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ajaxで取得したjsonをlodash製テンプレートに突っ込む

Last updated at Posted at 2016-02-16

#仕様
jsonの情報をテンプレートに突っ込んで、回数分DOMを生成したい。

#HTML
データ属性を付けた<div>内にloading.gifを入れておく。
入れとかなくても問題は無い。でもクルクルしてた方が格好いいと思う。

htnl
<div data-target-list="list">
	<img src="img/loading.gif">
</div>

#JSON

JSON
[
  {
    "title": "kuma",
    "url": "http://kuma.com",
    "img": "kuma.png",
  },
  {
    "title": "tama",
    "url": "http://tama.com",
    "img": "tama.png",
  },
  {
    "title": "kiso",
    "url": "http://kiso.com",
    "img": "kiso.png",
  },
]

#JS

  1. ajaxでjsonを取得。
  2. ajaxはdeferredオブジェクトを持っているので、宣言なしで.doneできる。
  3. lodashの_.templateでテンプレートを変数に格納。
  4. loading.gifを隠す。
  5. forで回して、テンプレートにjsonの情報を突っ込みつつ、DOM描画。

という流れ。

javascript
var list = function(){
	var $target = $('[data-target-list]');
	$.ajax({
		type: 'get',
		url: 'data/list.json',
		dataType: 'json'
	})
	.done(function(data){
		var temp = _.template(
			'<div class="primaryBox">' +
				'<a href="<%= url %>">' +
					'<%= title %>' +
				'</a>' +
				'<img src="<%= img %>">' +
			'</div>'
		);
		$('.loading').css('display','none');
		for (var i=0; i < data.length; i++){
			$target.append(temp(data[i]));
		}
	});
};
$(function(){
	list();
});
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