Question

java.lang.NumberFormatException: For input string: ""

org.apache.jasper.JasperException: Exception in JSP: /dd_del_save.jsp:40

37: outbeizhu=request.getParameter("outbeizhu");
38: if(request.getParameter("id")!=null)
39: num=request.getParameter("id");
40: int a=Integer.parseInt(shuliang.trim());
41: int b=Integer.parseInt(outshuliang.trim());
42: if(a>b)
43: {

Stacktrace:
org.apache.jasper.servlet.JspServletWrapper.handleJspException(JspServletWrapper.java:451)
org.apache.jasper.servlet.JspServletWrapper.service(JspServletWrapper.java:373)
org.apache.jasper.servlet.JspServlet.serviceJspFile(JspServlet.java:329)
org.apache.jasper.servlet.JspServlet.service(JspServlet.java:265)
javax.servlet.http.HttpServlet.service(HttpServlet.java:729)

root cause

java.lang.NumberFormatException: For input string: ""
java.lang.NumberFormatException.forInputString(Unknown Source)
java.lang.Integer.parseInt(Unknown Source)
java.lang.Integer.parseInt(Unknown Source)
org.apache.jsp.dd_005fdel_005fsave_jsp._jspService(dd_005fdel_005fsave_jsp.java:91)
org.apache.jasper.runtime.HttpJspBase.service(HttpJspBase.java:98)
javax.servlet.http.HttpServlet.service(HttpServlet.java:729)
org.apache.jasper.servlet.JspServletWrapper.service(JspServletWrapper.java:331)
org.apache.jasper.servlet.JspServlet.serviceJspFile(JspServlet.java:329)
org.apache.jasper.servlet.JspServlet.service(JspServlet.java:265)
javax.servlet.http.HttpServlet.service(HttpServlet.java:729)

Answer 1

int a=Integer.parseInt(shuliang.trim());
这一句里的变量shuliang接收的值是""，当你trim()之后还是的值还是""，Integer.parseInt(shuliang.trim());这个方法里面的字符串参数必须全部是数字才得行，楼主应该检查一下传值的问题，看看参数值的传递情况

Answer 2

这个错误是由于Integer.parseInt 接受到了空字符串的原因，错误提示很明确java.lang.NumberFormatException: For input string: ""；
确保要转型的字符串里只包含数字类型的字符就可解决。

Answer 3

int a=Integer.parseInt(shuliang.trim());

试着用System.out.println(shuliang);打印一下数量，看看是否正确的传入参数。

Answer 4

int a=Integer.parseInt(shuliang.trim())