我一直试图得到一个TR元素的索引,它有两个类,第一个将它识别为一个组,第二个作为一个子组,然后我尝试获取特定组的索引和它返回给我一个小组-1 ...有谁知道我做错了什么?
以下是一个例子:
<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.12.4/jquery.min.js"></script>
<table>
<tr class="trusure1 trrstatusC"></tr>
<tr class="trusure2 trrstatusC"></tr>
<tr class="trusure2 trrstatusC"></tr>
<tr class="trusure2 trrstatusR"></tr>
<tr class="trusure2 trrstatusR"></tr>
<tr class="trusure3 trrstatusC"></tr>
<tr class="trusure3 trrstatusC"></tr>
</table>
<div id="console"></div>
<script>
$('#console').append('<br> The tr with the class trusure2 and the lastest with class trrstatusR is on the position '+$('tr.trusure2.trrstatusR:last-child').index());
<script>
https://jsfiddle.net/nod9xbsh/
谢谢!
答案 0 :(得分:2)
您应该使用<?xml version="1.0" encoding="utf-8"?>
<manifest xmlns:android="http://schemas.android.com/apk/res/android" android:versionCode="1" android:versionName="1.0" package="com.random.testapp.testappandroid">
<uses-sdk android:minSdkVersion="19" android:targetSdkVersion="19" />
<application android:allowBackup="true" android:icon="@mipmap/icon" android:label="@string/app_name"></application>
<service android:name=".TestService" android:process=":test_service" android:enabled="true"></service>
</manifest>
代替