使用Hello,World Gridview教程示例,我试图在单击时使图像全屏,而不是在数组中显示图像的位置.由于我对Android不熟悉,这是我第一次尝试使用它,我很茫然.我虽然熟悉Java,但我尝试过这样的事情(显然不起作用):
public void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.main);
GridView gridview = (GridView) findViewById(R.id.gridview);
gridview.setAdapter(new ImageAdapter(this));
gridview.setOnItemClickListener(new OnItemClickListener() {
public void onItemClick(AdapterView<?> parent, View v, int position, long id) {
showImage(v, position);
Toast.makeText(HelloAndroid.this, "" + parent.getId(), Toast.LENGTH_SHORT).show();
}
});
}
private void showImage(View view, int position) {
ImageView imgView = (ImageView) view;
imgView.setImageResource(position);
}
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但应用程序崩溃(强制关闭).有任何想法吗?
我已经跟踪了一个奇怪的错误,undefined method `run_callbacks' for nil:NilClass并且能够使用此示例代码重现它.
基本上问题是活动记录正在超时(默认为5s)但抛出一个未定义的方法异常,这对我来说似乎不对.
但无论如何,处理这个问题的正确方法是什么?在我的真实代码中,我有一堆忙于做实际工作的线程但偶尔会遇到这个错误.所以想象一下,这puts是真正的代码.我希望现有的线程在发生这种情况时继续工作.
threads = []
10.times do |n|
threads << Thread.new {
ActiveRecord::Base.connection_pool.with_connection do |conn|
puts "#{n} #{conn}"
res = conn.execute("select sleep(6)", :async => true)
end
}
end
# block and wait for all threads to finish
threads.each { |t| puts "joined" ; t.join }
rescue Exception => e
puts $!, $@
end
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如果我按原样运行此代码,我会得到异常.如果我将睡眠减少到4s,我就不会.这是6s睡眠的输出.
joined
0 #<ActiveRecord::ConnectionAdapters::Mysql2Adapter:0xb73c6380>
1 #<ActiveRecord::ConnectionAdapters::Mysql2Adapter:0xb73c5548>
2 #<ActiveRecord::ConnectionAdapters::Mysql2Adapter:0xb73c4fe4>
3 #<ActiveRecord::ConnectionAdapters::Mysql2Adapter:0xb73c4a80>
4 #<ActiveRecord::ConnectionAdapters::Mysql2Adapter:0xb73c451c>
joined
joined
joined
joined
joined
undefined …Run Code Online (Sandbox Code Playgroud) mysql activerecord multithreading exception-handling ruby-on-rails
我可以(run-program "/usr/ls" '())在SBCL中运行.在任何Scheme实现中是否存在等价物?
我正在编写一个简单的绘图应用程序来了解HTML 5画布.问题是我似乎无法在canvas元素中获得正确的鼠标位置.我已经查看了stackoverflow上的其他问题,就像在这里用javascript在画布中获取鼠标位置来解决这个问题,但是他们的解决方案没有好像在帮助我.
这是我的代码:
<!DOCTYPE HTML>
<html>
<head>
<meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1">
<title>Untitled Document</title>
<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.4.4/jquery.min.js" type="text/javascript"></script>
<style type="text/css">
#test {
border: solid black 1px;
width: 500px;
height: 500px;
}
</style>
<script type="text/javascript">
$(function(){
var canvas=document.getElementById('test');
if(canvas.getContext){
var ctx =canvas.getContext('2d');
var draw = false;
ctx.strokeStyle = "rgb(200,0,0)";
ctx.lineWidth = 3;
ctx.lineCap = "round";
$('#test').mousedown(function(){draw=true;});
$('#test').mouseup(function(){draw=false;});
$('#test').mousemove(function(e){
if(draw){
var x , y;
x = e.layerX;
y = e.layerY;
ctx.moveTo(x,y);
ctx.lineTo(x+1,y+1);
ctx.stroke();
}
});
}
});
</script>
</head>
<body> …Run Code Online (Sandbox Code Playgroud) 我正在使用PayPal Pay API,具有自适应(链式)付款.我试图将用户转发到paypal,然后回到我预定义的return_url.
问题是:我需要在return-url中有一个PayKey.原因是:我需要调用PaymentDetail API来审核return_url中的付款.并且,我不想使用IPN,因为我需要在返回的Url上使用某些令牌进行验证.
我遇到的问题是:PayKey正在使用所有参数生成,包括return-url(因此在我构建实际数组之后,我从中得到我的$响应.我不能将PayKey放在return-Url中因为此时尚未生成.
//Create request payload with minimum required parameters
$bodyparams = array ("requestEnvelope.errorLanguage" => "en_US",
"actionType" => "PAY",
"currencyCode" => "USD",
"cancelUrl" => "http://www.paypal.com",
"returnUrl" => $return_url . "&payKey=${payKey}", **// Does not work - PAYKEY NEEDED TO ADD???**
"receiverList.receiver(0).email" => "account1@hotmail.com", //TODO
"receiverList.receiver(0).amount" => $price, //TODO
"receiverList.receiver(0).primary" => "true", //TODO
"receiverList.receiver(1).email" => "account2@hotmail.com", //TODO
"receiverList.receiver(1).amount" => $receiver_gets, //TODO
"receiverList.receiver(1).primary" => "false" //TODO
);
// convert payload array into url encoded query string
$body_data = http_build_query($bodyparams, "", …Run Code Online (Sandbox Code Playgroud) 是否有可能知道函数的名称?
main()->
myFunction().
myFunction()->
io:write("I am ~s !",[????]).
Run Code Online (Sandbox Code Playgroud) 当我在.m2/settings.xml文件中指定我的存储库时,我无法让Maven下载依赖项.但是,当我将存储库名称添加到我的pom时,Maven会下载这些依赖项.
具体来说,我正在尝试编译一些hibernate示例项目,并且我已经在说明中读到了我应该将以下存储库添加到我的pom或settings.xml中:
<repositories>
<repository>
<id>jboss-public-repository-group</id>
<name>JBoss Public Repository Group</name>
<url>http://repository.jboss.org/nexus/content/groups/public/</url>
<layout>default</layout>
<releases><enabled>true</enabled><updatePolicy>never</updatePolicy></releases>
<snapshots><enabled>true</enabled><updatePolicy>never</updatePolicy></snapshots>
</repository>
</repositories>
<pluginRepositories>
<pluginRepository>
<id>jboss-public-repository-group</id>
<name>JBoss Public Repository Group</name>
<url>http://repository.jboss.org/nexus/content/groups/public/</url>
<releases><enabled>true</enabled></releases>
<snapshots><enabled>true</enabled></snapshots>
</pluginRepository>
</pluginRepositories>
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当我将片段放在项目的pom.xml中时,一切正常,但是当我尝试将它放在settings.xml中时,我收到此错误:
org.hibernate的POM:hibernate-core:jar:3.6.1.Final缺失,没有可用的依赖信息
我有什么想法可能做错了吗?
是什么区别raw_input(),并input()在python3.x?
我有一个结构,你看到如下:
typedef struct _List {
Person *person; // pointer for people list
DoList *do; // Kinda timer, for checking list in some intervals
} List;
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有没有必要释放这个结构?如果是这样,我怎么能释放它?
我使用以下代码片段以特定模式振动手机,但它会抛出和ArrayIndexOutOfBoundsException.
vibrator.vibrate(new long[] { selectedDuration, CONSTANT_DELAY }, REPEAT);
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但
vibrator.vibrate(VIBRATE_DURATION);
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工作良好.有什么指针吗?