我的系统上安装了Java 1.6和Tomcat 5.5.
但是Tomcat 5.5访问Java 1.5,因此结果我在Bad version number in .class file使用JSP执行java代码时遇到错误.
如何将Tomcat版本更改为Java 1.6？
UPDATE
我尝试更改tomcat5w.exe指向1.6版本的JVM,现在我没有Bad version in .class file出错.但现在,我收到以下错误.
exception
org.apache.jasper.JasperException
org.apache.jasper.servlet.JspServletWrapper.handleJspException(JspServletWrapper.java:498)
org.apache.jasper.servlet.JspServletWrapper.service(JspServletWrapper.java:411)
org.apache.jasper.servlet.JspServlet.serviceJspFile(JspServlet.java:308)
org.apache.jasper.servlet.JspServlet.service(JspServlet.java:259)
javax.servlet.http.HttpServlet.service(HttpServlet.java:729)
root cause
java.lang.NullPointerException
myfirst.SearchLink.checkURL(SearchLink.java:20)
org.apache.jsp.Test_jsp._jspService(Test_jsp.java:52)
org.apache.jasper.runtime.HttpJspBase.service(HttpJspBase.java:98)
javax.servlet.http.HttpServlet.service(HttpServlet.java:729)
org.apache.jasper.servlet.JspServletWrapper.service(JspServletWrapper.java:369)
org.apache.jasper.servlet.JspServlet.serviceJspFile(JspServlet.java:308)
org.apache.jasper.servlet.JspServlet.service(JspServlet.java:259)
javax.servlet.http.HttpServlet.service(HttpServlet.java:729)
可能是根本原因？
我编写了下面的java代码,它执行另一个名为"Newsworthy_RB"的java程序.
Newsworthy_RB.java包含System.out.printlln()和System.err.println()语句.
我希望在命令提示符控制台中打印输出.
为了获得相同的东西,必须做些什么.
下面的程序只打印out.println()语句而不是err.println()语句.
请告诉我下面的代码是否会像我期望的那样发挥作用？
command = "java -cp .:../sqljdbc.jar SetHash Newsworthy_RB";
Process child1 = Runtime.getRuntime().exec(command);
InputStream in1 = child1.getErrorStream();
InputStream in2 = child2.getInputStream();
while ((c = in1.read()) != -1 || (c = in2.read()) != -1) {
System.out.print((char)c);
}
public static void main(String argv[]) {
try {
String date = new java.text.SimpleDateFormat("MM-dd-yyyy")
.format(new java.util.Date());
File inFolder = new File("Output/" + date + "_4D");
File outFolder = new File("Output/" + date + "_4D" + ".zip");
ZipOutputStream out = new ZipOutputStream(new BufferedOutputStream(
new FileOutputStream(outFolder)));
BufferedInputStream in = null;
byte[] data = new byte[1000];
String files[] = inFolder.list();
for (int i = 0; i < files.length; i++) {
in = new BufferedInputStream(new FileInputStream(
inFolder.getPath() + "/" + files[i]), 1000);
out.putNextEntry(new ZipEntry(files[i]));
int …Resultset rs=stmt.executeQuery("select count(*) from feedsca group by score order by score");
使用上面的上述java代码,我从名为feedsCA的表中检索行数.
虽然试图检索使用rs.getInt计数(1),rs.getInt(2),rs.getInt(3),I结束与一个错误如下说
Exception in thread "main" com.microsoft.sqlserver.jdbc.SQLServerException: The result set has no current row.
at com.microsoft.sqlserver.jdbc.SQLServerException.makeFromDriverError(Unknown Source)
at com.microsoft.sqlserver.jdbc.SQLServerResultSet.verifyResultSetHasCurrentRow(Unknown Source)
at com.microsoft.sqlserver.jdbc.SQLServerResultSet.getterGetColumn(Unknown Source)
at com.microsoft.sqlserver.jdbc.SQLServerResultSet.getInt(Unknown Source)
at SimpleMail.main(SimpleMail.java:151)
更新:
上述例外已经解决.
但我得到以下异常,我不知道原因.请指教.
