I am trying to write a webservice class which actually handles multipart requests. T client will try to upload zip or tar.gz files which may be upto 1GB in size.
I dont want to validate the extension i client side.
I want to validate the file extension in server side and discard the request before the file is uploaded.
How can I do this?
The JSP page is opening but the problem is if l click on the ADD or EDIT button l receive the 405 error Iam using MySQL and glassfish server
web.xml
<?xml version="1.0" encoding="UTF-8"?>
<web-app version="3.1" xmlns="http://xmlns.jcp.org/xml/ns/javaee" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="http://xmlns.jcp.org/xml/ns/javaee http://xmlns.jcp.org/xml/ns/javaee/web-app_3_1.xsd">
[Code].....
I have a program that allows the user to select what type (i.e. tif, jpg, sid) of file they have to load. I use a JCombo box for the selection type. I send this JCombo box string to my ElevationConvert class. What I want to do is check to see if that extension the user selected exists prior to converting it. If the file extension does not exist, I want to pop up a window (JOptionPane) stating is does not exit. I have tried suggestions found on this site, but does not seem to work for my application. The program works with (Process cN = Runtime), if I try not to check for the extension. But when I add the check for (If, else), it bombs. This executable runs in the main folder. The file extensions are in a subfolder called Ned.
/*
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* To change this template file, choose Tools | Templates
* and open the template in the editor.
*/
package terraindbbuilder1;
import java.awt.Component;
import java.io.*;
import java.io.IOException;
import javax.swing.JOptionPane;
[Code] .....
I have a GUI that has three jTextFiled's. The first on gets filled in with a file that I choose and shows the path to that file with the file name
i.e. - c:
s34bil.exe
un estrun.i
I want to change the file extension from .i to .r in one jTextField and then .i to .o in another jTextField.
I am slowly learning Java. I have this coded in VB, but not sure how to do it in java.
Here is the VB script :
Dim thefile As String = txtInput.Text
Dim fn1 As String = My.Computer.FileSystem.GetName(thefile)
Dim fn2 As String = fn1.Replace(".i", "")
[Code] ....
i want to find an embedded file extension i.e suppose if word file is there in that excel is also included then is should detect both file types word n excel and is is it possible to find an embedded file type.
View Replies View RelatedI am currently working on an application with java and I want to do is upload and download details in .xlsx file. Uploading process is successfully worked but the problem is that when I download the file .xlsx with details at opening the file it gives me the following message :
Excel cannot open the file '(filename)'.xlsx' because the file format or file extension is not valid. Verify that the file has not been corrupted and that the file extension matches the format of the file.
I need to be able to select a file (compressed), change its file extension (to zip) and load all its content, preferably as objects. If it makes things easier one can assume all these files to be of the same type.
The content of the zip file are placed in folders, known Beforehand which might make things easier.
I'm trying to make a search engine that will search my computer given a path with any options i decide. So if i decides to look for the file name "resume" with extension "pdf", it will give me the files in the directory, including subfiles.
import java.io.*;
import java.util.*;
import java.text.*;
public class mySearch
{
private final static int path = 0;
private final static int name = 1;
[Code] .....
I have the following questions:
1. Does a GET HTTP request contain a request body? If yes what is contained in it? Are the request headers also part of the request body?
2. Is it possible to send a byte array as part of the GET request in its body?
3. Is there a size limitation on the data that can be sent via a GET request?
I am working in JSF and I've forgot few of the basics :'I use eclipse IDE to develope JSF projects. When I used to create a JSF file, it's extension is basically .jsp and when I ran that file in server .jsf extension used to be displayed in url. But recently, when I did the same thing I got .jsp extension in browser. in web.xml
when url-pattern is the following, .jsp extension is shown, eg: FileName.jsp
<servlet-mapping>
<servlet-name>Faces Servlet</servlet-name>
<url-pattern>/faces/*</url-pattern>
</servlet-mapping>
but when I change the url-pattern as per following, .jsf extention is shown , eg: FileName.jsf
<servlet-mapping>
<servlet-name>Faces Servlet</servlet-name>
<url-pattern>*.jsf</url-pattern>
</servlet-mapping>
even if I change the url-pattern to abc or xyz, the webpage url is shown in that format. eg: FileName.abc this is the actual scenario to specify extension of a jsf page, or is this an error.
recently I noticed that most of the Sun site pages have the extension .jspa, some even have .xml.So just curious about it because I don't know about jspa but as far as I know of xml, it is not rendered by browser like the way we see on Sun site.
View Replies View RelatedI would like to know the details about how a server handles a multipart request. I know that a webserver comprises of a HTTP adapter(which is responsible for receiving http requests and sending http responses) and a container(which is responsible for handling dynamic requests).
So, when a client sends a request, the HTTP adapter receives it. Then transfers the request to the container if the http adapter cannot handle it(jsps, servlets for example).
Suppose, a client sends a multipart request which has an avi file of size upto 100mb. For example, lets assume that the client takes 5 minutes to upload the file.
Lets also assume that my application is only interested in flv files. Is there any way to stop/terminate the multipart request before the file is transferred to the server? So that the client's time will not be wasted?
I think that the HTTP adapter will transfer the request to the container only after receiving the entire request, i.e. the servlet will be called only after the file is transferred to the server from client. Is it right?
If its right then there is no easy way to terminate the multipart request unless the entire file is transferred to the server. Is this right?
I want to programmatically change the default program which an extension is opened in. So .resantic extensions for example to run in my jar file when I double-click on it. I know i need a bat file or exe file because i can't directly run the jar file. But i was wondering if its possible to run it if its an executable jar file. Else i can just make a bat file that runs my program. Also how can i on double click send the path to the file that was clicked in the args[] array so i can open the actual file aswell instead of just opening the program?
View Replies View RelatedI need to built a file upload service which should be memory effective. I should avoid loading the entire file into memory,Since I may have multiple http request which will pile up the Heap memory. Any effective way to upload a large file(For ex:1GB file) using http Streaming. I need to do the file upload on a single http call. Let's consider a scenario where 1 GB file to be uploaded using 512MB Heap memory. Not Sure If practically I can achieve this or not.
View Replies View RelatedIn SOAPUI, how can I add a new property in a HTTP Test Request? I konw that this there isn´t, but should be somethink like that:
testRunner.testCase.getTestStepByName("myHttpTestRequest").addProperty("newProperty","newValue");
I am trying to generate this request with my Java program, but I just can't seem to figure it out: curl -v -H "Authorization:Basic MY_AUTH_CODE" -H "Accept:application/xml" -H "app_id:MY_APP_ID" -H "app_key:MY_APP_KEY" -X GET "THIRD_PARTY_URL"
I tried to generate a query string, but I keep getting 403 errors. I also tried to input my parameters using URLConnection.getRequestProperty("name", "value"), but that didn't work either.
I have been trying to implement custom request method in HTTP header while posting my data to the server URL. My application specific URL accepts -X parameter and -d for the data and it is mandatory for that url. Basically I am trying to dump JSON data into my influx DB using CURL command which is working fine from the shell. But the issue is, if I am implementing the same in java with proper approach, it is not supported or working. My CURL command is :
curl -X POST -d '<my_json_data>' '<my_url>'
How can I implement the same in java using HttpUrlConnection or other available approach?
i am using netbeans6.7.1 version i implemented one program i.e web application ..when am excuting this one i got one error http status 500...
this is the code:
SumServlet.java
package com.sum;
import java.io.IOException;
import java.io.PrintWriter;
import javax.servlet.ServletConfig;
import javax.servlet.ServletException;
import javax.servlet.http.HttpServlet;
[code]....
i have one html page through which i want to go to servlet page but after submit it give the http 404 error.
i am using tomcat apache server.
correct this error
WishSrv.java
package com.wish;
import java.io.IOException;
import java.io.PrintWriter;
import java.util.Calendar;
import javax.servlet.ServletException;
[code]...
in my web.xml I declared a resource under the block
<user-data-constraint>
<transport-guarantee>CONFIDENTIAL</transport-guarantee>
</user-data-constraint>
This resource is about a login page.But when I logout or when I visit some pages like Contacts or Home Page, I wouldn't use HTTPS protocol.At the moment, HTTPS remain in the url even if I declared this protocol only for that resource..This is my (little) web.xml
<?xml version="1.0" encoding="UTF-8"?>
<web-app xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns="http://java.sun.com/xml/ns/javaee" xmlns:web="http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd" xsi:schemaLocation="http://java.sun.com/xml/ns/javaee
[code]....
I am trying to use hidden variable in project.When I launch my project i am able to get the welcome page.But when submit login values i am getting HTTP 404 error- Resource not found error.
`My home Page/Login Page
<body>
<form action="<%=request.getContextPath() %>/LoginServlet" method="get">
USERNAME<input type="text" name="uname"><br>
PASSWORD<input type="password" name="pass"><br>
<input type="submit" name="submit" value="submit">
My LoginServlet
protected void doGet(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
// TODO Auto-generated method stub
System.out.println("sandeep");
PrintWriter out=response.getWriter();
String userName=request.getParameter("uname");
[code].....
So I have a screen I click on. The "webPage" comes back as the requested resource unavailable.
What I don't get is that when I run the server locally on my machine it works just fine. But when this is deployed out to server I get the error. The screen comes up but the data doesn't show up. Where should I look to troubleshoot this?
@RequestMapping(value = "/webPage_data", method = RequestMethod.GET)
public @ResponseBody
DataTableDto pastControllerDataFeed(Model model) {
Below is my Servletclass code
import java.io.IOException;
import java.io.PrintWriter;
import javax.servlet.ServletException;
import javax.servlet.http.HttpServlet;
import javax.servlet.http.HttpServletRequest;
import javax.servlet.http.HttpServletResponse;
public class MyFirstServlet extends HttpServlet {
[Code] .....
I have configured tomcat7 in my ecclipse ide...
When does HTTP Session object is created in web application. Suppose I have a website. Home page of website is HTTP page which contains details of company and link to Login page.
Consider below mentioned user journey as scenario:
a. user arrives at home page of website
b. user click on Login page
c. user fill in login details on login page and click on Submit
d. user is successfully authenticated and authorized from back end
e. User specific page is shown
f. user click on logout link
g. user is successfully logged out from website
h. user is redirected to home page
i. user closes browser
In the above mentioned user journey,
a. at which step does HTTP session starts (means at which steps does HTTP Session object is created ? )
b. at which step does HTTP session ends ?
In case required, assume tech stack to be Java 7, Servlet 2.5, JSP, Tomcat 7, Apache web server (for static web contents).....
I'm trying to learn java by making a login page using java, servlet, javascript, html and mysql. I can login with username and password, I can get all the information from database in the edit page. However, when I edit and click on "Submit" in EditPage.jsp, it gives HTTP Status 404. Same screen appears when I click on "Sign Up" from first page (NewFile.jsp) and click on "Submit" button after filling up user information.
I am posting all my code here but probably important ones are:
EditPage.jsp, Edit.java, EditDetails.java, UpdateUser.java and for Sign Up SignUpPage.jsp and Registration.java.
Firstly here are my error message and ss from my project explorer:
Here is my code:
LoginServlet.java
package com.amzi.servlets;
import java.io.IOException;
import java.io.PrintWriter;
import javax.servlet.RequestDispatcher;
import javax.servlet.ServletException;
[Code] ....