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how to upload using html or javascript?

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posted 17 years ago

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hi i have code of type

what i require is suppose i press submit button. the file selected using browse should be uploaded in some location.

can anybody give hints like what i can do??

Ulf Dittmer

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posted 17 years ago

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Since you know about the proper HTML element already, I suppose this question is about the server side? If the receiving end is a Java servlet, then you can use a library like Apache Commons FileUpload to deal with it; it's quite easy to use.

J radolf

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this is my front end html code

this is my jsp code (server side)

<%@ page import="java.util.List" %>
 <%@ page import="java.util.Iterator" %>
 <%@ page import="java.io.File" %>
 <%@ page import="org.apache.commons.fileupload.servlet.ServletFileUpload"%>
 <%@ page import="org.apache.commons.fileupload.disk.DiskFileItemFactory"%>
 <%@ page import="org.apache.commons.fileupload.*"%>
 <%@ page contentType="text/html;charset=UTF-8" language="java" %>
 <center><table border="2">
 <tr><td><h1>Your files uploaded </h1></td></tr>
 <%
 boolean isMultipart = ServletFileUpload.isMultipartContent(request);
 if (!isMultipart) {
 } else {
	 FileItemFactory factory = new DiskFileItemFactory();
	 ServletFileUpload upload = new ServletFileUpload(factory);
	 List items = null;
	 try {
		 items = upload.parseRequest(request);
	 } catch (FileUploadException e) {
		 e.printStackTrace();
	 }
	 Iterator itr = items.iterator();
	 while (itr.hasNext()) {
	 FileItem item = (FileItem) itr.next();
	 if (item.isFormField()) {
	 } else {
		 try {
			 String itemName = item.getName();
			 File savedFile = new File(config.getServletContext
().getRealPath("/")+"uploadedFiles/"+itemName);
			 item.write(savedFile);

out.println("<tr><td>Your file has been saved 
at the loaction:</td></tr><tr><td>"+config.getServletContext().getRealPath
("/")+"uploadedFiles"+"\\"+itemName+"</td></tr>");
		 } catch (Exception e) {
			 e.printStackTrace();
		 }
	 }
	 }
 }
 %>
 </table>
 </center>

this is the error i am getting.
any idea why is it so?

org.apache.jasper.JasperException: Unable to compile class for JSP:

An error occurred at line: 9 in the generated java file
Only a type can be imported. org.apache.commons.fileupload.servlet.ServletFileUpload resolves to a package

An error occurred at line: 10 in the generated java file
Only a type can be imported. org.apache.commons.fileupload.disk.DiskFileItemFactory resolves to a package

An error occurred at line: 34 in the jsp file: /upload_file_multipale.jsp
String literal is not properly closed by a double-quote
31: item.write(savedFile);
32:
33:
34: out.println("<tr><td>Your file has been saved
35: at the loaction:</td></tr><tr><td>"+config.getServletContext().getRealPath
36: ("/")+"uploadedFiles"+"\\"+itemName+"</td></tr>");
37: } catch (Exception e) {

J radolf

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posted 17 years ago

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i think this is a issue realted to adding some external jar to be added can anyone tell me which jar has to be added?

J radolf

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posted 17 years ago

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am using tomacat server does anybody knows which are the external jars that has to be added and which one's to keep in lib folder?

Ulf Dittmer

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posted 17 years ago

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Is your first name really "january"? If not, we have a policy on screen names here at JavaRanch. Basically, it must consist of a real-sounding first name and last name, and not be obviously fictitious. If yours does not conform with it, please take a moment to change it, which you can do using the "My Profile" link at the top of the page.

Obviously, you need to have the FileUpload jar file in the WEB-INF/lib folder of your web app. It's usually called something like "commons-fileupload-1.2.1.jar".

It's also bad design to keep file I/O code in your JSP pages. It makes proper error handling (amongst other things) hard to do. You should move that code to a servlet instead.

J radolf

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posted 17 years ago

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yes my first name is january.

After adding the jar file to the LIB folder also am not able to get the files uploaded to the given path

The Jsp is executing and just showing

your files uploaded..

but acutal uploading is not happening.

Bauke Scholtz

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posted 17 years ago

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Do not write Java code in JSP files. Scriptlets indicate a bad design and are receipt for disaster. They are highly discouraged.

Write Java code in Java classes. Servlets are highly suitable for your requirement.

Read the 'User Guide' and 'Frequently Asked Questions' sections of http://commons.apache.org/fileupload how to use it properly.

Bear Bibeault

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posted 17 years ago

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Since this is not about HTML, and should be done in a servlet, I've moved this to the Servlets forum.

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J radolf

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i have done everything possible by me

the following are the changes i did

added the following jar in my LIB folder

commons-fileupload1.2.1.jar
commons-io1.4.jar

still when i send a request tp the jsp program

i get the JSP executed and get

YOUR FILES UPLOADED
with no errors or other trace
but my selected files are not uploaded.

i know its not a good design as you said earlier but my team wants this design . so have to do it??

anybody knows what else can i do to resolve this issue?

Bauke Scholtz

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Have you read the appserver logs?

J radolf

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hi thank you for your previous posts.
rest of the problem have been solved.

Now am Trying to find how can you retrieve Text values in a jsp in a upload file program.

Because request.getParameter("name") always gives null.

How can i come over this problem??

Bear Bibeault

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posted 17 years ago

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That greatly depends upon which upload package you are using. Read its documentation.

