org.apache.jasper.JasperException: Exception in JSP: /index.jsp:90请问名位高手,怎么解决呢,急

来源：百度知道 编辑：UC知道 时间：2024/05/27 19:21:45
87: <%
88:
89: ResultSet rs = con.getRs("SELECT top 5 a.jobid,b.name,b.school,b.specialty,b.knowledge,a.job,a.emolument FROM tb_sjob AS a,tb_student AS b WHERE a.sname=b.sname and getdate()<=atime ORDER BY ptime DESC");
90: while (rs.next())
91: {
92: %>
93: <tr>

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.NullPointerException
org.apache.jsp.index_jsp._jspService(index_jsp.java:158)
org.apache.jasper.runtime.HttpJspBase.service(HttpJspB

应该是连接数据库没连上，导致sql语句异常，rs结果集里没有东西，默认为null。你再仔细看看sql语句，那里错了很冤的

另外，你光这样发，一般人是不知道你到底错哪里的，关键还得你自己找
可以采取在某个地方取值的方式来查看错误地点

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)

HttpServlet.service好像说的是服务请求异常.\
首先确定你的数据库有没有问题.
看数据据连接有没有问题.