我正在使用谷歌地图Android SDK 11.6.2(也试过15.0.1),但我在地图显示之前遇到了崩溃.已经检查过清单中的API密钥,它可用,但仍会出现此问题.我有targetSDk版本为28.这是导致此问题.
java.lang.NoClassDefFoundError: Failed resolution of: Lorg/apache/http/ProtocolVersion;
at el.b(:com.google.android.gms.dynamite_mapsdynamite@12848063@12.8.48 (100408-196123505):3)
at ek.a(:com.google.android.gms.dynamite_mapsdynamite@12848063@12.8.48 (100408-196123505):4)
at em.a(:com.google.android.gms.dynamite_mapsdynamite@12848063@12.8.48 (100408-196123505):51)
at com.google.maps.api.android.lib6.drd.ap.a(:com.google.android.gms.dynamite_mapsdynamite@12848063@12.8.48 (100408-196123505):11)
at dw.a(:com.google.android.gms.dynamite_mapsdynamite@12848063@12.8.48 (100408-196123505):16)
at dw.run(:com.google.android.gms.dynamite_mapsdynamite@12848063@12.8.48 (100408-196123505):61)
Caused by: java.lang.ClassNotFoundException: Didn't find class "org.apache.http.ProtocolVersion" on path: DexPathList[[zip file "/system/priv-app/PrebuiltGmsCorePi/app_chimera/m/MapsDynamite.apk"],nativeLibraryDirectories=[/data/user_de/0/com.google.android.gms/app_chimera/m/00000036/MapsDynamite.apk!/lib/armeabi-v7a, /data/user_de/0/com.google.android.gms/app_chimera/m/00000036/MapsDynamite.apk!/lib/armeabi, /system/lib]]
at dalvik.system.BaseDexClassLoader.findClass(BaseDexClassLoader.java:126)
at java.lang.ClassLoader.loadClass(ClassLoader.java:379)
at ad.loadClass(:com.google.android.gms.dynamite_dynamiteloader@12848063@12.8.48 (100408-196123505):25)
at java.lang.ClassLoader.loadClass(ClassLoader.java:312)
at el.b(:com.google.android.gms.dynamite_mapsdynamite@12848063@12.8.48 (100408-196123505):3)
at ek.a(:com.google.android.gms.dynamite_mapsdynamite@12848063@12.8.48 (100408-196123505):4)
at em.a(:com.google.android.gms.dynamite_mapsdynamite@12848063@12.8.48 (100408-196123505):51)
at com.google.maps.api.android.lib6.drd.ap.a(:com.google.android.gms.dynamite_mapsdynamite@12848063@12.8.48 (100408-196123505):11)
at dw.a(:com.google.android.gms.dynamite_mapsdynamite@12848063@12.8.48 (100408-196123505):16)
at dw.run(:com.google.android.gms.dynamite_mapsdynamite@12848063@12.8.48 (100408-196123505):61)
我需要限制用户,仅访问特定的模式表.所以我尝试了以下查询并以user1_gmail_com登录.但是当我尝试浏览任何架构表时出现以下错误.
我的查询:
SELECT clone_schema('my_application_template_schema','user1_gmail_com');
CREATE USER user1_gmail_com WITH PASSWORD 'myloginpassword';
REVOKE ALL ON ALL TABLES IN SCHEMA user1_gmail_com FROM PUBLIC;
GRANT SELECT ON ALL TABLES IN SCHEMA user1_gmail_com TO user1_gmail_com;
SQL错误:
ERROR: permission denied for schema user1_gmail_com at character 46
In statement:
SELECT COUNT(*) AS total FROM (SELECT * FROM "user1_gmail_com"."organisations_table") AS sub
更新的工作查询:
SELECT clone_schema('my_application_template_schema','user1_gmail_com');
CREATE USER user1_gmail_com WITH PASSWORD 'myloginpassword';
REVOKE ALL ON ALL TABLES IN SCHEMA user1_gmail_com FROM PUBLIC;
GRANT USAGE ON SCHEMA user1_gmail_com TO user1_gmail_com;
GRANT …我的Android应用程序中有一项服务,它总是运行.现在我通过GCM从我的服务器获得设置并将这些设置更新到我的服务.我把我的设置放在了oncreate of service上.所以我需要重新启动我的服务以获取最新设置.如何重启我的服务？
Google Maps V3部分加载在左上角.我尝试了以下方法:
google.maps.event.trigger(map, 'resize');地图初始化后添加.