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J radolf

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posted 17 years ago

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Am using apache uploading package

J radolf

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<%@page import="org.apache.commons.fileupload.*,org.apache.commons.fileupload.disk.*,org.apache.commons.fileupload.servlet.*,java.util.*,java.io.*"%>
<%
try{
FileItemFactory factory = new DiskFileItemFactory();
ServletFileUpload upload = new ServletFileUpload(factory);
boolean isMultipart = ServletFileUpload.isMultipartContent(request);

if(isMultipart)
{

List items=upload.parseRequest(request); 
Iterator iterator=items.iterator();
out.println("Files Uploaded successfully");
while(iterator.hasNext())
{
FileItem fitem=(FileItem)iterator.next();
if(fitem.isFormField()){ /********why ********************/
if(fitem.getFieldName().equals("documentname")){ /*******these lines are not executing*************/
String DocumentName=fitem.getString(); /************this too ***********************/
out.println("Document name is" +DocumentName); /*************this too rest are fine****/
}
}
StringBuffer s2=new StringBuffer(fitem.getName());
out.println("Files Saved are: " +s2 +"");
if(s2!=null && s2.length()>0)
{
int separatorPos=s2.lastIndexOf("\\");
String filename=s2.substring(separatorPos+1,s2.length());
String appPath = application.getRealPath("/");
String temp=appPath+filename;
//out.println("File upload path is" +temp);
File fl=new File(temp); 
fitem.write(fl);
}
}
}
else
out.println("Not a multipart request");
}
catch(Exception e)
{
out.println(e.toString());
}
%>

Ulf Dittmer

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posted 17 years ago

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If you have code inside of an if-block not executing, then the first guess would be that the if condition evaluates to false. Do you understand what "isFormField" checks?

J radolf

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posted 17 years ago

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yes. its basically a text field. but in my case isFormField always returns false and Document Name textField is never read.

Ulf Dittmer

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The HTML of the documentname field is malformed; that's probably what causes the trouble.

A tool like the Firefox extension LiveHttpheaders can help troubleshoot problems like this one by showing what exactly gets transmitted in the HTTP request and the response.

J radolf

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Thnkx for the reply..

so should i change my front end style.. i mean HTML tags.. because as per my requirement more styling is not required..!!

Bear Bibeault

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Who said anything about style? Your HTML is invalid.

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J radolf

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Hi All

when i saved the above malfunctioned file as jsp and clicked submit i was not able to read the text field value

but when the same thing i saved as HTML and submitted values i was able to read the text value entered.

can anyone tell me why this happened??

J radolf

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posted 17 years ago

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this is my final 2 files
FIRST FILE

SECOND FILE

if(isMultipart)
{

List items=upload.parseRequest(request); 
Iterator iterator=items.iterator();

while(iterator.hasNext())
{

FileItem fitem=(FileItem)iterator.next();
String fldName=fitem.getFieldName();
if(!fitem.isFormField()){
StringBuffer s2=new StringBuffer(fitem.getName()); 
if(s2!=null && s2.length()>0)
{
int separatorPos=s2.lastIndexOf("\\");
String filename="";
if(separatorPos>0)
filename=s2.substring(separatorPos+1,s2.length());
else
filename=s2.toString(); 
File fl=new File("e:\\"+filename);
fitem.write(fl);
}
}
else{
String fldValue=fitem.getString();
out.println("Fld name in html="+fldName+" value is "+fldValue);
}
} 
}
else 
{
out.println("Not a Multipart request");
}
}

catch(Exception e)
{
out.println(e.toString());
}
%>

now my question when i saved the FIRST file here as jsp and clicked submit i was not able to get Document name from the second file. but when i saved the first file as HTML and clicked submit i got the DOCUMENT name from the second file.

So can any anybody expalin why the difference?

Bauke Scholtz

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posted 17 years ago

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Again, JSP is a view technology. You should not write raw Java code in a JSP file. Java code belongs in Java classes. In your specific case, a Servlet is perfectly suitable for this.

Apart from your problem, the HTML <center> tag is deprecated since HTML 4.01 in 1998. You shouldn't be using it as it is subject to removal from the future HTML specifications. The same applies to the HTML tag, if you wasn't aware about it. Both are pure styling tags and thus totally not semantic. You should be using CSS for styling.

J radolf

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Thanks for your reply.

But i was asked to design it using jsp only by my project manager so did it.

he does not like servlets i think. :roll:

J radolf

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if(isMultipart)
{

List items=upload.parseRequest(request); 
Iterator iterator=items.iterator();

while(iterator.hasNext()) { THIS LINE

FileItem fitem=(FileItem)iterator.next();
String fldName=fitem.getFieldName();
if(!fitem.isFormField()){
StringBuffer s2=new StringBuffer(fitem.getName()); 
if(s2!=null && s2.length()>0)
{
int separatorPos=s2.lastIndexOf("\\");
String filename="";
if(separatorPos>0)
filename=s2.substring(separatorPos+1,s2.length());
else
filename=s2.toString(); 
out.println("Filename is" +filename+" ");
String appPath=application.getRealPath("/");
String filePath=appPath+filename;
File fl=new File(filePath);
fitem.write(fl);
}
}
else{
String fldValue=fitem.getString();
out.println("Fld name in html="+fldName+" value is "+fldValue+" ");
}
} 
}
else 
{
out.println("Not a Multipart request");
}
}

catch(Exception e)
{
out.println(e.toString());
}
%>

/**** can i retrieve these values without using Iterator?? *********/ i want the two files attached in separate variables

J radolf

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Like in LINE30 the variable filename is in loop and i have 2 files which are coming from a client
so i want to keep these files stored in 2 different variables,rather then in a loop so what can i do??