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자바 서블릿 java.lang.NumberFormatException의 :이 널 (null)JSP로 업로드 이미지 및 비디오 java.lang.NumberFormatException의
안녕하세요, 저는 JSP와 입력 텍스트 정보와 비디오 및 이미지 데이터에 서블릿을 코딩 한 데이터 베이스.
는 기본적 AddDataInfo.jsp에서 AddDataInfoServlet.java
의 제어 패스 그러나 여기 AddDataInfoServlet.java
public void doGet(HttpServletRequest request, HttpServletResponse response)
throws ServletException, IOException {
processRequest(request, response);
}
public void doPost(HttpServletRequest request, HttpServletResponse response)
throws ServletException, IOException {
processRequest(request, response);
}
public void processRequest(HttpServletRequest request,
HttpServletResponse response) throws ServletException, IOException{
try{
HttpSession session= request.getSession();
DataManagerDao dao = new DataManagerDao();
DataInfo dataInfo = new DataInfo();
if(ServletFileUpload.isMultipartContent(request)){
try{
for(FileItem item : multiparts){
switch(item.getFieldName()){
case "tabsId":
tabId = item.getString();
System.out.println("tabId " + tabId);
break;
case "categoryId":
System.out.println("categoryId ABC");
categoryId = item.getString();
System.out.println("categoryId " + categoryId);
break;
case "subCategoryId":
subCategoryId = item.getString();
System.out.println("subCategoryId " + subCategoryId);
break;
case "attributeId":
attributeId = item.getString();
System.out.println("attributeId " + attributeId);
break;
case "information":
Information = item.getString();
System.out.println("Information " + Information);
break;
case "imageName":
ImageName = item.getString();
System.out.println("ImageName " + ImageName);
break;
case "imageDescription":
ImageDescription = item.getString();
System.out.println("ImageDescription " + ImageDescription);
break;
case "videoName":
videoName = item.getString();
System.out.println("videoName " + videoName);
break;
case "videoDescription":
videoDescription = item.getString();
System.out.println("videoDescription " + videoDescription);
break;
default:
System.out.println("INVALID parameter");
}
dataInfo.setDataTabId(Integer.parseInt(tabId));
..........
}
}
}
}
}
대한 AddDataInfo.jsp
위한 코드<body>
<form action="addDataInfoServlet" method="post" enctype="multipart/form-data">
<%
String tabId = request.getParameter("tabId");
out.println("tabId " + tabId);
out.println("<br/>");
String categoryId = request.getParameter("categoryId");
out.println("categoryId :" + categoryId);
out.println("<br/>");
String subCategoryId = request.getParameter("subCategoryId");
out.println("subCategoryId " + subCategoryId);
out.println("<br/>");
String attributeId = request.getParameter("attributeId");
out.println("attributeId " + attributeId);
out.println("<br/>");
%>
<%
if(attributeId!=null){
%>
<input type="hidden" name="attributeId" value=<%=attributeId%>>
<br/>
<%=attributeId %>
<br/>
<%
}
%>
<input type="hidden" name="tabsId" value=<%=tabId%>>
<input type="hidden" name="categoryId" value=<%=categoryId%>>
<input type="hidden" name="subCategoryId" value=<%=subCategoryId%>>
Information: <input type="text" name="information" /><br/>
Image Name: <input type="text" name="imageName"/><br/>
Image Description: <input type="text" name="imageDescription" /><br/>
Image: <input type="file" name="image" />
<br />
Video Name: <input type="text" name="videoName"/><br/>
Video Description: <input type="text" name="videoDescription"/><br/>
<!-- Video: <input type="file" name="video" />-->
Video: <input type="file" name="video" />
<input type="submit" value="Upload" />
</form>
</body>
코드 java.lang.NumberFormatException가 발생합니다. null
at t 그의 라인 : dataInfo.setDataTabId (Integer.parseInt (tabId));
내가 JSP 파일에서 값을 두 번 확인, 그들은
가 탭을 잡고
Java에서 multiparts는 어디에 설정합니까? – rob