小编zon*_*ono的帖子

我正在尝试制作我的第一款Android应用.我注意到,SQLiteOpenHelper.onCreate()如果数据库不存在,则不会调用该方法来创建表.但是,onCreate()即使我尝试调试,该方法也无法正常工作.

请查看下面的代码,并给我任何建议.任何帮助将不胜感激.

public class NameToPinyinActivity extends Activity {

    DatabaseOpenHelper helper = null;

    @Override
    public void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        setContentView(R.layout.nametopinyin);

        Button searchButton = (Button) findViewById(R.id.search);
        searchButton.setOnClickListener(new ButtonClickListener());

        helper = new DatabaseOpenHelper(NameToPinyinActivity.this);
    }

public class DatabaseOpenHelper extends SQLiteOpenHelper {

    /** DB Name */
    private static final String DB_NAME = "pinyin";

    /** CREATE TABLE SQL */
    private static final String CREATE_TABLE_SQL = "CREATE TABLE UNICODE_PINYIN"
            + "(ID INTEGER PRIMARY KEY AUTOINCREMENT, "
            + "UNICODE TEXT NOT NULL, PINYIN …

springframework.beans.BeanUtils复制对象非常有用,我经常使用"ignoreProperties"选项.但是,有时我只想复制特定对象(基本上,与"忽略属性"相反).有谁知道我该怎么做？任何帮助将不胜感激.

import org.springframework.beans.BeanUtils;

public class Sample {    
    public static void main(String[] args) {    
        DemoADto demoADto = new DemoADto();
        demoADto.setName("Name of Demo A");
        demoADto.setAddress("Address of Demo A");

        DemoBDto demoBDto = new DemoBDto();

        // This is "ignoreProperties" option
        // But I want to know how I can copy only name field by using BeanUtils or something.
        BeanUtils.copyProperties(demoADto, demoBDto, new String[] {"address"});

        System.out.println(demoBDto.getName());
        System.out.println(demoBDto.getAddress());    
    }    
}

public class DemoADto {    
    private String name;    
    private String address;

    public String getName() {
        return name; …

我没有任何想法.你能给我任何线索(比如参考站点).任何帮助将不胜感激.

Model1: GROUP(id, name)
Model2: USER_GROUP_CMB(id, user_id, group_id)

预期的SQL语句:

SELECT * 
FROM groups AS g LEFT OUTER JOIN user_group_cmbs AS cmb 
            ON g.id = cmb.group_id
WHERE cmb.user_id = 1

我尝试在下面建立协会,但我不知道在此之后该怎么办.

class Group < ActiveRecord::Base
  has_many :user_group_cmb
end

class UserGroupCmb < ActiveRecord::Base
  has_many :group
end

Rails版本:3.1.1

我的目标是比较两列并添加结果列.R使用ifelse,但我需要知道熊猫的方式.

[R

> head(mau.payment)
  log_month user_id install_month payment
1   2013-06       1       2013-04       0
2   2013-06       2       2013-04       0
3   2013-06       3       2013-04   14994

> mau.payment$user.type <-ifelse(mau.payment$install_month == mau.payment$log_month, "install", "existing")
> head(mau.payment)
  log_month user_id install_month payment user.type
1   2013-06       1       2013-04       0  existing
2   2013-06       2       2013-04       0  existing
3   2013-06       3       2013-04   14994  existing
4   2013-06       4       2013-04       0  existing
5   2013-06       6       2013-04       0  existing
6   2013-06       7       2013-04       0  existing

熊猫

>>> maupayment
user_id  log_month  install_month
1 …

我试图了解Spring MVC的架构.但是,我完全被@SessionAttributes的行为搞糊涂了.

请看下面的SampleController,它是由SuperForm类处理post方法.事实上,只是SuperForm类的字段只能像我预期的那样绑定.

但是,在我将@SessionAttributes放入Controller后,处理方法作为SubAForm进行绑定.任何人都可以解释我在这个绑定中发生了什么.