重新排列<script>索引文件中的标记
但没有一个适合我.如何解决这个问题.是否存在此问题的官方错误和解决方案？
的index.html
<!DOCTYPE html>
<html lang="en">
<head>
<meta charset="utf-8">
<title>Bootstrap, from LayoutIt!</title>
<meta name="viewport" content="width=device-width, initial-scale=1.0">
<meta name="description" content="">
<meta name="author" content="">
<!-- Bootstrap css files -->
<link href="stylesheets/bootstrap.min.css" rel="stylesheet">
<link href="stylesheets/bootstrap-responsive.min.css" rel="stylesheet">
<link href="stylesheets/style.css" rel="stylesheet">
</head>
<body>
<div id="contentbar">
<div id="dashboard-content" class="contentbar-main">
Some Content
</div>
<div id="summary-content" class="contentbar-main" style="display:none;">
<div class="container-fluid">
<div class="row-fluid">
<div class="span12">
<div class="row-fluid">
<div class="span7" id="sidebarid">
<p>Responsive web design is an approach, I often call it …我尝试将我的HelloWorld应用程序holo主题更改为Material.Light.DarkActionBar主题(正如新的Android开发工具会话所述).但我得到了以下错误.我尝试更改目标SDK版本为21.但我们在SDK Manager中没有21 SDK.在那次会议中,他们说,为v-21设置样式xml.
价值观/ styles.xml
<resources>
<!-- Base application theme. -->
<style name="AppTheme" parent="android:Theme.Holo.Light.DarkActionBar">
<!-- Customize your theme here. -->
</style>
</resources>
值-V21/styles.xml
<?xml version="1.0" encoding="utf-8"?>
<resources>
!-- Base application theme. -->
<style name="AppTheme" parent="android:Theme.Material.Light.DarkActionBar">
<!-- Customize your theme here. -->
</style>
</resources>
的build.gradle
apply plugin: 'com.android.application'
android {
compileSdkVersion 20
buildToolsVersion "20.0.0"
defaultConfig {
applicationId "com.ramapps.helloworld"
minSdkVersion 15
targetSdkVersion 20
versionCode 1
versionName "1.0"
}
buildTypes {
release {
runProguard false
proguardFiles getDefaultProguardFile('proguard-android.txt'), 'proguard-rules.pro'
}
}
}
dependencies { …我的数据库表中有超过500行,具有特定日期.
查询具有特定日期的行.
select * from msgtable where cdate='18/07/2012'
这将返回500行.
如何逐步查询这500行×10行.查询前10行并在浏览器中显示,然后查询下10行并在浏览器中显示？
如何使用Jquery在列表的第一个位置附加一个listitem？
<ul id="mylist">
<li>First</li>
<li>Second</li>
<li>Third</li>
<li>Fourth</li>
</ul>
是否可以在第一次<li>使用jquery 之前附加一个列表项.我试着$('#mylist li:eq(1)').before("<li></li>").但它不起作用？
我有两个24小时的时间值,并希望使用PHP进行比较.
我尝试过以下方法:
$time="00:05:00"; //5 minutes
if($time1<='00:03:00')
{
//do some work
}
else
{
//do something
}
这是使用PHP比较2个时间值的正确方法吗？
我正在使用PostgreSQL数据库.我想通过排除另一个表中存在的值来从表中获取列值.
select id from mytable where exclude(select id from another table)
在第一个表中可用的id:
101,102,103,104,105
在第二个表中可用的id:
101,104
我想要结果:
102,103,105 (excluded values exist in second table)
如何写查询？
当我尝试从scala Playmework发送邮件时,我收到了以下错误,
[ERROR] [10/10/2013 13:31:16.263] [play-akka.actor.default-dispatcher-75] [TaskInvocation] Sending the email to the following server failed : smtp.gmail.com:25
org.apache.commons.mail.EmailException: Sending the email to the following server failed : smtp.gmail.com:25
at org.apache.commons.mail.Email.sendMimeMessage(Email.java:1242)
at org.apache.commons.mail.Email.send(Email.java:1267)
at com.typesafe.plugin.CommonsMailer.send(MailerPlugin.scala:241)
at com.typesafe.plugin.MailerBuilder$class.sendHtml(MailerPlugin.scala:204)
at com.typesafe.plugin.CommonsMailer.sendHtml(MailerPlugin.scala:215)
at models.SignUpProcess$$anonfun$models$SignUpProcess$$sendEmail$1.apply$mcV$sp(SignUpProcess.scala:261)
at akka.actor.DefaultScheduler$$anon$8.run(Scheduler.scala:193)
at akka.dispatch.TaskInvocation.run(AbstractDispatcher.scala:137)
at scala.concurrent.forkjoin.ForkJoinTask$AdaptedRunnableAction.exec(ForkJoinTask.java:1417)
at scala.concurrent.forkjoin.ForkJoinTask.doExec(ForkJoinTask.java:262)
at scala.concurrent.forkjoin.ForkJoinPool$WorkQueue.runTask(ForkJoinPool.java:975)
at scala.concurrent.forkjoin.ForkJoinPool.runWorker(ForkJoinPool.java:1478)
at scala.concurrent.forkjoin.ForkJoinWorkerThread.run(ForkJoinWorkerThread.java:104)
Caused by: javax.mail.AuthenticationFailedException
at javax.mail.Service.connect(Service.java:319)
at javax.mail.Service.connect(Service.java:169)
at javax.mail.Service.connect(Service.java:118)
at javax.mail.Transport.send0(Transport.java:188)
at javax.mail.Transport.send(Transport.java:118)
at org.apache.commons.mail.Email.sendMimeMessage(Email.java:1232)
... 12 more
如何解决这个错误？
SignUpProcess.scala
private def sendEmail(subject: String, recipient: String, …