java HTTP 状态 500 - 空
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HTTP Status 500 - null
提问by Tweety
I wrote a java servlet program but when i run it, It was showing the Error HTTP Status 500 - null What it means ; My source code is on this page.
我写了一个 java servlet 程序，但是当我运行它时，它显示错误 HTTP 状态 500 - null 是什么意思；我的源代码在这个页面上。
The error was
错误是
HTTP Status 500 - null
type Exception report
message null
description The server encountered an internal error that prevented it from fulfilling this request.
exception
java.lang.NumberFormatException: null
java.lang.Integer.parseInt(Unknown Source)
java.lang.Integer.parseInt(Unknown Source)
skypark.Registration.doPost(Registration.java:56)
javax.servlet.http.HttpServlet.service(HttpServlet.java:647)
javax.servlet.http.HttpServlet.service(HttpServlet.java:728)
Please help me to resolve this....
请帮我解决这个问题......
回答by Jigar Joshi
possibly
可能
int phone=Integer.parseInt(phone1);
phone1is nullpassed
phone1被null通过
回答by Nipun Jain
hey correct ur html there is no input type =number just correct as follows
嘿正确你的 html 没有输入类型 = 数字正确如下
Phone Number
</td><td>
<input type="text" name="phone">
</td></tr><tr><td>
回答by Matin Kh
From the error I can say that there is something other than a numberpassed to your function and this exception is raised.
从错误中，我可以说除了number传递给您的函数之外还有其他内容，并且引发了此异常。
Something like this:
像这样的东西：
int a = Integer.ParseInt(aStr);
where the aStris abc
其中，aStr为abc
回答by Tweety
You should have attribute name in the input tag.
您应该在输入标签中有属性名称。
<input type="text" name="xyz">
In d servletaccess the value after parsing it, as follows.
在d中servlet访问解析后的值，如下。
int x=Integer.parseInt(request.getParameter("xyz"));
