org.apache.jasper.JasperException: java.lang.NumberFormatException: null
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/7474322/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
org.apache.jasper.JasperException: java.lang.NumberFormatException: null
提问by inggar
I got exception: org.apache.jasper.JasperException: java.lang.NumberFormatException: nullwhen trying this code:
尝试此代码时出现异常：org.apache.jasper.JasperException：java.lang.NumberFormatException：null：
<%@page contentType="text/html" pageEncoding="UTF-8"%>
<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN"
"http://www.w3.org/TR/html4/loose.dtd">
<html>
<head>
<meta http-equiv="Content-Type" content="text/html; charset=UTF-8">
<title>JSP Page</title>
</head>
<body>
<h3>Tanggal sekarang:</h3>
<%= new java.util.Date() %>
<% //begin scriplet
String nama = request.getParameter( "namadepan" );
String id = request.getParameter("idUser");
int idInt = Integer.parseInt(id);
if(nama!=null) {
%> <%-- end scriplet to insert fixed template data --%>
<h3>Halo <%= nama %>, <br/>
int idInt = Integer.parseInt(idUser);
Selamat datang di web Koperasi Aneka Usaha!
id anda :<%=idInt %>
</h3>
<% //continue scriptlet
}//end if
else {
%> <%-- end scriplet to insert fixed template data --%>
<form name="login" action="index.jsp" method="post">
<table width="200" border="1">
<tr>
<td width="56">Nama </td>
<td width="128"><input type="text" name="namadepan"></td>
</tr>
<tr>
<td>ID</td>
<td><input type="text" name="idUser"></td>
</tr>
<tr>
<td>Umur</td>
<td><input type=int name="umur"></td>
</tr>
<tr>
<td colspan="2"><input type="submit" name="Submit" value="Submit"></td>
</tr>
</table>
</form>
<% //continue scriptlet
}//end else
%> <%-- end scriplet --%>
</body>
</html>
How to fix that exception? im sure the error is about convertint input type string (idUser) to integer.
如何修复该异常？我确定错误是关于将输入类型字符串（idUser）转换为整数的。
回答by nfechner
The problem is the call to Integer.parseInt(id);. If idis not a parsable number, a NumberFormatExceptionwill be thrown. In your case because idwas null. You should at least include a check for null, but it would be better do to something like:
问题是调用Integer.parseInt(id);. 如果id不是可解析的数字，NumberFormatException则将抛出 a。在你的情况下，因为id是null。您至少应该包括对 null 的检查，但最好执行以下操作：
int idInt = 0; // Or a different default value
try {
idInt = Integer.parseInt(id);
} catch(NumberFormatException e) {
// log the error or ignore it
}
