[英]Trying to find a more efficient way to do multiple replaces without repeating the same code in jQuery
我有一堆用#selector1
, #selector2
, #selector3
等包裹的<select>
。我也有一堆用#data1
, #data2
, #data3
包裹的h1
。
每次更改每個#selector
的下拉菜單時,它都會替換其各自的#data
的文本。 因此,例如,更改#selector1
的下拉列表將更改#data1
的文本。
為了更好的主意,這是我的代碼:
$("#selector1 .choose").on('change', function(e) { var titleName = $('#selector1 .choose').find(":selected").text(); $("#data1 .title").text(titleName); e.preventDefault(); }); $("#selector2 .choose").on('change', function(e) { var titleName = $('#selector2 .choose').find(":selected").text(); $("#data2 .title").text(titleName); e.preventDefault(); }); $("#selector3 .choose").on('change', function(e) { var titleName = $('#selector3 .choose').find(":selected").text(); $("#data3 .title").text(titleName); e.preventDefault(); });
.title { font-size: 16px;}
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script> <div id="selector1"> <select class="choose"> <option value="1">Test 1</option> <option value="2">Test 2</option> <option value="3">Test 3</option> </select> </div> <div id="selector2"> <select class="choose"> <option value="1">Test 1</option> <option value="2">Test 2</option> <option value="3">Test 3</option> </select> </div> <div id="selector3"> <select class="choose"> <option value="1">Test 1</option> <option value="2">Test 2</option> <option value="3">Test 3</option> </select> </div> <!-- etc ... --> <div id="data1"> <h1 class="title"> Test 0 </h1> </div> <div id="data2"> <h1 class="title"> Test 0 </h1> </div> <div id="data3"> <h1 class="title"> Test 0 </h1> </div> <!-- etc ... -->
如您所見,jQuery非常低效。 因為我必須一遍又一遍地重復相同的代碼,所以只需更改每個選擇器的編號。
什么是更好,更有效的方法?
無需復制這樣的代碼,只需使其通用,按類綁定,並通過$(this)...
訪問值。 您可以將數據屬性用作h1
的選擇器。
您可以像這樣改善它:
$(".choose").on('change', function(e) { var id = $(this).data('id'); var text = $(this).find(":selected").text(); $('#' + id + ' .title').text(text); e.preventDefault(); });
.title { font-size: 16px;}
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script> <div id="selector1"> <select class="choose" data-id="data1"> <option value="1">Test 1</option> <option value="2">Test 2</option> <option value="3">Test 3</option> </select> </div> <div id="selector2"> <select class="choose" data-id="data2"> <option value="1">Test 1</option> <option value="2">Test 2</option> <option value="3">Test 3</option> </select> </div> <div id="selector3"> <select class="choose" data-id="data3"> <option value="1">Test 1</option> <option value="2">Test 2</option> <option value="3">Test 3</option> </select> </div> <!-- etc ... --> <div id="data1"> <h1 class="title"> Test 0 </h1> </div> <div id="data2"> <h1 class="title"> Test 0 </h1> </div> <div id="data3"> <h1 class="title"> Test 0 </h1> </div> <!-- etc ... -->
$(".choose").on('change', function(e) { var id = e.target.getAttribute('data-id'); var titleName = $(this).find(":selected").text(); $("#"+id+" .title").text(titleName); e.preventDefault(); });
.title { font-size: 16px;}
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script> <div id="selector1"> <select class="choose" data-id="data1"> <option value="1">Test 1</option> <option value="2">Test 2</option> <option value="3">Test 3</option> </select> </div> <div id="selector2" > <select class="choose" data-id="data2"> <option value="1">Test 1</option> <option value="2">Test 2</option> <option value="3">Test 3</option> </select> </div> <div id="selector3" > <select class="choose" data-id="data3"> <option value="1">Test 1</option> <option value="2">Test 2</option> <option value="3">Test 3</option> </select> </div> <!-- etc ... --> <div id="data1"> <h1 class="title"> Test 0 </h1> </div> <div id="data2"> <h1 class="title"> Test 0 </h1> </div> <div id="data3"> <h1 class="title"> Test 0 </h1> </div> <!-- etc ... -->
$('.choose').change(function() {
var this_nr=$(this).parent().attr('id').substr(-1); //get curent nr from parent id
var this_val=$(this).val(); //get value from curent select
$('#data'+this_nr+' .title').text(this_val); //add value to output
});
聲明:本站的技術帖子網頁,遵循CC BY-SA 4.0協議,如果您需要轉載,請注明本站網址或者原文地址。任何問題請咨詢:yoyou2525@163.com.