java.lang.NumberFormatException: null的解決方案__java.lang.NumberForm

    1  首先附上My Code：

 Integer reviewid=Integer.valueOf(request.getParameter("reviewid")); System.out.println(reviewid);Integer articleid=Integer.valueOf(request.getParameter("articleid")); if(ybReviewService.deleteReview(reviewid)>0){ System.out.println("加油啊");response.sendRedirect("YbReviewServlet?method=list&articleid="+articleid); }else{     request.setAttribute("msg","刪除評論失敗。<a href=\"JavaScript:window.history.back()\">返回</a>"); }}

    2  出現這個報錯，一般要考慮

Integer.valueOf(request.getParameter("articleid"));

中擷取的值，可能是null；當然，也可以通過設定斷點或是像我一樣插入

 System.out.println("加油啊");

這種語句來判斷；

    3  所以，要檢查你擷取的資料是否為空白值；

    4  我改過的代碼如下，當然，不會報錯了；

protected void delete(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException{ Integer reviewid=Integer.valueOf(request.getParameter("reviewid")); System.out.println(reviewid); Connection con=null; con=DBUtil.getDb(); PreparedStatement pstmt=null; ResultSet rs=null; YbReview ybReview=null; String sql="select article_id from yb_review where review_id=?"; try{  pstmt=con.prepareStatement(sql);      pstmt.setInt(1,reviewid);    rs=pstmt.executeQuery();  while(rs.next()){  ybReview=new YbReview();  ybReview.setArticleid(rs.getInt("article_id"));  } }catch(SQLException ex){ System.out.println(ex.getMessage()); }Integer articleid=ybReview.getArticleid(); if(ybReviewService.deleteReview(reviewid)>0){     System.out.println("加油啊");response.sendRedirect("YbReviewServlet?method=list&articleid="+articleid); }else{     request.setAttribute("msg","刪除評論失敗。<a href=\"JavaScript:window.history.back()\">返回</a>"); }}