关于WEB开发的一个错误，大家来研究下

[yuanrongchao]

严重: Servlet.service() for servlet jsp threw exception
java.lang.NumberFormatException: For input string: "userid"
at java.lang.NumberFormatException.forInputString(NumberFormatException.java:48)
at java.lang.Integer.parseInt(Integer.java:447)
at java.lang.Integer.parseInt(Integer.java:497)
at javax.el.ArrayELResolver.coerce(ArrayELResolver.java:153)
at javax.el.ArrayELResolver.getValue(ArrayELResolver.java:45)
at javax.el.CompositeELResolver.getValue(CompositeELResolver.java:53)
at org.apache.el.parser.AstValue.getValue(AstValue.java:97)
at org.apache.el.ValueExpressionImpl.getValue(ValueExpressionImpl.java:186)
at org.apache.jasper.runtime.PageContextImpl.proprietaryEvaluate(PageContextImpl.java:923)
at org.apache.jsp.MyJsp_jsp._jspService(MyJsp_jsp.java:112)
at org.apache.jasper.runtime.HttpJspBase.service(HttpJspBase.java:70)
at javax.servlet.http.HttpServlet.service(HttpServlet.java:803)
at org.apache.jasper.servlet.JspServletWrapper.service(JspServletWrapper.java:393)
at org.apache.jasper.servlet.JspServlet.serviceJspFile(JspServlet.java:320)
at org.apache.jasper.servlet.JspServlet.service(JspServlet.java:266)
at javax.servlet.http.HttpServlet.service(HttpServlet.java:803)
at org.apache.catalina.core.ApplicationFilterChain.internalDoFilter(ApplicationFilterChain.java:290)
at org.apache.catalina.core.ApplicationFilterChain.doFilter(ApplicationFilterChain.java:206)
at org.apache.catalina.core.ApplicationDispatcher.invoke(ApplicationDispatcher.java:654)
at org.apache.catalina.core.ApplicationDispatcher.processRequest(ApplicationDispatcher.java:445)
at org.apache.catalina.core.ApplicationDispatcher.doForward(ApplicationDispatcher.java:379)
at org.apache.catalina.core.ApplicationDispatcher.forward(ApplicationDispatcher.java:292)
at org.apache.struts.chain.commands.servlet.PerformForward.handleAsForward(PerformForward.java:113)
at org.apache.struts.chain.commands.servlet.PerformForward.perform(PerformForward.java:96)
at org.apache.struts.chain.commands.AbstractPerformForward.execute(AbstractPerformForward.java:54)
at org.apache.struts.chain.commands.ActionCommandBase.execute(ActionCommandBase.java:51)
at org.apache.commons.chain.impl.ChainBase.execute(ChainBase.java:190)
at org.apache.commons.chain.generic.LookupCommand.execute(LookupCommand.java:304)
at org.apache.commons.chain.impl.ChainBase.execute(ChainBase.java:190)
at org.apache.struts.chain.ComposableRequestProcessor.process(ComposableRequestProcessor.java:283)
at org.apache.struts.action.ActionServlet.process(ActionServlet.java:1913)
at org.apache.struts.action.ActionServlet.doGet(ActionServlet.java:449)
at javax.servlet.http.HttpServlet.service(HttpServlet.java:690)
at javax.servlet.http.HttpServlet.service(HttpServlet.java:803)

[yuanrongchao]

这个错误本来是接受参数的时候强转错误，，可是我没接受，，是查询全表

[bunrise]

数据库出来的是string 你把他转换成int
看看自己代码

[yuanrongchao]

我根本没有写转换的代码

[bunrise]

数据库字段类型和你前台接收类型一样吗？

[huxiaowi]

看API
Thrown to indicate that the application has attempted to convert a string to one of the numeric types, but that the string does not have the appropriate format.


你的userid格式不合法，就是说你的userid不能转成数字。
比如： "12a" ==> ERROR

[laker_914]

   看tomcat/work OR 你的工程下的work目录 org.apache.jsp.MyJsp_jsp._jspService(MyJsp_jsp.java:112)
 
   找到MyJsp_jsp.java 看112行，是哪个地方报错

[kala197]

.toString(); 比如你把StringBuffer变量的值传到前台 用String变量来获得 就会报java.lang.NumberFormatException: For input string: "userid"
就是你的代码类型转换要一致 且注意传值和接值的类型

[cuiqunhao]

看看是不是越界了.....

[masterjames]

肯定是转换类型错了,看看函数体里有问题没有.

[沐希]

就是类型转换的问题，或者说你pojo中对应的USERID是INT型，而你FORMBEAN中的是STRING型，你看看吧。肯定是这2个不对应

[crazylaa]

类型转换错误，这个应该最容易排查。

[lhzx_zjg]

看下你的EL表达式是不是转换出错

[VilenZYP]

[Quote=引用 8 楼 cuiqunhao 的回复:]
看看是不是越界了.....
[/Quote]

这个。。。。。非常强大。。。

[huxiaowi]

这么多说类型转换异常。。。
请问你们java.lang.NumberFormatException 和 java.lang.ClassCaseException是一个东东么？