Exception in thread "main" com.microsoft.sqlserver.jdbc.SQLServerException: The index 2 is out of range.
at com.microsoft.sqlserver.jdbc.SQLServerException.makeFromDriverError(Unknown Source)
at com.microsoft.sqlserver.jdbc.SQLServerResultSet.verifyValidColumnIndex(Unknown Source)
at com.microsoft.sqlserver.jdbc.SQLServerResultSet.getterGetColumn(Unknown Source)
at com.microsoft.sqlserver.jdbc.SQLServerResultSet.getInt(Unknown Source)
at SimpleMail.main(SimpleMail.java:152)
这就是我更新程序的方式.找到一个合乎逻辑的方法,因为我可以理解,下面的循环将无法按要求工作.
rs=stmt.executeQuery("select count(*) from feedsca group by score order by score");
while(rs.next()){ …import java.io.*;
import java.net.URL;
import java.net.URLConnection;
import java.sql.*;
public class linksfind{
public static void main(){
String html = "http://www.apple.com/pr/";
Document document = Jsoup.parse(html); // Can also take an URL.
for (Element element : document.getElementsByTag("a")) {
System.out.println(element.attr("href"));
}
}
}
伙计们,在上面的程序中,在执行时我发现了这些错误.怎么解决？我已经在我的文件夹位置下载了Jsoup.jar文件.我还该怎么办？
linksfind.java:8: cannot find symbol
symbol : class Document
location: class linksfind
Document document = Jsoup.parse(html); // Can also take a
^
linksfind.java:8: cannot find symbol
symbol : variable Jsoup
location: class linksfind
Document document = Jsoup.parse(html); // Can also take …为什么会发生以下情况？如何解决？
Exception in thread "main" java.lang.NoClassDefFoundError: org/apache/commons/lang/Validate
at org.jsoup.DataUtil.load(DataUtil.java:47)
at org.jsoup.Jsoup.parse(Jsoup.java:57)
at linksfind.main(linksfind.java:12)
Caused by: java.lang.ClassNotFoundException: org.apache.commons.lang.Validate
at java.net.URLClassLoader$1.run(Unknown Source)
at java.security.AccessController.doPrivileged(Native Method)
at java.net.URLClassLoader.findClass(Unknown Source)
at java.lang.ClassLoader.loadClass(Unknown Source)
at sun.misc.Launcher$AppClassLoader.loadClass(Unknown Source)
at java.lang.ClassLoader.loadClass(Unknown Source)
... 3 more
在Action.java文件中,我使用以下代码.
request.setAttribute("TAREWEIGHT", tareWeightList);
request.setAttribute("BARCODE", barcodeList);
return (mapping.findForward(target));
tareWeightList和barcodeList实际上只包含很少的值.将列表值设置为属性后,java文件将内容转发到JSP文件.
在JSP文件中,我可以使用下面的行获取内容,
<%=request.getAttribute("TAREWEIGHT")%>
<%=request.getAttribute("BARCODE") %>
我的要求是该列表的内容应以表格格式显示.
第一列中的条形码值及其在第二列中的相应Tareweight值.
建议我在JSP文件中编写代码,以便以列表格式显示内容.
import java.net.URL;
import java.net.URLConnection;
import java.sql.*;
public class searchlink{
public static void main(String args[]) throws Exception {
//String link="http://hosted.ap.org";
Connection con=null;
Statement stmt=null;
Statement stmtR=null;
if(con==null){
SQLConnection.setURL("jdbc:sqlserver://192.168.2.53\\SQL2005;user=sa;password=365media;DatabaseName=LN_ADWEEK");
con=SQLConnection.getNewConnection();
stmt=con.createStatement();
stmtR=con.createStatement();
}
ResultSet rs;
rs=stmt.executeQuery("select url from urls where url='http://www.topix.com/rty/elyria-oh'");
while(rs.next()){
String mem=rs.getString(1);
System.out.println("Result is "+mem);}
}
}
如果查询返回一行,上面的程序将打印输出.如果查询未返回任何内容,程序将停止而不打印任何内容.
而不是在没有打印任何东西的情况下停止它,我希望程序识别出查询没有返回任何内容并打印输出,如下所示"SQL查询执行后没有返回任何内容".
如何识别使用某些方法或变量来执行查询而不返回任何行？
import org.apache.commons.net.ftp.FTPClient;
import org.apache.commons.net.ftp.FTPReply;
import org.apache.commons.net.ftp.FTPFile;
import java.io.*;
public class FTPUpload{
public static boolean uploadfile(String server,String username,String Password,String source_file_path,String dest_dir){
FTPClient ftp=new FTPClient();
try {
int reply;
ftp.connect(server);
ftp.login(username, Password);
System.out.println("Connected to " + server + ".");
System.out.print(ftp.getReplyString());
reply = ftp.getReplyCode();
if(!FTPReply.isPositiveCompletion(reply)) {
ftp.disconnect();
System.err.println("FTP server refused connection.");
return false;
}
System.out.println("FTP server connected.");
InputStream input= new FileInputStream(source_file_path);
ftp.storeFile(dest_dir, input);
System.out.println( ftp.getReplyString() );
input.close();
ftp.logout();
} catch(Exception e) {
System.out.println("err");
e.printStackTrace();
return false;
} finally {
if(ftp.isConnected()) {
try { …