Java 无法将字符串转换为整数
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Can't convert String to Integer
提问by Salini L
I am a beginner. I know this is basic. In my project I am using Java and MySQL workbench. I am reading a data from user by using select box which is coming from database by using ajax. The code for select box is mentioned below
我是初学者。我知道这是基本的。在我的项目中，我使用 Java 和 MySQL 工作台。我正在通过使用 ajax 来自数据库的选择框从用户读取数据。下面提到了选择框的代码
<%
String a =request.getParameter("course");
if(a!=null)
{
ResultSet rs=s.selectsub(a);
String Query="select * from subject where course_id='"+a+"'";
%>
<select name="subject" id="subject">
<option>Select Subject</option>
<%
while(rs.next())
{
%>
<option value="<% out.println(rs.getString("subject_id")); %>">
<% out.println(rs.getString("subject")); %></option>
<% } %>
</select>
<%
}
%>
and the subject id is passed to another page using post method and trying this code
并且使用 post 方法将主题 ID 传递到另一个页面并尝试此代码
String subject=request.getParameter("subject");
int subjectid=Integer.parseInt(subject);
But the line Integer conversion is not working. Error is showing. The error is
但是行整数转换不起作用。错误显示。错误是
org.apache.jasper.JasperException: An exception occurred processing JSP page /saveuser.jsp at line 29
26: String email=request.getParameter("email");
27: String designation=request.getParameter("designation");
28: String subject=request.getParameter("subject");
29: int subjectid=Integer.parseInt(subject);
30: String institute=request.getParameter("institute");
31: String inemail=request.getParameter("inemail");
32: String uname=request.getParameter("uname");
采纳答案by Juned Ahsan
Integer.parseIntmethod will convert a valid integer value string to an int. Otherwise it will throw an exception. So make sure you pass a valid integer value to it. Sometimes the trailing space in the string causes exception. So call trim method on the input string to avoid that:
Integer.parseInt方法会将有效的整数值字符串转换为 int。否则会抛出异常。因此，请确保将有效的整数值传递给它。有时字符串中的尾随空格会导致异常。所以在输入字符串上调用 trim 方法来避免这种情况：
int subjectid=Integer.parseInt(subject.trim());
Also make sure that subject is not null. So this seems better:
还要确保主题不为空。所以这看起来更好：
String subject=request.getParameter("subject");
int subjectid = 0;
if(subject !=null && !subject.isEmpty())
subjectid=Integer.parseInt(subject.trim());
回答by vikingsteve
It looks like you are mixing up your data types, trying to convert getParameter("subject")to an Integer.
看起来您正在混淆数据类型，试图转换getParameter("subject")为整数。
Perhaps you have a parameter named "subject_id", and could this work?
也许您有一个名为 的参数"subject_id"，这可以工作吗？
String subject_id_string=request.getParameter("subject_id");
int subjectid=Integer.parseInt(subject_id_string);
回答by Sri Harsha Chilakapati
Use Integer.parseInt(String)method.
int subjectid = Integer.parseInt(subject.trim());
From the documentation
从文档
Parses the string argument as a signed decimal integer. The characters in the string must all be decimal digits, except that the first character may be an ASCII minus sign '-' ('\u002D') to indicate a negative value or an ASCII plus sign '+' ('\u002B') to indicate a positive value. The resulting integer value is returned, exactly as if the argument and the radix 10 were given as arguments to the parseInt(java.lang.String, int) method.
将字符串参数解析为有符号十进制整数。字符串中的字符必须都是十进制数字，除了第一个字符可以是 ASCII 减号 '-' ('\u002D') 表示负值或 ASCII 加号 '+' ('\u002B')表示正值。返回结果整数值，就像参数和基数 10 作为参数提供给 parseInt(java.lang.String, int) 方法一样。
