我有一些虚拟XML文件:
<Week number="2013-W45">
<Day dow="1" templateDay="Monday">
<Job name="wake up" >
<Job name="get dressed" >
<Job name="prepare breakfast" >
<Job name="eat breakfast" > </Job>
</Job>
</Job>
</Job>
<Job name="work 9-5" >
</Job>
</Day>
<Day dow="2" templateDay="Tuesday" >
<Job name="wake up" >
<Job name="get dressed" >
<Job name="prepare breakfast" >
<Job name="eat breakfast" > </Job>
</Job>
</Job>
</Job>
<Job name="work 9-5" >
</Job>
<Job name="football" >
</Job>
</Day>
<Day dow="3" templateDay="Wednesday" >
<Job name="wake up" >
<Job name="get dressed" >
<Job name="prepare breakfast" >
<Job name="eat breakfast" > </Job>
</Job>
</Job>
</Job>
<Job name="work 9-5" >
</Job>
</Day>
<Day dow="4" templateDay="Thursday" >
<Job name="wake up" >
<Job name="get dressed" >
<Job name="prepare breakfast" >
<Job name="eat breakfast" > </Job>
</Job>
</Job>
</Job>
<Job name="work 9-5" >
</Job>
<Job name="football" >
</Job>
</Day>
<Day dow="5" templateDay="Friday" >
<Job name="go to pub" >
</Job>
</Day>
<Day dow="6" templateDay="Saturday" >
<Job name="work 9-5" >
</Job>
</Day>
<Day dow="7" templateDay="Sunday" >
<!-- nothing to do on sunday -->
</Day>
</Week>
使用此库http://code.google.com/p/x2js/将其转换为json,并转换为变量myData
{
"Week" : {
"Day" : [{
"Job" : [{
"Job" : {
"Job" : {
"Job" : {
"_name" : "eat breakfast"
},
"_name" : "prepare breakfast"
},
"_name" : "get dressed"
},
"_name" : "wake up"
}, {
"_name" : "work 9-5"
}
],
"_dow" : "1",
"_templateDay" : "Monday"
}, {
"Job" : [{
"Job" : {
"Job" : {
"Job" : {
"_name" : "eat breakfast"
},
"_name" : "prepare breakfast"
},
"_name" : "get dressed"
},
"_name" : "wake up"
}, {
"_name" : "work 9-5"
}, {
"_name" : "football"
}
],
"_dow" : "2",
"_templateDay" : "Tuesday"
}, {
"Job" : [{
"Job" : {
"Job" : {
"Job" : {
"_name" : "eat breakfast"
},
"_name" : "prepare breakfast"
},
"_name" : "get dressed"
},
"_name" : "wake up"
}, {
"_name" : "work 9-5"
}
],
"_dow" : "3",
"_templateDay" : "Wednesday"
}, {
"Job" : [{
"Job" : {
"Job" : {
"Job" : {
"_name" : "eat breakfast"
},
"_name" : "prepare breakfast"
},
"_name" : "get dressed"
},
"_name" : "wake up"
}, {
"_name" : "work 9-5"
}, {
"_name" : "football"
}
],
"_dow" : "4",
"_templateDay" : "Thursday"
}, {
"Job" : {
"_name" : "go to pub"
},
"_dow" : "5",
"_templateDay" : "Friday"
}, {
"Job" : {
"_name" : "work 9-5"
},
"_dow" : "6",
"_templateDay" : "Saturday"
}, {
"_dow" : "7",
"_templateDay" : "Sunday"
}
],
"_number" : "2013-W45"
}
}
Day可以有任意数量的Jobs,Jobs可以嵌套并且包含任意数量的其他Jobs。
现在使用此代码
<p ng-repeat="day in myData.Week.Day">
{{day._dow}} - {{day._templateDay}}
</p>
我可以列出有效的日期。我希望使用以下代码
<p ng-repeat="day in myData.Week.Day">
{{day._dow}} - {{day._templateDay}}
<span ng-repeat="job in day.Job">
{{job._name}}
<span/>
</p>
我可以列出当天的工作日和最高级的工作,但这是行不通的。(要列出嵌套的Jobs,将是下一个任务,而不是现在询问该任务)。
那么,如何至少列出顶级职位?同样在json格式中,我看到一些JobsObjects
和一些are Arrays
。如何处理两种情况?
PS .:使用角度1.2.0-rc.3