小编Ara*_*yan的帖子

如何分割字符串,分隔符保留在结果的末尾?

我有几个分隔符.例如{del1,del2,del3}.假设我有文本:Text1 del1 text2 del2 text3 del3

我想以这种方式拆分字符串:

  1. Text1 del1
  2. text2 del2
  3. text3 del3

当数组的每个元素都是texti deli时,我需要获取字符串数组.我怎么能在C#中做到这一点?

c# string split delimiter

2
推荐指数
1
解决办法
4840
查看次数

Tablerow高度没变?

例如,我以编程方式创建了4个表格,我希望每个表格高度为屏幕的1/4大小.每个桌子高度是屏幕尺寸的一半.我尝试了不同的方法,但是桌子高度并没有改变.我管理更改tablerow宽度,但高度没有.这是我的代码:

@SuppressLint("NewApi")
public class GameActivity extends Activity {

    private Context context;

    public void drawTable(){
        int i;
        TableLayout table = new TableLayout(context);

        table.setLayoutParams(new TableLayout.LayoutParams(TableLayout.LayoutParams.MATCH_PARENT, TableLayout.LayoutParams.MATCH_PARENT));
        //table.setStretchAllColumns(true);
        //table.setShrinkAllColumns(true);
        //table.setWeightSum(1.0F);
        TableRow[] rows = new TableRow[5];
        for(i = 0; i < 4; ++i){
            rows[i] = new TableRow(context);
            rows[i].setBackgroundResource(R.drawable.images214);
            rows[i].setWeightSum((float)1.0);
            rows[i].setLayoutParams(new TableRow.LayoutParams(TableRow.LayoutParams.MATCH_PARENT, 10));
            //rows[i].setScaleY(0.25F);
            table.addView(rows[i]);
        }

        LinearLayout l = (LinearLayout) findViewById(R.id.linear);
        l.addView(table);
    }
    @Override
    public void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        setContentView(R.layout.activity_game);

        context = this;
//        TableLayout table = new TableLayout(this);
//        TableRow row1 = new …
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android tablerow layoutparams android-4.0-ice-cream-sandwich android-tablelayout

2
推荐指数
1
解决办法
6138
查看次数

如何在Listbox中知道单击了哪个ListItem按钮?

我有Listbox:

  <ListBox x:Name="FriendsRequestList">
            <ListBox.ItemTemplate>
                <DataTemplate>
                    <StackPanel Orientation="Horizontal">
                        <StackPanel>
                            <TextBlock Text="{Binding FullName}" Foreground="#FF316DCB"/>
                            <TextBlock Text="{Binding RequestText}" />
                            <StackPanel Orientation="Horizontal">
                                <Button Name="Accept" Content="Accept" Click="Accept_Click"  Foreground="#FF28901F" Background="#FFB4D8BA"/>
                                <Button Name="Decline" Content="Decline" Click="Decline_Click"  Foreground="#FF28901F" Background="#FFB4D8BA"/>
                            </StackPanel>
                        </StackPanel>
                    </StackPanel>
                </DataTemplate>
            </ListBox.ItemTemplate>
  </Listbox>
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我在代码中尝试这些:

  private void Accept_Click(object sender, RoutedEventArgs e)
    {
        Button clickedButton = sender as Button;
        StackPanel st1 = clickedButton.Parent as StackPanel;
        StackPanel st2 = st1.Parent as StackPanel;
        StackPanel st3 = st2.Parent as StackPanel;
        object parentControl = st3.Parent;
        object obj = FriendsRequestList.Items[3];
        int index1 = FriendsRequestList.Items.IndexOf(obj); …
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silverlight listbox selectedindex controltemplate windows-phone-7

0
推荐指数
1
解决办法
3415
查看次数

传输sendMessage不会抛出异常

我发送消息到无效的电子邮件地址,但我没有任何异常.它的工作就像一切都很好,并且在调试器中它显示一切正常.

private MailService() {
    Properties props = new Properties();
    props.put("mail.transport.protocol", "smtps");
    props.put("mail.smtps.host", HOST);
    props.put("mail.smtps.auth", "true");
    props.put("mail.smtp.from", FROM);
    props.put("mail.smtps.quitwait", "false");

    mailSession = Session.getDefaultInstance(props);
    mailSession.setDebug(true);
}

public static void sendMessage(String recipient, String subject,
        String message) throws MessagingException {

    if (mailService == null) {
        mailService = new MailService();
    }

    MimeMessage mimeMessage = new MimeMessage(mailSession);

    mimeMessage.setFrom(new InternetAddress(FROM));
    mimeMessage.setSender(new InternetAddress(FROM));
    mimeMessage.setSubject(subject);
    mimeMessage.setContent(message, "text/plain");

    mimeMessage.addRecipient(Message.RecipientType.TO, new InternetAddress(recipient));

    Transport transport = mailSession.getTransport("smtps");
    transport.connect(HOST, PORT, USER, PASSWORD);

    transport.sendMessage(mimeMessage,
            mimeMessage.getRecipients(Message.RecipientType.TO));
    transport.close();

}
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我尝试了不同的电子邮件地址,并且尽管它没有抛出异常.我错了什么?

java sendmail jakarta-mail transport

0
推荐指数
1
解决办法
6340
查看次数

如何获取jsf f所选项目的索引:selectItems?

我有seleconeradio,例如:

<h:selectOneRadio value="#{myBean.selectedValue}" layout="pageDirection">
    <f:selectItems value="#{myBean.myList}" var="a" itemValue="#{a}" itemLabel="#{a}"/>
</h:selectOneRadio>
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其中myList是整数列表,例如1,3,2,4.如果用户在myBean selectedValue中选择第二个元素(即3),那么我想得到selectItems项的索引.

我应该在f:selectItems itemValue标签中写什么?或者这是不可能的?

PS我可以通过创建一个新类来实现它,在该类中我有索引属性并创建该类的新列表,给出正确的索引.但这是非常糟糕的解决方案.

jsf jsf-2 selectoneradio

0
推荐指数
1
解决办法
4474
查看次数