下載共享資料文件上傳

Question

我正在嘗試使用apache commons文件上傳庫從jsp/ajax上傳文件。下面是代碼：

JSP：

function importFileListForm() 
{ 
var html = "<div><h3>Add File</h3></div>"; 
html += '<table style="margin-top:20px;" width="100%"><tbody>'; 
html += '<tr><td>Choose File</td><td><input id="fileId" type="file" name="fileName" required size="50"/></td><td></td></tr>'; 
html += '</tbody></table>'; 

$("#dialog-modal-content").html(html); 
$(function() { 
    $("#dialog-modal").dialog({ 
     height: 300, 
     width: 600, 
     modal: true, 
     buttons: { 
     "Cancel": function() { 
      $(this).dialog("close"); 
     }, 
     "Submit": function() { 
      importFileConfirm($("#fileId").val()); 
     } 
     } 
    }); 
}); 
} 

function importFileConfirm(file) 
{ 
var url = "/config/fileManageAction.do?method=importFile&file=" +file; 
var req = new FormData(); 
req.append("f", file); 
$.ajax({ 
    type: 'POST', 
    data: req, 
    url: url, 
    processData: false, 
    contentType: false, 
    success: function(data) {console.log(data);} 

}); 
} 

我的動作類方法：

 DiskFileItemFactory factory = new DiskFileItemFactory(); 
     factory.setSizeThreshold(maxSize); 
     factory.setRepository(new File("/production/cpu/temp.xlsx")); 

     ServletFileUpload upload = new ServletFileUpload(factory); 
     // maximum file size to be uploaded. 
     upload.setSizeMax(maxSize); 
     try{ 
      List fileItem = upload.parseRequest(request); 
      java.lang.System.out.println("List "+fileItem.size()); 

    // Part filePart = request.getPart("file"); 
    // java.lang.System.out.println("File Part "+filePart); 

     // Process the uploaded file items 
      Iterator i = fileItem.iterator(); 
      while(i.hasNext()){ 
       FileItem fi = (FileItem)i.next(); 
       if(!fi.isFormField()){ 
        java.lang.System.out.println("Fi "+fi); 
        String fieldname = fi.getFieldName(); 
        java.lang.System.out.println("FieldName "+fieldname); 
        String filename = FilenameUtils.getName(fi.getName()); 
        java.lang.System.out.println("File Name "+fi.getString()); 
        InputStream filecontent = fi.getInputStream(); 
       }else{ 
        java.lang.System.out.println("not a form field"); 
        java.lang.System.out.println("Fi "+fi); 
        String fileName = fi.getName(); 
        java.lang.System.out.println("File Name: "+fileName); 
       } 


      } 
     }catch(Exception e){ 
      java.lang.System.out.println("Exception occured "+e); 
      e.printStackTrace(); 
     } 

// } 

} 

但在我的文件項目我只得到字段名和fi.getName()正在恢復空值。有人能說出我缺少的東西嗎？我不能用getParts()函數，因爲我的tomcat不支持它。

Answer 1

您必須將File對象添加到FormData對象。 $("#fileId").val()給你的文件名，以獲得文件使用$("#fileId")[0].files[0]

req.append("f", $("#fileId")[0].files[0]);