【发布时间】:2016-12-26 14:11:38
【问题描述】:
我有一个如下 JSON 响应:
var res =
{
"response": {
"data": {
"profilesearchsnippet": [
[
{
"profileInfo": {
"firstname": "Sundar",
"lastname": "v",
"gender": "male",
"country": "Afghanistan",
"state": "Badakhshan",
"city": "Eshkashem",
"pincode": "",
"plancode": "T001",
"userid": 13
},
"roleInfo": {
"defaultphotoid": 94
}
}
],
[
{
"profileInfo": {
"firstname": "ghg",
"lastname": "vbhvh",
"gender": "male",
"state": "Badakhshan",
"city": "Eshkashem",
"pincode": "454",
"plancode": "T001",
"userid": 22
},
"roleInfo": {
"defaultphotoid": 171
}
}
]
]
}
}
}
我想获取表中的所有名字,国家/地区城市。我尝试将 profilesearchsn-p 值分配给变量 var SearchData 并尝试使用 profileinfo 获取名字,因为它的对象。我在某处遗漏了一些需要帮助的东西。
HTML:
<tr ng-repeat= "item in searchData">
<td>{{item.profileInfo.firstname}}</td>
<td> {{item.profileInfo.country}}</td>
</tr>
JS:
var searchData = res.response.data.profilesearchsnippet[0];
【问题讨论】:
-
您是否已将 searchData 分配给控制器或范围之类的东西?
-
searchData 不是您要查找的可迭代对象,您将其定义为
var searchData = res.response.data.profilesearchsnippet[0];这只是搜索结果之一。尝试不使用[0]。