小编Bru*_*nco的帖子

我有这个问题,REPLACE功能不适用于eclipselink版本2.5.2.

这是我的代码:

String sSql = " SELECT e FROM br.com.megasoft.protocolo.entity.Assunto e  WHERE  ( REPLACE(REPLACE(REPLACE( UPPER(e.titulo), '/', ''), '-', ''), '.', '') LIKE  UPPER('A') )";
TypedQuery<?> query = getEntityManager().createQuery(sSql, Class.forName(this.tabela));

第二个参数的值为: class br.com.megasoft.protocolo.entity.Assunto

它给出了这个例外:

Caused by: Exception [EclipseLink-0] (Eclipse Persistence Services - 2.5.2.v20140319-9ad6abd): org.eclipse.persistence.exceptions.JPQLException
Exception Description: Syntax error parsing [ FROM br.com.megasoft.protocolo.entity.Assunto e  WHERE  ( REPLACE(REPLACE(REPLACE( UPPER(e.titulo), '/', ''), '-', ''), '.', '') LIKE  UPPER( :valorPesq10) ) ]. 
[54, 150] The expression is not a valid conditional expression.
    at …

我想通过使用HttpURLConnection进行POST.我在两个方面尝试这个,但在做的时候我总是得到一个例外:conn.getOutputStream();

我在两种情况下得到的例外是:

java.net.SocketException:操作超时:connect:可能是由于地址无效

功能1:

public void makePost(String title, String comment, File file) {
    try {
        URL servlet = new URL("http://" + "www.server.com/daten/web/test/testupload.nsf/upload?CreateDocument");            
        HttpURLConnection conn=(HttpURLConnection)servlet.openConnection();
        conn.setDoOutput(true);
        conn.setDoInput(true);
        conn.setUseCaches(false);
        String boundary = "---------------------------7d226f700d0";
        conn.setRequestProperty("Content-type","multipart/form-data; boundary=" + boundary);
        //conn.setRequestProperty("Referer", "http://127.0.0.1/index.jsp");
        conn.setRequestProperty("Cache-Control", "no-cache");

        OutputStream os = conn.getOutputStream(); //exception throws here!
        DataOutputStream out = new DataOutputStream(os);
        out.writeBytes("--" + boundary + "\r\n");
        writeParam(INPUT_TITLE, title, out, boundary);
        writeParam(INPUT_COMMENT, comment, out, boundary);
        writeFile(INPUT_FILE, file.getName(), out, boundary);
        out.flush();
        out.close();

        InputStream stream = conn.getInputStream();
        BufferedInputStream in = new BufferedInputStream(stream); …

我在从MySql数据库中获取数据时遇到问题.我的代码如下:

        try {
        Connection con = datasource.getConnection();
        Statement stmt = con.createStatement();

        ResultSet rs;
        rs = stmt.executeQuery("SELECT tittle,date,path " +
                "FROM announcement "+
                "ORDER BY date");
        String tittle=rs.getString("tittle");
        String date=rs.getString("date");
        String text=rs.getString("path");


     if (!rs.isBeforeFirst())
     {
        out.println("<p>No data !</p>");
    }
     else
     {            
        out.println("<table class=\"data\">");
        out.println("<tr ><td class=\"sectionheader\"> tittle?</td><td            class=\"sectionheader\">date</td><td class=\"sectionheader\">text</td></tr>");

        while (rs.next()) {
            String row="";
            row  += "<td><a href=\"\"> "+tittle+ "</td>";
            row  += "<td>" + date + "</td>";
            row +="<td>"+text+"</td>";

            row +="</tr>";
            out.println(row);

        }

我得到的错误是"java.sql.SQLException: Before start of result …

我想要一个添加ToolTip到自定义MapMarker的JMapViewer。但是，取消搜索并不能帮助我解决这个问题。

自定义MapMarker为：

public class MapMarkerUnit extends MapObjectImpl implements MapMarker

绘制方法的覆盖是

public void paint(Graphics g, Point position, int radio) {
    String filename = "marker.png";
    //System.out.print(filename);
    BufferedImage x = null;
    try {
        x = ImageIO.read(getClass().getResource(filename));
    } catch (IOException ex) {
        Logger.getLogger(MapMarkerUnit.class.getName()).log(Level.SEVERE, null, ex);
    }

    g.drawImage(x, position.x-16, position.y-37,null);

    //if(getLayer()==null||getLayer().isVisibleTexts()) paintText(g, new       Point(position.x+20,position.y));
}

感谢您提供的任何帮助。

嗯,我找不到错误:

package udemy;

public class ClassOne {

private int positiveNum;

public ClassOne (int positiveNum){


    if (positiveNum< 0 ){
        throw new IllegalArgumentException("positiveNum should be greater than 0");
    }

    this.positiveNum= positiveNum;
}

public int getNum(){
    return positiveNum;
}

}

在我的主要课程中,我得到了这段代码:

package udemy;

public class ClassTwo {
public static void main (String[] args) {


    ClassOne number= new ClassOne(-15);
    try {
        System.out.printf("%s%n",number.getNum());
    } 
    catch(IllegalArgumentException e) 
    {
        System.out.printf("%s",e.getMessage());
    }

}

}

但我得到以下输出:

Exception in thread "main" java.lang.IllegalArgumentException:positiveNum should be greater than 0
 at udemy.ClassOne.<init>(ClassOne.java:11)
at …