问题
I want to upload a file to a server, for which I am writing a servlet program.The location of the directory where the document would be uploaded should be fetched from a parameter in web.xml. I have not use web.xml before and only know that it makes entries for each servlet. I am not able to see this file in my web application project that i am making in netbeans. Please help me with this. Thank you.
回答1:
It should be located in YOURPROJECT\web\WEB-INF folder, so the full path will be: YOURPROJECT\web\WEB-INF\web.xml
EDIT (Aug 21, 2015)
Got a downvote with a comment from duffymo, that my answer is not correct.
And I decided to illustrate my answer with step-by-step pictures to avoid any misunderstanding.
I am going to illustrate Netbeans behaviour in Linux (Ubuntu) and Windows (Windows 7) operating systems.
Linux:
- Let's create a simple Java Web project with default settings.
- We're going to the project folder to inspect the contents of this folder:
note, that web folder is there.
- Navigating further:
You can create a web.xml file in this folder manually or do it using Netbeans via project context menu "New -> Create -> Other":
Now, we're navigating YOURPROJECT\web\WEB-INF\ folder to see, that web.xml is there:
The same rules are correct for windows operating system, check the pictures below:
You can create web.xml here:
or use Netbeans, as I described above.
回答2:
I know it's so late but I had the same problem, so here's the solution below:
To create web.xml:
- Right click on your project
- Choose New
- Choose Other
- Box New file is opened, in Filter search web.xml
- You will get the file you want web.xml, then click next...then finish
(Tested on Netbean 7.4 JDK 7)
回答3:
web.xml is optional in Java EE 6. So, by default it is not loaded in Netbeans. You need to manually load web.xml from Netbeans.
回答4:
You need to firstly create a servlet page then web.xml will be generated in WEB.INF/
回答5:
Try to right click the project and select New -> Other-> Web -> Standard Deployment Descriptor (web.xml) -> Next -> Finish. Follow that and it will be created in Configuration Files.
Video tutorial: https://www.youtube.com/watch?v=UAMOeHtPwrc
回答6:
You could use context-parameters in your web.xml
In you normal java class you read this this static fields.
<?xml version="1.0" encoding="UTF-8"?>
<web-app ...>
...
<context-param>
<description>directory where the document would be uploaded</description>
<param-name>directory</param-name>
<param-value>/tmp</param-value>
</context-param>
...
</web-app>
And you can access this context parameter with ServletContext.getInitParameter function.
If you are using Servlet 3.0 specification you can use annotations(http://docs.oracle.com/javaee/6/api/javax/servlet/annotation/package-summary.html).
I think that @WebInitParam is what are you looking for.
来源：https://stackoverflow.com/questions/19317410/unable-to-find-web-xml-in-netbeans-7-0-1