Question

我正在处理以下代码。为什么我无法基于selected数组更新chekbox状态？

var selected = [134,135,136,137,138,139,140,141,142,143,144];
$.each(selected, function(index, value){
  $("input[type='checkbox'][value"+ value.member_key+ "]").prop("checked", true);
});

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<input type="checkbox" name="vehicle" value="133">133<br>
<input type="checkbox" name="vehicle" value="134">134<br>
<input type="checkbox" name="vehicle" value="135">135<br>
<input type="checkbox" name="vehicle" value="136">136<br>
<input type="checkbox" name="vehicle" value="137">137<br>
<input type="checkbox" name="vehicle" value="138">138<br>
<input type="checkbox" name="vehicle" value="139">139<br>
<input type="checkbox" name="vehicle" value="140">140<br>
<input type="checkbox" name="vehicle" value="141">141<br>
<input type="checkbox" name="vehicle" value="142">142<br>
<input type="checkbox" name="vehicle" value="143">143<br>
<input type="checkbox" name="vehicle" value="144">144<br>
<input type="checkbox" name="vehicle" value="145">145<br>
<input type="checkbox" name="vehicle" value="146">146<br>

Answer 1

传递的member_key上没有名为value的属性。由于您是在数字数组上实现.each()，因此在每次迭代中，回调函数都会将数字本身作为第二个参数。因此，只需使用value：

var selected = [134,135,136,137,138,139,140,141,142,143,144];
 $.each(selected, function(index, value){
  $("input[type='checkbox'][value="+value+"]").prop("checked", true);
})

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<input type="checkbox" name="vehicle" value="133">133<br>
<input type="checkbox" name="vehicle" value="134">134<br>
<input type="checkbox" name="vehicle" value="135">135<br>
<input type="checkbox" name="vehicle" value="136">136<br>
<input type="checkbox" name="vehicle" value="137">137<br>
<input type="checkbox" name="vehicle" value="138">138<br>
<input type="checkbox" name="vehicle" value="139">139<br>
<input type="checkbox" name="vehicle" value="140">140<br>
<input type="checkbox" name="vehicle" value="141">141<br>
<input type="checkbox" name="vehicle" value="142">142<br>
<input type="checkbox" name="vehicle" value="143">143<br>
<input type="checkbox" name="vehicle" value="144">144<br>
<input type="checkbox" name="vehicle" value="145">145<br>
<input type="checkbox" name="vehicle" value="146">146<br>

Answer 2

var selected = [134,135,136,137,138,139,140,141,142,143,144];
 $.each(selected, function(index, value){
                                       //include the = and fix the value being compared
              $("input[type='checkbox'][value="+ value+ "]").prop("checked", true);
         })

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<input type="checkbox" name="vehicle" value="133">133<br>
<input type="checkbox" name="vehicle" value="134">134<br>
<input type="checkbox" name="vehicle" value="135">135<br>
<input type="checkbox" name="vehicle" value="136">136<br>
<input type="checkbox" name="vehicle" value="137">137<br>
<input type="checkbox" name="vehicle" value="138">138<br>
<input type="checkbox" name="vehicle" value="139">139<br>
<input type="checkbox" name="vehicle" value="140">140<br>
<input type="checkbox" name="vehicle" value="141">141<br>
<input type="checkbox" name="vehicle" value="142">142<br>
<input type="checkbox" name="vehicle" value="143">143<br>
<input type="checkbox" name="vehicle" value="144">144<br>
<input type="checkbox" name="vehicle" value="145">145<br>
<input type="checkbox" name="vehicle" value="146">146<br>

Answer 3

其他答案都在

上发表评论

select  
    a.*, b.address 
from 
    customer_details  a 
left join 
    address_details b on a.street_no = b.street_no 
where 
    b.case_id = 1

我认为这是将代码复制到stackoverflow中的错误。我猜您在实际代码中缺少的部分是单词“值”与其实际值之间的等号。

您的代码将产生

value.member_key

您需要的地方

[value133]

如何基于数组更新复选框状态？

3 个答案: