此代码无效
var next = $("#orders").find(".next");
if (next.length == 1) {
var address = $(next[0]).find(".directionsAddress");
var destination = $(address[0]).text();
}
假设找到一个带有" next"的等级的div。我知道它存在于页面上,然后在结果集数组的那一项中,将有一个div,其类名为directionsAddress。 "下一个"数组以1的长度返回,所以看起来问题出在我的$(next [0])。找到因为地址数组返回0长度而我正在制作某种类型的语法错误不明白。
答案 0 :(得分:0)
$(document).ready(function() {
alert($('#orders div.next:first div.directionsAddress:first').text());
});

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<div id = "orders">
<div class="next">
<div class="directionsAddress">THIS IS WHAT I WANT</div>
<div class="directionsAddress">This is not the one I want</div>
</div>
<div class="next">
<div class="directionsAddress">This is not the one I want</div>
</div>
<div class="next">
<div class="directionsAddress">This is not the one I want</div>
</div>
</div>
&#13;