Question

我正在使用一系列数据在我的网页上存储数据。

    <div style="display: none;">
       <span class="data_location" data-name="Joe Bloggs" data-description="Lorem ipsum" data-location="CA"></span>
       <span class="data_location" data-name="Jane Doe" data-description="Ipsom erum" data-location="AN"></span>
       <span class="data_location" data-name="John Doe" data-description="Dorem noloy" data-location="CZ"></span>
       <span class="data_location" data-name="William Gates" data-description="Lorem ipsum" data-location="AG"></span>
       <span class="data_location" data-name="Henry Kissinger" data-description="Nuymt calum" data-location="AN"></span>
    </div>

我想循环遍历每个跨度，读取我的数据属性。下面的概念证明代码总是返回＆＃34; undefined＆＃34;。

        $(document).ready(function () {
            $.each(".data_location", function () {
                var location = $(this).data("location")
                alert(location)
            });
        });

我觉得问题在于$（这个），但我无法识别。任何人都可以帮助我吗？

Answer 1

select t.ID, c.name, sum(t.amount) from table1 t left join customers c on table1.ID = customers.ID2 group by t.ID, c.name having sum(t.amount) between 1000 and 10000函数被错误判断。你可以这样改变array#each。

为什么未定义？

您正在应用每种类型的数组方法。但是$(".data_location").each(function () {不是数组。元素的对象

$(".data_location")

$(document).ready(function () {
            $(".data_location").each(function () {
                var location = $(this).data("location")
                console.log(location)
            });
        });

Answer 2

您可以参考以下代码。

-- DO THIS FOR TICKET #3
UPDATE  records
SET     statue = True
    -- No need to flip it if it is already True
WHERE status <> True
    -- New condition for ticket #3
AND true_criteria IN ( 'something')
    -- regression ticket #1
AND this_criteria NOT IN( 'that' )
    -- regression ticket #2
AND another_criteria NOT IN ('whatever')
    ;

 $(document).ready(function () {
         $.each($(document).find(".data_location"), function () {
                var location = $(this).data("location");
                alert(location);
            });
            });

Answer 3

您的每个函数都是正确的，但回调方法必须包含2个参数1作为索引，另一个参数包含元素。

$(document).ready(function () {
    $.each($(".data_location"), function (index, element) {
       // Your $(this) equivalent will be $(element)
       var location = $(element).data("location");
       console.log(location);
    });
});

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<div style="display: none;">
       <span class="data_location" data-name="Joe Bloggs" data-description="Lorem ipsum" data-location="CA"></span>
       <span class="data_location" data-name="Jane Doe" data-description="Ipsom erum" data-location="AN"></span>
       <span class="data_location" data-name="John Doe" data-description="Dorem noloy" data-location="CZ"></span>
       <span class="data_location" data-name="William Gates" data-description="Lorem ipsum" data-location="AG"></span>
       <span class="data_location" data-name="Henry Kissinger" data-description="Nuymt calum" data-location="AN"></span>
    </div>

Answer 4

问题是因为$.each()逻辑this内没有引用迭代中的元素。如果要使用此模式，则需要使用传入处理函数的第二个参数，该函数是当前元素。试试这个：

$(document).ready(function() {
  $.each($(".data_location"), function(i, obj) {
    var location = $(obj).data("location")
    console.log(location)
  });
});

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<div style="display: none;">
  <span class="data_location" data-name="Joe Bloggs" data-description="Lorem ipsum" data-location="CA"></span>
  <span class="data_location" data-name="Jane Doe" data-description="Ipsom erum" data-location="AN"></span>
  <span class="data_location" data-name="John Doe" data-description="Dorem noloy" data-location="CZ"></span>
  <span class="data_location" data-name="William Gates" data-description="Lorem ipsum" data-location="AG"></span>
  <span class="data_location" data-name="Henry Kissinger" data-description="Nuymt calum" data-location="AN"></span>
</div>

然而，在这种情况下使用$('.data_location').each(fn)并在该处理函数中维护this引用更合适。

在使用.each（）循环时获取数据属性将返回undefined

4 个答案: