我有以下 JSON,我想从中提取技能是真的。
[
{
"_id":"5de9f351baca28556c6a4b71",
"Name":"Harsha",
"Age":20,
"Gender":"M",
"Skills":{
"Java":"",
"Mule":true,
"Angular":""
}
},
{
"_id":"5de9f358baca28556c6a4b72",
"Name":"Anji",
"Age":21,
"Gender":"M",
"Skills":{
"Java":"",
"Mule":true,
"Angular":true
}
},
{
"_id":"5dea110297c2b65298b136e4",
"Name":"Abhi",
"Age":25,
"Gender":"M",
"Skills":{
"Java":"",
"Mule":true,
"Angular":""
}
}
]
Run Code Online (Sandbox Code Playgroud)
我可以使用以下代码打印其余数据
<table *ngIf="formTemplate">
<tr>
<th *ngFor="let header of questionTitleArray" >{{header}}</th>
</tr>
<tr *ngFor="let data of surveyDataFromDB">
<ng-container *ngFor="let head of questionTitleArray">
<td>{{data[head]}}</td>
</ng-container>
</tr>
</table>
Run Code Online (Sandbox Code Playgroud)
(这里的 JSON 是“surveyDataFromDB”)
下面是我得到的输出
Name Age Gender Skills
Harsha 20 M [object Object]
Anji 21 M [object Object]
Abhi 25 M [object Object]
Run Code Online (Sandbox Code Playgroud)
我想要真正的技能代替 [object Object]。请帮忙。
<table>
<tr>
<th *ngFor="let header of questionTitleArray">{{ header }} </th>
</tr>
<tr *ngFor="let data of surveyDataFromDB">
<ng-container *ngFor="let head of questionTitleArray">
<span *ngIf="checkType(data[head]) else elsePart">
<span *ngFor="let j of data[head] | keyvalue">
<td *ngIf="j.value==true">
{{j.key}}
</td>
</span>
</span>
<ng-template #elsePart>
<td>{{data[head]}}</td>
</ng-template>
</ng-container>
</tr>
</table>
Run Code Online (Sandbox Code Playgroud)
在 ts 中:
checkType(Ob:any)
{
if(typeof (Ob) === 'object')
return true;
else
return false;
}
Run Code Online (Sandbox Code Playgroud)
| 归档时间: |
|
| 查看次数: |
317 次 |
| 最近记录: |