我正在使用骨干,把手,jquery for app。以下代码行无效
this.$el.find('.quiz-container').append(quiz.tmplt.result(this.collection.toJSON()));
模板
<h2 class="score">Congratulation ABCD! You have scored {{score}} !!!</h2>
<table>
<thead>
<tr>
<th>Question</th>
<th>Your Answer</th>
<th>Correct Answer</th>
</tr>
</thead>
<tbody>
{{#each qtext}}
<tr>
<td>{{this}}</td>
<td>
{{#each ../answer}}
<span>{{this}}</span>
{{/each}}
</td>
</tr>
{{/each}}
</tbody>
</table>
这是我的JSON文件
[{
"qid": "1001",
"qtext": "This is question one?",
"options": [{"id":"opt1", "str":"15%"},{"id":"opt2", "str": "16%"},{"id":"opt3", "str": "17%"},{"id":"opt4", "str": "19%"}],
"answer": ["opt2"]},
{
"qid": "1002",
"qtext": "This is question two?",
"options": [{"id":"opt1", "str":"item0"},{"id":"opt2", "str": "item1"},{"id":"opt3", "str": "item2"},{"id":"opt4", "str": "item3"}],
"answer": ["opt1"]},
{
"qid": "1003",
"qtext": "This is question three?",
"options": [{"id":"opt1", "str":"5%"},{"id":"opt2", "str": "2.5%"},{"id":"opt3", "str": "11%"},{"id":"opt4", "str": "0%"}],
"answer": ["opt3"]}
]
我知道,我可以拆分这个模板并使用collection.each方法我可以从JSON文件渲染每个对象,但我想保留它第一个模板并直接从集合数据渲染。
提前致谢
维克拉姆