2

Not a very common implementation, but using Processing as a Java Servlet has been discussed in previous posts before (1, 2). I have been developing a program in Processing to use as a Java servlet. However, when I have code (such as like the first linked example) in a Processing sketch, I get the error message

The package 'javax.servlet' does not exist.  You might be missing a library.

I have Tomcat 5.5 service running on XP, and my environment variables are as follows

CLASSPATH = C:\tomcat\common\lib\servlet-api.jar
CATALINA_HOME = C:\tomcat
JAVA_HOME = C:\Program Files\Java\jdk1.6.0_21
JRE_HOME  = C:\Program Files\Java\jre6

I have been trying to diagnose why the javax.servlet packages are not recognized by Processing for many hours and there is nothing online that seems to yield an explanation other than fixing the environment variables to recognize the tomcat libraries. I have also tried to put the servlet-api.jar and jsp-api.jar files in Processing as a kind of custom Processing library but they are not recognized there either, nor when I try to run the program within the tomcat/webapps folder.

I also have tried 

javap -classpath my;class;path javax.servlet.Servlet

on the cmd and it gave me the same error.

I'm not that good with Tomcat yet so please forgive me if this is a result of my unfamiliarity. If anyone more knowledgeable than I can shed some light as to why Processing cannot recognize this package would be tremendous. Thanks so much~

4

5 回答 5

1

CLASSPATH = C:\tomcat\common\lib\servlet-api.jar

这就是问题 。你的类路径应该是一个级别，即

类路径 = C:\tomcat\common\lib\

这应该可以正常工作。

于 2011-01-13T09:23:20.007 回答
0

右击我的电脑->属性->高级->环境变量->

设置一个新的用户变量名作为classpath，并给出你的servlet-api.jar文件所在的变量值（适用于tomcat webserver），例如：->

变量名：类路径

变量值：D:\Tomcat\lib\servlet-api.jar

现在你可以运行你的反汇编程序来找出servlet类和接口信息

于 2013-11-19T09:08:30.873 回答
0

好的！.. 我买了一台新笔记本电脑，遇到了同样的问题，我希望这也能解决你的 Windows 10 设备上的问题。

首先要知道错误只是由于路径或类路径不正确或不完整。安装 Tomcat 和 JDK 后，设置环境变量如下（位置可能因您的安装而异）：

JAVA_HOME = C:\Program Files (x86)\Java\jdk1.7.0_80

JRE_HOME = C:\Program Files (x86)\Java\jre7

CATALINA_HOME = C:\Program Files (x86)\Apache Software Foundation\Tomcat 7.0

PATH = C:\Program Files (x86)\Java\jdk1.7.0_80\bin; C:\Program Files (x86)\Apache Software Foundation\Tomcat 7.0\bin;

CLASSPATH = C:\Program Files (x86)\Apache Software Foundation\Tomcat 7.0\lib\servlet-api.jar；C:\Program Files (x86)\Java\jdk1.7.0_80\lib\tools.jar；JAVA_HOME\lib

确保您的类路径指向提到的 jar 文件，因为其中包含 javax（尝试使用 winrar 检查内部内容）。

保存设置后，使用 javap javax.servlet.Servlet 进行测试

好吧，如果您仍然遇到任何问题，请分享错误消息和屏幕截图。

于 2017-01-22T21:12:13.667 回答
0

如果您无法加载 servlet 包，请尝试将环境变量 CLASSPATH 设置为 .;C:\tomcat\common\lib\servlet-api.jar。那应该有帮助。

于 2016-09-21T17:32:41.193 回答
-1

可能是因为 Tomcat 不在 Java Build Path 中。尝试将 Tomcat 添加到您的库中

1）右键单击您的项目文件夹>构建路径>配置构建路径

2）单击库选项卡>单击添加库按钮

3）选择服务器运行时>单击下一步按钮

4）选择您的服务器>单击 完成按钮

于 2018-04-06T07:43:05.697 回答