-------------------------------------------------------

@Controller
@SessionAttributes("form")
@RequestMapping(value = "/sample")
public class SampleController {

    @RequestMapping(method = RequestMethod.GET)
    public String getCreateForm(Model model) {
        model.addAttribute("form", new SubAForm());
        return "sample/input";
    }

    @RequestMapping(method = RequestMethod.POST)
    public String register(@ModelAttribute("form") SuperForm form, Model model) {
        return "sample/input";
    }
}

-------------------------------------------------------

public class SuperForm {

    private Long superId;

    public Long getSuperId() {
        return superId;
    }

    public void setSuperId(Long superId) {
        this.superId = superId;
    }

}

-------------------------------------------------------

public class SubAForm extends SuperForm {

    private Long subAId;

    public Long …

有时候,我找到了一些类名,包括感知,如ApplicationContextAware和MessageSourceAware(春季).这个Aware有什么特别的含义还是一个着名的规则？

我正在尝试使用来自GCE(Google Compute Engine)实例的Google Cloud SQL over SSL.我的问题是我无法通过SSL连接到Cloud SQL实例.

mysql命令正常工作.我可以使用认证文件连接到Cloud SQL实例.

mysql -uroot -p -h [IP Address] --ssl-ca=/home/user/.cert/server-ca.pem --ssl-cert=/home/user/.cert/client-cert.pem  --ssl-key=/home/user/.cert/client-key.pem

但是,当我从PHP程序访问时,我收到了警告和致命错误.

<?php
$pdo = new PDO('mysql:host=[IP Address];dbname=testdb', 'root', 'test', array(
    PDO::MYSQL_ATTR_SSL_KEY    =>'/home/user/.cert/client-key.pem',
    PDO::MYSQL_ATTR_SSL_CERT=>'/home/user/.cert/client-cert.pem',
    PDO::MYSQL_ATTR_SSL_CA    =>'/home/user/.cert/server-ca.pem'
    )
);
$stmt = $pdo->query("SHOW TABLES;");
$row = $stmt->fetch(PDO::FETCH_ASSOC);
print_r($row);
?>?

PHP Warning:  PDO::__construct(): Peer certificate CN=`[GCP project name]:[Cloud SQL instance name]' did not match expected CN=`[IP Address]' in /tmp/mysql.php on line 7
Fatal error: Uncaught exception 'PDOException' with message 'SQLSTATE[HY000] [2002] ' in /tmp/mysql.php on …

我想在用户进入某个区域时发布通知.但是,由于同样的两个CLLocationManagerDelegate方法,我很困惑.我该如何正确使用这两种方法？

有人说如果应用程序在该区域开始,则需要"didDetermineState"方法来开始区域观察.

谢谢,

- (void)locationManager:(CLLocationManager *)manager
         didEnterRegion:(CLRegion *)region
{
    [self sendNotification:@"didEnterRegion"];
}

- (void)locationManager:(CLLocationManager *)manager
      didDetermineState:(CLRegionState)state
              forRegion:(CLRegion *)region
{
    switch (state) {
        case CLRegionStateInside:
         [self sendNotification:@"didEnterRegion"];
            break;
        case CLRegionStateOutside:
            break;
        case CLRegionStateUnknown:
            break;
        default:
            break;
    }
} 

我使用一个名为"ECOTree"的梦幻般的javascript组织树生成器,但是当我放置时绘图不起作用<!DOCTYPE html>.盒子中的标签没有正确放入盒子中.

您可以在下面的网站上的sample1.htm中看到错误.任何人都可以建议我如何解决. http://www.codeproject.com/script/Articles/ViewDownloads.aspx?aid=16192

(为了遵循我的项目规则,我必须定义<!DOCTYPE html>)

我正在寻找一种从应用程序ID获取应用程序图标的方法.你知道怎么做吗？请分享方式.谢谢.

例如Instagram,我正在寻找的id是:id389801252

https://itunes.apple.com/jp/app/instagram/id389801252?mt=8

我想得到这个图片: