Casual Leave'
},
currentProject:'Venon,',
initialDate:'2017-07-03',
finalDate:'2017-07-04',
workingDays:2,
employee:{
id:206,
name:'malya'
},
actionItems:[
{
id:2,
name:'APPROVE'
},
{
id:3,
name:'REJECT'
},
{
id:4,
name:'CANCEL'
}
],
halfDay:false
},
{
id:285,
leaveType:{
id:1,
name:'Casual Leave'
},
currentProject:'ferrar,',
initialDate:'2017-09-25',
finalDate:'2017-09-25',
workingDays:1,
employee:{
id:68,
name:'Geena'
},
actionItems:[
{
id:2,
name:'APPROVE'
},
{
id:3,
name:'REJECT'
},
{
id:4,
name:'CANCEL'
}
],
halfDay:false
},
{
id:286,
leaveType:{
id:1,
name:'Casual Leave'
},
currentProject:'ole,',
initialDate:'2017-09-26',
finalDate:'2017-09-26',
workingDays:1,
employee:{
id:68,
name:'harry'
},
actionItems:[
{
id:2,
name:'APPROVE'
},
{
id:3,
name:'REJECT'
},
{
id:4,
name:'CANCEL'
}
],
halfDay:false
}
]
以上是一个简单的json数组请建议我循环删除json数组的下面部分。下面的对象必须从数组中完全删除。我正在处理一个react-native项目所以代码必须反应js。
{
id:4,
name:'CANCEL'
}
我的预期输出低于
[
{
id:265,
leaveType:{
id:1,
name:'Casual Leave'
},
currentProject:'parlo,',
initialDate:'2017-07-06',
finalDate:'2017-07-06',
workingDays:1,
employee:{
id:211,
name:'appu'
},
actionItems:[
{
id:2,
name:'APPROVE'
},
{
id:3,
name:'REJECT'
}
],
halfDay:false
},
{
id:277,
leaveType:{
id:1,
name:'Casual Leave'
},
currentProject:'Venon,',
initialDate:'2017-07-03',
finalDate:'2017-07-04',
workingDays:2,
employee:{
id:206,
name:'malya'
},
actionItems:[
{
id:2,
name:'APPROVE'
},
{
id:3,
name:'REJECT'
}
],
halfDay:false
},
{
id:285,
leaveType:{
id:1,
name:'Casual Leave'
},
currentProject:'ferrar,',
initialDate:'2017-09-25',
finalDate:'2017-09-25',
workingDays:1,
employee:{
id:68,
name:'Geena'
},
actionItems:[
{
id:2,
name:'APPROVE'
},
{
id:3,
name:'REJECT'
}
],
halfDay:false
},
{
id:286,
leaveType:{
id:1,
name:'Casual Leave'
},
currentProject:'ole,',
initialDate:'2017-09-26',
finalDate:'2017-09-26',
workingDays:1,
employee:{
id:68,
name:'harry'
},
actionItems:[
{
id:2,
name:'APPROVE'
},
{
id:3,
name:'REJECT'
}
],
halfDay:false
}
]
答案 0 :(得分：2)
这是一个使用for循环的简单解决方案。
var json = [ { id:265, leaveType:{ id:1, name:'Casual Leave' }, currentProject:'parlo,', initialDate:'2017-07-06', finalDate:'2017-07-06', workingDays:1, employee:{ id:211, name:'appu' }, actionItems:[ { id:2, name:'APPROVE' }, { id:3, name:'REJECT' }, { id:4, name:'CANCEL' } ], halfDay:false }, { id:277, leaveType:{ id:1, name:'Casual Leave' }, currentProject:'Venon,', initialDate:'2017-07-03', finalDate:'2017-07-04', workingDays:2, employee:{ id:206, name:'malya' }, actionItems:[ { id:2, name:'APPROVE' }, { id:3, name:'REJECT' }, { id:4, name:'CANCEL' } ], halfDay:false }, { id:285, leaveType:{ id:1, name:'Casual Leave' }, currentProject:'ferrar,', initialDate:'2017-09-25', finalDate:'2017-09-25', workingDays:1, employee:{ id:68, name:'Geena' }, actionItems:[ { id:2, name:'APPROVE' }, { id:3, name:'REJECT' }, { id:4, name:'CANCEL' } ], halfDay:false }, { id:286, leaveType:{ id:1, name:'Casual Leave' }, currentProject:'ole,', initialDate:'2017-09-26', finalDate:'2017-09-26', workingDays:1, employee:{ id:68, name:'harry' }, actionItems:[ { id:2, name:'APPROVE' }, { id:3, name:'REJECT' }, { id:4, name:'CANCEL' } ], halfDay:false } ]
;
for(var i =0;i<json.length;i++){
actionItems = json[i].actionItems;
for(var j =0;j<actionItems.length;j++){
if(actionItems[j].name === 'CANCEL'){
json[i].actionItems.splice(j,1);
}
}
}
console.log(json);
答案 1 :(得分：0)
您可以映射外部数组，然后根据actionItem name property
var data = [
{
id:265,
leaveType:{
id:1,
name:'Casual Leave'
},
currentProject:'parlo,',
initialDate:'2017-07-06',
finalDate:'2017-07-06',
workingDays:1,
employee:{
id:211,
name:'appu'
},
actionItems:[
{
id:2,
name:'APPROVE'
},
{
id:3,
name:'REJECT'
},
{
id:4,
name:'CANCEL'
}
],
halfDay:false
},
{
id:277,
leaveType:{
id:1,
name:'Casual Leave'
},
currentProject:'Venon,',
initialDate:'2017-07-03',
finalDate:'2017-07-04',
workingDays:2,
employee:{
id:206,
name:'malya'
},
actionItems:[
{
id:2,
name:'APPROVE'
},
{
id:3,
name:'REJECT'
},
{
id:4,
name:'CANCEL'
}
],
halfDay:false
},
{
id:285,
leaveType:{
id:1,
name:'Casual Leave'
},
currentProject:'ferrar,',
initialDate:'2017-09-25',
finalDate:'2017-09-25',
workingDays:1,
employee:{
id:68,
name:'Geena'
},
actionItems:[
{
id:2,
name:'APPROVE'
},
{
id:3,
name:'REJECT'
},
{
id:4,
name:'CANCEL'
}
],
halfDay:false
},
{
id:286,
leaveType:{
id:1,
name:'Casual Leave'
},
currentProject:'ole,',
initialDate:'2017-09-26',
finalDate:'2017-09-26',
workingDays:1,
employee:{
id:68,
name:'harry'
},
actionItems:[
{
id:2,
name:'APPROVE'
},
{
id:3,
name:'REJECT'
},
{
id:4,
name:'CANCEL'
}
],
halfDay:false
}
]
var newData = data
.map((ele)=>{
let newElement = Object.assign({}, ele);
newElement.actionItems = newElement.actionItems.filter(actionItem => actionItem.name !== 'CANCEL');
return newElement
})
console.log(newData);
答案 2 :(得分：0)
首先，您需要确定数组中的匹配索引。 当你搜索一系列项目时，你看起来需要两个循环。
找到要删除的部分后，只需使用Array.splice(index, 1)完全删除数组中的密钥。
假设您要删除此部分
{
id:4,
name:'CANCEL'
}
来自子对象的所有actionItems属性。
首先循环遍历您的对象集合，然后循环遍历您的操作集合
for (let i = 0; i < yourMainArray.length; i++) {
let indexToDelete = -1
for (let j = 0; j < yourMainArray[i].actionItems.length; j++) {
if (yourMainArray[i].actionItems[j].name === 'CANCEL') {
indexToDelete = j
}
}
if (indexToDelete !== -1) {
yourMainArray[i].actionItems.splice(indexToDelete, 1)
}
}
由于您希望在同时循环时修改数组，您需要保留索引的引用以删除和删除循环外的项目。